Datasets:

pkuAI4M

/

AOPS

like 1

Follow

AI4M 17

[ "Suppose we require $a$ $7$ s, $b$ $77$ s, and $c$ $777$ s to sum up to $7000$ $a,b,c \\ge 0$ ). Then $7a + 77b + 777c = 7000$ , or dividing by $7$ $a + 11b + 111c = 1000$ . Then the question is asking for the number of values of $n = a + 2b + 3c$\nManipulating our equation, we have $a + 2b + 3c = n = 1000 - 9(b + 12c) \\Longrightarrow 0 \\le 9(b+12c) < 1000$ . Thus the number of potential values of $n$ is the number of multiples of $9$ from $0$ to $1000$ , or $112$\nHowever, we forgot to consider the condition that $a \\ge 0$ . For a solution set $(b,c): n=1000-9(b+12c)$ , it is possible that $a = n-2b-3c < 0$ (for example, suppose we counted the solution set $(b,c) = (1,9) \\Longrightarrow n = 19$ , but substituting into our original equation we find that $a = -10$ , so it is invalid). In particular, this invalidates the values of $n$ for which their only expressions in terms of $(b,c)$ fall into the inequality $9b + 108c < 1000 < 11b + 111c$\nFor $1000 - n = 9k \\le 9(7 \\cdot 12 + 11) = 855$ , we can express $k$ in terms of $(b,c): n \\equiv b \\pmod{12}, 0 \\le b \\le 11$ and $c = \\frac{n-b}{12} \\le 7$ (in other words, we take the greatest possible value of $c$ , and then \"fill in\" the remainder by incrementing $b$ ). Then $11b + 111c \\le 855 + 2b + 3c \\le 855 + 2(11) + 3(7) = 898 < 1000$ , so these values work.\nSimilarily, for $855 \\le 9k \\le 9(8 \\cdot 12 + 10) = 954$ , we can let $(b,c) = (k-8 \\cdot 12,8)$ , and the inequality $11b + 111c \\le 954 + 2b + 3c \\le 954 + 2(10) + 3(8) = 998 < 1000$ . However, for $9k \\ge 963 \\Longrightarrow n \\le 37$ , we can no longer apply this approach.\nSo we now have to examine the numbers on an individual basis. For $9k = 972$ $(b,c) = (0,9)$ works. For $9k = 963, 981, 990, 999 \\Longrightarrow n = 37, 19, 10, 1$ , we find (using that respectively, $b = 11,9,10,11 + 12p$ for integers $p$ ) that their is no way to satisfy the inequality $11b + 111c < 1000$\nThus, the answer is $112 - 4 = \\boxed{108}$", "To simplify, replace all the $7$ ’s with $1$ ’s. \nBecause the sum is congruent to $n \\pmod 9$ and \\[1000 \\equiv 1 \\pmod 9 \\implies n \\equiv 1 \\pmod 9\\] Also, $n \\leq 1000$ . There are $\\big\\lfloor \\tfrac{1000}{9} \\big\\rfloor + 1 = 112$ positive integers that satisfy both conditions i.e. $\\{1, 10, 19, 28, 37, 46, . . . , 1000\\}.$\nFor $n = 1, 10, 19$ , the greatest sum that is less than or equal to $1000$ is \\[6 \\cdot 111 + 1 = 677 \\implies 112-3 = 109.\\]\nThus $n \\geq 28$ and let $S = \\{28, 37, 46, . . . , 1000\\}$\nNote that $n=28$ is possible because $9 \\cdot 111+1 \\cdot 1 = 1000$\nWhen $n = 37$ , the greatest sum that is at most $1000$ is $8 \\cdot 111+6\\cdot 11+1 \\cdot 1 = 955$\nAll other elements of $S$ are possible because if any element $n$ of $S$ between $46$ and $991$ is possible, then $(n+ 9)$ must be too.\n$\\textbf{Case 1:}$ Sum has no $11$ 's\nIt must have at least one $1$ . \nIf it has exactly one $1$ , there must be nine $111$ ’s and $n = 28$ . \nThus, for $n \\geq 46$ , the sum has more than one $1$ , so it must have at least $1000 - 8 \\cdot 111 = 112$ number of $1$ ’s.\nFor $n \\leq 1000$ , at least one $111$ . \nTo show that if $n$ is possible, then $(n + 9)$ is possible, replace a $111$ with $1 + 1 + 1$ , replace eleven $(1 + 1)$ ’s with eleven $11$ ’s, and include nine new $1$ ’s as $+1$ ’s. The sum remains $1000$\n$\\textbf{Case 2:}$ Sum has at least one $11$\nReplace an $11$ with $1 + 1$ , and include nine new $1$ ’s as $+1$ ’s.\nNow note that $46$ is possible because $8 \\cdot 111 + 10 \\cdot 11 + 2 \\cdot 1 = 1000$ .\nThus all elements of $S$ except $37$ are possible.\nThus there are $\\boxed{108}$ possible values for $n$" ]

[ "Let $M$ be the desired mean. Then because $\\dbinom{2015}{1000}$ subsets have 1000 elements and $\\dbinom{2015 - i}{999}$ have $i$ as their least element, \\begin{align*} \\binom{2015}{1000} M &= 1 \\cdot \\binom{2014}{999} + 2 \\cdot \\binom{2013}{999} + \\dots + 1016 \\cdot \\binom{999}{999} \\\\ &= \\binom{2014}{999} + \\binom{2013}{999} + \\dots + \\binom{999}{999} \\\\ & + \\binom{2013}{999} + \\binom{2012}{999} + \\dots + \\binom{999}{999} \\\\ & \\dots \\\\ & + \\binom{999}{999} \\\\ &= \\binom{2015}{1000} + \\binom{2014}{1000} + \\dots + \\binom{1000}{1000} \\\\ &= \\binom{2016}{1001}. \\end{align*} Using the definition of binomial coefficient and the identity $n! = n \\cdot (n-1)!$ , we deduce that \\[M = \\frac{2016}{1001} = \\frac{288}{143}.\\] The answer is $\\boxed{431}.$", "Each 1000-element subset $\\left\\{ a_1, a_2,a_3,...,a_{1000}\\right\\}$ of $\\left\\{1,2,3,...,2015\\right\\}$ with $a_1<a_2<a_3<...<a_{1000}$ contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\\left\\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ . There are $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<...<a_{1000}+1$ $k$ can be anything from $1$ to $a_1$ inclusive). Thus, the number of ways to choose the set $\\left\\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ is equal to the sum. But choosing a set $\\left\\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ is the same as choosing a 1001-element subset from $\\left\\{1,2,3,...,2016\\right\\}$\nThus, the average is $\\frac{\\binom{2016}{1001}}{\\binom{2015}{1000}}=\\frac{2016}{1001}=\\frac{288}{143}$ . Our answer is $p+q=288+143=\\boxed{431}$", "Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$ ). We can easily find the answers for smaller values of $p$ and $q$\nFor $p = 2$ and $q = 2$ , the answer is $1$\nFor $p = 3$ and $q = 2$ , the answer is $\\frac43$\nFor $p = 4$ and $q = 2$ , the answer is $\\frac53$\nFor $p = 3$ and $q = 3$ , the answer is $1$\nFor $p = 4$ and $q = 3$ , the answer is $\\frac54$\nFor $p = 5$ and $q = 3$ , the answer is $\\frac32$\nAt this point, we can see a pattern: our desired answer is always $\\frac{p+1}{q+1}$ . Plugging in $p = 2015$ and $q = 1000$ , the answer is $\\frac{2016}{1001}=\\frac{288}{143}$ , so $288 + 143 = \\boxed{431}$" ]

[ "Assume that $5 \\in \\{a_1, a_2, a_3\\}$ $m \\neq 5$ , and WLOG, $\\max{(a_1, a_2, a_3)} = 5$ . Then we know that the other two medians in $\\{a_1, a_2, a_3\\}$ and the smallest number of rows 1, 2, and 3 are all less than 5. But there are only 4 numbers less than 5 in $1, 2, 3, \\dots, 9$ , a Contradiction. Thus, if $5 \\in \\{a_1, a_2, a_3\\}$ , then $m = 5$\nWLOG, assume $5$ is in the upper left corner. One of the two other values in the top row needs to be below $5$ , and the other needs to be above $5$ . This can be done in $4\\cdot4\\cdot2=32$ ways.\nThe other $6$ can be arranged in $6!=720$ ways.\nFinally, accounting for when $5$ is in every other space, our answer is $32\\cdot720\\cdot9$ , which is $207360$ . But we only need the last $3$ digits, so $\\boxed{360}$ is our answer.", "(Complementary Counting with probability)\nNotice that m can only equal 4, 5, or 6, and 4 and 6 are symmetric.\nWLOG let $m=4$\n1. There is a $\\frac{15}{28}$ chance that exactly one of 1, 2, 3 is in the same row with 4.\nThere are 3 ways to select which of the smaller numbers will get in the row, and then 5\nways to select the number larger than 4.\n$\\frac{\\dbinom{3}{1}\\cdot\\dbinom{5}{1}}{\\dbinom{8}{2}} = \\frac{15}{28}$\n2. There is a $\\frac{2}{5}$ chance that the other two smaller numbers end up in the same row.\nThere are 2 ways to select the row that the two smaller number are in, and then $\\dbinom{3}{2}$ ways\nto place the smaller numbers in the row.\n$\\frac{\\dbinom{2}{1}\\cdot\\dbinom{3}{2}}{\\dbinom{6}{2}} = \\frac{2}{5}$\n$9!(1-2*\\frac{15}{28}*\\frac{2}{5})=362880*\\frac{4}{7}=207\\boxed{360}$", "We will make sure to multiply by $3!$ in the end to account for all the possible permutation of the rows.\nWLOG, let $5$ be present in the Row # $1$\nNotice that $5$ MUST be placed with a number lower than it and a number higher than it.\nThis happens in $4\\cdot4$ ways. You can permutate Row # $1$ in $3!$ ways.\nNow, take a look at Row $2$ and Row $3$\nBecause there are $6$ numbers to choose from now, you can assign #'s to Row's #2&3 in\n$\\frac{\\binom{6}{3}\\cdot\\binom{3}{3}}{2}$ ways. There are $3!\\cdot3!$ ways to permute the numbers in the individual Rows.\nHence, our answer is $3!(4\\cdot4\\cdot3!\\cdot{10}\\cdot{3!}\\cdot{3!})=3!(16\\cdot60\\cdot36)=3!(34560)\\implies{207\\boxed{360}$", "We see that if one of the medians is 5, then there are two remaining numbers greater than 5 and two less than 5, so it follows that $m=5$ . There are 3 ways to choose which row to have 5 in, $4\\cdot 4=16$ ways to choose the other two numbers in that row, $3!=6$ ways to arrange the numbers in that row, and $6!=720$ ways for the remaining numbers, for our answer is $3\\cdot 16\\cdot 6\\cdot 720=207\\boxed{360}$ .\n-Stormersyle", "We take the grid, and we do a bunch of stuff with it. First, we sort each row, smallest on the left, largest on the right, then we arrange these 3 rows such that the middle #s are increasing, from top to bottom. Thus, we get that the cell in the very center of the grid must be 5. (We need to multiply by 1296 at the end)\nLet S mean a number that is < 5, L mean a number > 5.\nBobby.jpg\nIn the 2 blank corners, one can be S, and the other one has to be L.\nIf the top right corner is S, there are 16 ways, otherwise, there are 144 ways. (This is left as an exercise to the reader).\nThus, there are 160 * 1296 total configurations, which gets us an answer of $207\\boxed{360}$", "We note that if $5$ is a median of one of the rows, then $m=5$ . First, focus on the row with $5$ in it. There are $4^2$ ways to choose the other numbers in that row and then $3!$ ways to order it. Now, clearly, there are $6!$ ways to put the other $6$ numbers into the remaining slots so $Q=6!\\cdot3!\\cdot4^2\\cdot3=207360$ . Hence, our answer is $\\boxed{360}$" ]

[ "The discriminant of the quadratic is $b^2 - 4c$ . Since the quadratic has real roots, \\[b^2 - 4c \\ge 0\\] \\[b^2 \\ge 4c\\] If $b = 6$ , then $c$ can be from $1$ to $6$ . If $b = 5$ , then $c$ can also be from $1$ to $6$ . If $b=4$ , then $c$ can be from $1$ to $4$ . If $b=3$ , then $c$ can be $1$ or $2$ . If $b=2$ , then $c$ can only be $1$ . If $b = 1$ , no values of $c$ in the set would work.\nThus, there are a total of $19$ equations that work. The answer is $\\boxed{19}$" ]

[ "We have \\begin{align*} |f(f(800))-f(f(400))| &\\leq \\frac12|f(800)-f(400)| &&(\\bigstar) \\\\ &\\leq \\frac12\\left|\\frac12|800-400|\\right| \\\\ &= 100, \\end{align*} from which we eliminate answer choices $\\textbf{(D)}$ and $\\textbf{(E)}.$\nNote that \\begin{alignat*}{8} |f(800)-f(300)| &\\leq \\frac12|800-300| &&= 250, \\\\ |f(800)-f(900)| &\\leq \\frac12|800-900| &&= 50, \\\\ |f(400)-f(300)| &\\leq \\frac12|400-300| &&= 50, \\\\ |f(400)-f(900)| &\\leq \\frac12|400-900| &&= 250. \\\\ \\end{alignat*} Let $a=f(300)=f(900).$ Together, it follows that \\begin{align*} |f(800)-a|&\\leq 50, \\\\ |f(400)-a|&\\leq 50. \\\\ \\end{align*} We rewrite $(\\bigstar)$ as \\begin{align*} |f(f(800))-f(f(400))| &\\leq \\frac12|f(800)-f(400)| \\\\ &= \\frac12|(f(800)-a)-(f(400)-a)| \\\\ &\\leq \\frac12|50-(-50)| \\\\ &=\\boxed{50} ~MRENTHUSIASM", "Denote $f(900)-f(600) = a$ .\nBecause $f(300) = f(900)$ $f(300) - f(600) = a$\nFollowing from the Lipschitz condition given in this problem, $|a| \\leq 150$ and \\[ f(800) - f(600) \\leq \\min \\left\\{ a + 50 , 100 \\right\\} \\] and \\[ f(400) - f(600) \\geq \\max \\left\\{ a - 50 , -100 \\right\\} . \\] Thus, \\begin{align*} f(800) - f(400) & \\leq \\min \\left\\{ a + 50 , 100 \\right\\} - \\max \\left\\{ a - 50 , -100 \\right\\} \\\\ & = 100 + \\min \\left\\{ a, 50 \\right\\} - \\max \\left\\{ a , - 50 \\right\\} \\\\ & = 100 + \\left\\{ \\begin{array}{ll} a + 50 & \\mbox{ if } a \\leq -50 \\\\ 0 & \\mbox{ if } -50 < a < 50 \\\\ -a + 50 & \\mbox{ if } a \\geq 50 \\end{array} \\right. . \\end{align*} Thus, $f(800) - f(400)$ is maximized at $a = 0$ $f(800)-f(600) = 50$ $f(400)-f(600)=-50$ , with the maximal value 100.\nBy symmetry, following from an analogous argument, we can show that $f(800) - f(400)$ is minimized at $a = 0$ $f(800)-f(600) = -50$ $f(400)-f(600)=50$ , with the minimal value $-100$\nFollowing from the Lipschitz condition, \\begin{align*} f(f(800)) - f(f(400)) & \\leq \\frac{1}{2} \\left| f(800) - f(400) \\right| \\\\ & \\leq 50 . \\end{align*} We have already construct instances in which the second inequality above is augmented to an equality.\nNow, we construct an instance in which the first inequality above is augmented to an equality.\nConsider the following piecewise-linear function: \\[ f(x) = \\left\\{ \\begin{array}{ll} \\frac{1}{2} \\left( x - 300 \\right) & \\mbox{ if } x \\leq 300 \\\\ -\\frac{1}{2} \\left( x - 300 \\right) & \\mbox{ if } 300 < x \\leq 400 \\\\ \\frac{1}{2} \\left( x - 600 \\right) & \\mbox{ if } 400 < x \\leq 800 \\\\ -\\frac{1}{2} \\left( x - 900 \\right) & \\mbox{ if } x > 800 \\end{array} \\right.. \\] Therefore, the maximum value of $f(f(800)) - f(f(400))$ is $\\boxed{50}$", "Divide both sides by $|x - y|$ to get $\\frac{|f(x) - f(y)|}{|x - y|} \\leq \\frac{1}{2}$ . This means that when we take any two points on $f$ , the absolute value of the slope between the two points is at most $\\frac{1}{2}$\nLet $f(300) = f(900) = c$ , and since we want to find the maximum value of $|f(800) - f(400)|$ , we can take the most extreme case and draw a line with slope $-\\frac{1}{2}$ down from $f(300)$ to $f(400)$ and a line with slope $\\frac{1}{2}$ up from $f(800)$ to $f(900)$ . Then $f(400) = c - 50$ and $f(800) = c + 50$ , so $|f(800) - f(400)| = |c + 50 - (c - 50)| = 100$ , and this is attainable because the slope of the line connecting $f(400)$ and $f(800)$ still has absolute value less than $\\frac{1}{2}$\nTherefore, $|f(f(800)) - f(f(400))| \\leq \\frac{1}{2}|f(800) - f(400)| = \\frac{1}{2}(100) = \\boxed{50}$", "Consider $g(x) = f(x)-f(300)$ . Then $g(x)$ satisfies all the conditions and $g(300) = g(900) = 0$ . We would want $g(400)$ and $g(800)$ as distant from each other as possible. So assign $g(400) = -50$ and $g(800) = 50$ , the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for $|g(g(800)) - g(g(400))| = |g(50) - g(-50 )| \\leq \\frac{1}{2}(100) = \\boxed{50}$ as the answer." ]

[ "Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that \\[P(-1)=-a+b-c+d=-9.\\] Let $a'=9-a$ and $c'=9-c.$ Note that both of $a'$ and $c'$ are integers from $0$ through $9.$ Moreover, the ordered quadruples $(a,b,c,d)$ and the ordered quadruples $(a',b,c',d)$ have one-to-one correspondence.\nWe rewrite the given equation as $(9-a)+b+(9-c)+d=9,$ or \\[a'+b+c'+d=9.\\] By the stars and bars argument, there are $\\binom{9+4-1}{4-1}=\\boxed{220}$ ordered quadruples $(a',b,c',d).$", "Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that $P(-1)=-a+b-c+d=-9,$ which rearranges to \\[b+d+9=a+c.\\] Note that $b+d+9$ is an integer from $9$ through $27,$ and $a+c$ is an integer from $0$ through $18.$ Therefore, both of $b+d+9$ and $a+c$ are integers from $9$ through $18.$ We construct the following table: \\[\\begin{array}{c|c|c|c||c} & & & & \\\\ [-2.5ex] \\boldsymbol{b+d} & \\boldsymbol{\\#}\\textbf{ of Ordered Pairs }\\boldsymbol{(b,d)} & \\boldsymbol{a+c} & \\boldsymbol{\\#}\\textbf{ of Ordered Pairs }\\boldsymbol{(a,c)} & \\boldsymbol{\\#}\\textbf{ of Ordered Quadruples }\\boldsymbol{(a,b,c,d)} \\\\ [0.5ex] \\hline & & & & \\\\ [-2ex] 0 & 1 & 9 & 10 & 1\\cdot10=10 \\\\ 1 & 2 & 10 & 9 & \\phantom{0}2\\cdot9=18 \\\\ 2 & 3 & 11 & 8 & \\phantom{0}3\\cdot8=24 \\\\ 3 & 4 & 12 & 7 & \\phantom{0}4\\cdot7=28 \\\\ 4 & 5 & 13 & 6 & \\phantom{0}5\\cdot6=30 \\\\ 5 & 6 & 14 & 5 & \\phantom{0}6\\cdot5=30 \\\\ 6 & 7 & 15 & 4 & \\phantom{0}7\\cdot4=28 \\\\ 7 & 8 & 16 & 3 & \\phantom{0}8\\cdot3=24 \\\\ 8 & 9 & 17 & 2 & \\phantom{0}9\\cdot2=18 \\\\ 9 & 10 & 18 & 1 & 10\\cdot1=10 \\end{array}\\] We sum up the counts in the last column to get the answer $2\\cdot(10+18+24+28+30)=\\boxed{220}.$" ]

[ "It never hurts to compute a few terms of the sequence in order to get a feel how it looks like. In our case, the definition is that $\\forall$ (for all) $n>1:~ a_n = a_{n-1}a_{n+1} - 1$ . This can be rewritten as $a_{n+1} = \\frac{a_n +1}{a_{n-1}}$ . We have $a_1=x$ and $a_2=2000$ , and we compute:\n\\begin{align*} a_3 & = \\frac{a_2+1}{a_1} = \\frac{2001}x \\\\ a_4 & = \\frac{a_3+1}{a_2} = \\frac{ \\dfrac{2001}x + 1 }{ 2000 } = \\frac{2001 + x}{2000x} \\\\ a_5 & = \\frac{a_4+1}{a_3} = \\frac{ \\frac{2001 + x}{2000x} + 1 }{ \\frac{2001}x } = \\frac{ \\frac{2001 + 2001x}{2000x} }{ \\frac{2001}x } = \\frac{1+x}{2000} \\\\ a_6 & = \\frac{a_5+1}{a_4} = \\frac{ \\frac{1+x}{2000} + 1 }{ \\frac{2001 + x}{2000x} } = \\frac{ \\frac{2001+x}{2000} }{ \\frac{2001 + x}{2000x} } = x \\\\ a_7 & = \\frac{a_6+1}{a_5} = \\frac{ x+1 }{ \\frac{1+x}{2000} } = 2000 \\end{align*}\nAt this point we see that the sequence will become periodic: we have $a_6=a_1$ $a_7=a_2$ , and each subsequent term is uniquely determined by the previous two.\nHence if $2001$ appears, it has to be one of $a_1$ to $a_5$ . As $a_2=2000$ , we only have four possibilities left. Clearly $a_1=2001$ for $x=2001$ , and $a_3=2001$ for $x=1$ . The equation $a_4=2001$ solves to $x = \\frac{2001}{2000\\cdot 2001 - 1}$ , and the equation $a_5=2001$ to $x=2000\\cdot 2001 - 1$\nNo two values of $x$ we just computed are equal, and therefore there are $\\boxed{4}$ different values of $x$ for which the sequence contains the value $2001$" ]

[ "Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$ . If a sequence ends in an $A$ , then it must have been formed by appending two $A$ s to the end of a string of length $n-2$ . If a sequence ends in a $B,$ it must have either been formed by appending one $B$ to a string of length $n-1$ ending in an $A$ , or by appending two $B$ s to a string of length $n-2$ ending in a $B$ . Thus, we have the recursions \\begin{align*} a_n &= a_{n-2} + b_{n-2}\\\\ b_n &= a_{n-1} + b_{n-2} \\end{align*} By counting, we find that $a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0$ \\[\\begin{array}{|r||r|r|||r||r|r|} \\hline n & a_n & b_n & n & a_n & b_n\\\\ \\hline 1&0&1& 8&6&10\\\\ 2&1&0& 9&11&11\\\\ 3&1&2& 10&16&21\\\\ 4&1&1& 11&22&27\\\\ 5&3&3& 12&37&43\\\\ 6&2&4& 13&49&64\\\\ 7&6&5& 14&80&92\\\\ \\hline \\end{array}\\] Therefore, the number of such strings of length $14$ is $a_{14} + b_{14} = \\boxed{172}$", "Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$\nAdditionally, let $t_n$ denote the total number of sequences of length $n$ . Then, $t_n=a_n+b_n$ , as the total amount of sequences of length $n$ consists of the sequences of length $n$ ending in $A$ and the sequences of length $n$ ending in $B$ \\begin{align*} a_n &= a_{n-2} + b_{n-2}\\\\ b_n &= a_{n-1} + b_{n-2} \\end{align*} The recursion for $a_n$ tells us that $a_n=a_{n-2}+b_{n-2}$ . However, this is also the definition for $t_{n-2}$ . Therefore, $a_n=t_{n-2}$\nWe also know from our recursion for $b_n$ that $b_n=a_{n-1}+b_{n-2}$ . Substituting for $a_n$ and $b_n$ into our recursion for $t_n$ gives us $t_n=t_{n-2}+a_{n-1}+b_{n-2}$\nFurthermore, note that since $a_n=t_{n-2}$ $a_{n-1}=t_{n-3}$ . Furthermore, using our definition for $t_{n-2}$ , we can rewrite $b_{n-2}$ as $t_{n-2}-a_{n-2}$ . Substituting for $a_{n-1}$ and $b_{n-2}$ into our recursion for $t_n$ gives us $t_n=t_{n-2}+t_{n-3}+t_{n-2}-a_{n-2}$\nFinally, note that since $a_n=t_{n-2}$ $a_{n-2}=t_{n-4}$ . Substituting for $a_{n-2}$ into our recursion for $t_n$ gives us $t_n=2t_{n-2}+t_{n-3}-t_{n-4}$ . We now have a recursion only in terms of $t$\nBy counting, we find that $t_1=1$ $t_2=1$ $t_3=3$ , and $t_4=2$ \\[\\begin{array}{|r|r||r|r|}\t \\hline\t n & t_n & n & t_n\\\\\t \\hline\t 1&1&8&16\\\\\t 2&1&9&22\\\\ 3&3&10&37\\\\\t 4&2&11&49\\\\\t 5&6&12&80\\\\\t 6&6&13&113\\\\\t 7&11&14&172\\\\\t \\hline \\end{array}\\]\nTherefore, the number of such sequences of length 14 is $\\boxed{172}$", "There must be an even amount of runs of consecutive $B$ s due to parity. Thus, we can split this sequence into the following cases: $A$ $BAAB$ $AABAAB$ $BAABAA$ $AABAABAA$ $BAABAABAAB$ $AABAABAABAAB$ $BAABAABAABAA$ , and $AABAABAABAABAA$ , in which the amount of letters in one run does not necessarily represent the amount of letters there can be.\nFor the first case and the last case, there is only one possible sequence of letters.\nFor all other cases, we can insert two of the same letter at a time into a run that has the exact same letter. For example, for the second case, we can insert two $A$ s and make the sequence $BAAAAB$ . There are three \"slots\" in which we can insert two additional letters in, and we must insert five groups of new letters. By stars and bars , the number of ways for the second case is $\\binom{7}{2}=21$\nApplying this logic to all of the other cases gives us $\\binom{7}{3}$ $\\binom{7}{3}$ $\\binom{7}{4}$ $\\binom{8}{6}$ $\\binom{8}{1}$ , and $\\binom{8}{1}$ . Adding 1+ $\\binom{7}{2}$ $\\binom{7}{3}$ $\\binom{7}{3}$ $\\binom{7}{4}$ $\\binom{8}{6}$ $\\binom{8}{1}$ $\\binom{8}{1}$ gives us the answer $\\boxed{172}$" ]

[ "Let $M_1=\\begin{bmatrix}a_1 & b_1 & c_1\\end{bmatrix}, M_2=\\begin{bmatrix}a_2 & b_2 & c_2\\end{bmatrix},$ and $M_3=\\begin{bmatrix}a_3 & b_3 & c_3\\end{bmatrix}.$\nWe wish to count the ordered triples $(M_1,M_2,M_3)$ of row matrices. We perform casework:\nThere are $9+3=12$ ordered triples $(M_1,M_2,M_3).$\nSimilarly, for each of $M_1+M_3=M_2$ and $M_2+M_3=M_1,$ there are $12$ ordered triples $(M_1,M_2,M_3).$\nIn this subcase, we have $\\boldsymbol{12\\cdot3=36}$ ordered triples $\\boldsymbol{(M_1,M_2,M_3).}$\nTogether, the answer is $8+168+126+36=\\boxed{338}.$", "We will use complementary counting and do casework on the equations.\nThere are $8$ possible equations:\nEquation 1: $0 = 0$\nEquation 2: $x = 0$\nEquation 3: $y = 0$\nEquation 4: $z = 0$\nEquation 5: $x + y = 0$\nEquation 6: $x + z = 0$\nEquation 7: $y + z = 0$\nEquation 8: $x + y + z = 0$\nWe will continue to refer to the equations by their number on this list.\n$8^3 = 512$ total systems. Note that no two equations by themselves can force $x = y = z = 0$ . Therefore no system with Equation 1 or with repeated equations can force $x = y = z = 0$\nCase 1: Equation 8 ( $x + y + z = 0$ ) is present.\nCase 1a: Equation 8, and two equations from $\\{5, 6, 7\\}$\nThere are $\\binom{3}{2} = 3$ ways to choose two equations from $\\{5, 6, 7\\}$ and $3! = 6$ ways to arrange each case. The number of options that force $x = y = z = 0$ is $3 \\cdot 3! = 18$\nCase 1b: Equation 8, one equation from $\\{5, 6, 7\\}$ , and one equation from $\\{2, 3, 4\\}$\nThere are $\\binom{3}{1} = 3$ ways to choose one equation from $\\{5, 6, 7\\}$ . WLOG let us choose Equation 7. Given $x + y + z = 0$ and $y + z = 0$ , we conclude that $x = 0$ . The third equation can be either $y = 0$ or $z = 0$ . There are $3!$ ways to arrange each case. The number of options that force $x = y = z = 0$ is $3 \\cdot 2 \\cdot 3! = 36$\nCase 1c: Equation 8, and two equations from $\\{2, 3, 4\\}$\nThere are $\\binom{3}{2} = 3$ ways to choose two equations from $\\{2, 3, 4\\}$ and $3! = 6$ ways to arrange each case. Each of these cases forces $x = y = z = 0$ $3 \\cdot 3! = 18$ total options.\nCase 2: Equation 8 is $\\textbf{not}$ present, at least one equation from $\\{5, 6, 7\\}$ is present.\nCase 2a: Equations $\\{5, 6, 7\\}$ are all present.\nThere are $3!$ ways to arrange the three equations. $6$ options.\nCase 2b: Two equations from $\\{5, 6, 7\\}$ are present. One equation from $\\{2, 3, 4\\}$ is present.\nThere are $\\binom{3}{2}$ ways to choose two equations from $\\{5, 6, 7\\}$ . WLOG let Equations 5 and 6 be in our system: $x + y = 0$ and $x + z = 0$ . Any equation from $\\{2, 3, 4\\}$ will force $x = y = z = 0$ . There are $3!$ ways to arrange the equations. The number of options that force $x = y = z = 0$ is $\\binom{3}{2} \\cdot \\binom{3}{1} \\cdot 3! = 54$\nCase 2c: One equation from $\\{5, 6, 7\\}$ is present. Two equations from $\\{2, 3, 4\\}$ are present.\nThere are $\\binom{3}{1}$ ways to choose one equation from $\\{5, 6, 7\\}$ . WLOG let Equation 5 ( $x + y = 0$ ) be present. One of the two equations from $\\{2, 3, 4\\}$ must be Equation 4, $z = 0$ , since it is the only equation that restricts $z$ . The last equation can be either 2 or 3. There are $3!$ ways to arrange the equations. The number of options that force $x = y = z = 0$ is $\\binom{3}{1} \\cdot \\binom{2}{1} \\cdot 3! = 36$\nCase 3: Only equations $\\{2, 3, 4\\}$ are present.\nThere are $3!$ ways to arrange the three equations. $6$ options.\nWe add up the cases: $18 + 36 + 18 + 6 + 54 + 36 + 6 = 174$ total systems force $x = y = z = 0$ . Thus $512 - 174 = \\boxed{338}$ do not.", "The total number of possible systems is $2^9 = 512$ , with $8$ possible sets of coefficients per equation. We will use complementary counting to find the number of systems which only have the solution $(0, 0, 0)$ and subtract that from the total. Similar to what is observed in Solution 2, if any equation is repeated or $0x + 0y + 0z = 0$ , there will only be two or fewer equations for three variables, making one unique solution impossible. Therefore, we must choose $3$ different equations from $7$ possible ones, giving $7 \\cdot 6 \\cdot 5 = 210$ systems. However, there are two exceptions to consider, which will have more than one solution. The first is of the form $x + y + z = 0$ $x + y = 0$ $z = 0$ ; the second is of the form $x = 0$ $y = 0$ $x + y = 0$ . In both cases, there are $3$ ways to choose the variables in the equations, and then $6$ ways to arrange them, giving $2 \\cdot 3 \\cdot 6 = 36$ exceptions. Subtracting this gives $210 - 36 = 174$ systems with only one solution, and the answer is then $512 - 174 = \\boxed{338}$", "Denote vector $\\overrightarrow{i} = \\left( i_1, i_2, i_3 \\right)^T$ for $i \\in \\left\\{ a, b, c \\right\\}$ .\nThus, we need to count how many vector tuples $\\left( \\overrightarrow{a} , \\overrightarrow{b} , \\overrightarrow{c} \\right)$ are linearly dependent.\nWe do complementary counting.\nFirst, the total number of vector tuples $\\left( \\overrightarrow{a} , \\overrightarrow{b} , \\overrightarrow{c} \\right)$ is $\\left( 2^3 \\right)^3 = 512$\nSecond, we count how many many vector tuples $\\left( \\overrightarrow{a} , \\overrightarrow{b} , \\overrightarrow{c} \\right)$ are linearly independent.\nTo meet this condition, no vector can be a zero vector $\\overrightarrow{0} = \\left( 0, 0, 0 \\right)^T$\nNext, we do the casework analysis.\nCase $1^c$ : Three vectors are all on axes.\nIn this case, the number of $\\left( \\overrightarrow{a} , \\overrightarrow{b} , \\overrightarrow{c} \\right)$ is $3!$\nCase $2^c$ : Two vectors are on axes and the third vector is not.\nWe construct such an instance in the following steps.\nStep 1: We determine which two vectors lie on axes.\nThe number of ways is $3$\nStep 2: For two vectors selected in Step 1, we determine which two axes they lie on.\nThe number of ways is $3 \\cdot 2$\nStep 3: For the third unselected vector, we determine its value.\nTo make three vectors linear independent, the third vector cannot be on the plane formed by the first two vectors.\nSo the number of ways is $3$\nFollowing from the rule of product, the number of $\\left( \\overrightarrow{a} , \\overrightarrow{b} , \\overrightarrow{c} \\right)$ in this case is $3 \\cdot 3 \\cdot 2 \\cdot 3$\nCase $3^c$ : One vector is on an axis and the other two are not.\nWe construct such an instance in the following steps.\nStep 1: We determine which vector lies on an axis.\nThe number of ways is $3$\nStep 2: For the selected vector, we determine which axis it lies on.\nThe number of ways is $3$\nStep 3: We determine the values of the two unselected vectors.\nFirst, to be linearly independent, these two vectors are distinct.\nSecond, to be linearly independent, we cannot have one vector $(1,1,1)$ and another one that is a diagonal vector on the plane that is perpendicular to the first selected vector.\nThus, the number or ways in this step is $4 \\cdot 3-2 = 10$\nFollowing from the rule of product, the number of $\\left( \\overrightarrow{a} , \\overrightarrow{b} , \\overrightarrow{c} \\right)$ in this case is $3 \\cdot 3 \\cdot 10$\nCase $(4.4)^c$ : No vector is on any axis.\nIn this case, any three distinct vectors are linearly independent.\nSo the number of $\\left( \\overrightarrow{a} , \\overrightarrow{b} , \\overrightarrow{c} \\right)$ in this case is $4 \\cdot 3 \\cdot 2$\nPutting all cases together, the number of vector tuples $\\left( \\overrightarrow{a} , \\overrightarrow{b} , \\overrightarrow{c} \\right)$ that are linearly independent is \\[ 8^3 - 3! - 3 \\cdot 3 \\cdot 2 \\cdot 3 - 3 \\cdot 3 \\cdot 10 - 4 \\cdot 3 \\cdot 2 = \\boxed{338}. \\] ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)" ]

[ "We can partition the $n^\\text{th}$ ring into $4$ rectangles: two containing $2n+1$ unit squares and two containing $2n-1$ unit squares.\nThere are $2(2n+1)+2(2n-1)=4n+2+4n-2=8n$ unit squares in the $n^\\text{th}$ ring.\nThus, the $100^\\text{th}$ ring has $8 \\times 100 = \\boxed{800}$ unit squares.", "We can make the $n^\\text{th}$ ring by removing a square of side length $2n-1$ from a square of side length $2n+1$\nThis ring contains $(2n+1)^2-(2n-1)^2=(4n^2+4n+1)-(4n^2-4n+1)=8n$ unit squares.\nThus, the $100^\\text{th}$ ring has $8 \\times 100 = \\boxed{800}$ unit squares.", "Notice that the first ring around the center square contains $8$ unit squares, the second ring contains $16$ unit squares, the third contains $24$ unit squares, and so on. The number of squares in the $n^\\text{th}$ ring is determined by the expression $8 \\times n$ . Thus, the number of unit squares in the $100^\\text{th}$ ring is equal to $8 \\times 100$ , which equals $\\boxed{800}$ unit squares." ]

[ "We apply casework to this problem. The only sets that contain two multiples of seven are those for which:\nEach case has $\\left\\lfloor\\frac{100}{7}\\right\\rfloor=14$ sets. Therefore, the answer is $14\\cdot3=\\boxed{42}.$", "We find a pattern. \\begin{align*} &\\{1,2,3,\\ldots,10\\}, \\\\ &\\{11,12,13,\\ldots,20\\},\\\\ &\\{21,22,23,\\ldots,30\\},\\\\ &\\vdots\\\\ &\\{991,992,993,\\ldots,1000\\}. \\end{align*} We can figure out that the first set has $1$ multiple of $7$ . The second set also has $1$ multiple of $7$ . The third set has $2$ multiples of $7$ . The fourth set has $1$ multiple of $7$ . The fifth set has $2$ multiples of $7$ . The sixth set has $1$ multiple of $7$ . The seventh set has $2$ multiples of $7$ . Calculating this pattern further, we can see (reasonably) that it repeats for each $7$ sets.\nWe see that the pattern for the number of multiples per $7$ sets goes: $1,1,2,1,2,1,2.$ So, for every $7$ sets, there are three sets with $2$ multiples of $7$ . We calculate $\\left\\lfloor\\frac{100}{7}\\right\\rfloor$ and multiply that by $3$ . (We also disregard the remainder of $2$ since it doesn't add any extra sets with $2$ multiples of $7$ .). We get $14\\cdot3= \\boxed{42}$", "Each set contains exactly $1$ or $2$ multiples of $7$\nThere are $\\dfrac{1000}{10}=100$ total sets and $\\left\\lfloor\\dfrac{1000}{7}\\right\\rfloor = 142$ multiples of $7$\nThus, there are $142-100=\\boxed{42}$ sets with $2$ multiples of $7$" ]

[ "We apply casework to this problem. The only sets that contain two multiples of seven are those for which:\nEach case has $\\left\\lfloor\\frac{100}{7}\\right\\rfloor=14$ sets. Therefore, the answer is $14\\cdot3=\\boxed{42}.$", "We find a pattern. \\begin{align*} &\\{1,2,3,\\ldots,10\\}, \\\\ &\\{11,12,13,\\ldots,20\\},\\\\ &\\{21,22,23,\\ldots,30\\},\\\\ &\\vdots\\\\ &\\{991,992,993,\\ldots,1000\\}. \\end{align*} We can figure out that the first set has $1$ multiple of $7$ . The second set also has $1$ multiple of $7$ . The third set has $2$ multiples of $7$ . The fourth set has $1$ multiple of $7$ . The fifth set has $2$ multiples of $7$ . The sixth set has $1$ multiple of $7$ . The seventh set has $2$ multiples of $7$ . Calculating this pattern further, we can see (reasonably) that it repeats for each $7$ sets.\nWe see that the pattern for the number of multiples per $7$ sets goes: $1,1,2,1,2,1,2.$ So, for every $7$ sets, there are three sets with $2$ multiples of $7$ . We calculate $\\left\\lfloor\\frac{100}{7}\\right\\rfloor$ and multiply that by $3$ . (We also disregard the remainder of $2$ since it doesn't add any extra sets with $2$ multiples of $7$ .). We get $14\\cdot3= \\boxed{42}$", "Each set contains exactly $1$ or $2$ multiples of $7$\nThere are $\\dfrac{1000}{10}=100$ total sets and $\\left\\lfloor\\dfrac{1000}{7}\\right\\rfloor = 142$ multiples of $7$\nThus, there are $142-100=\\boxed{42}$ sets with $2$ multiples of $7$" ]

[ "Substituting $y = Ax^2$ into the equation $y^2+3 = x^2+4y$ gives \\begin{align*}\\left(Ax^2\\right)+3 = x^2+4\\left(Ax^2\\right) &\\iff A^2x^4+3 = x^2+4Ax^2 \\\\ &\\iff A^2x^4-\\left(4A+1\\right)x^2+3 = 0 \\\\ &\\iff x^2 = \\frac{4A+1 \\pm \\sqrt{4A^2+8A+1}}{2A^2} \\\\ &\\text{(using the quadratic formula)}.\\end{align*} Now observe that since $A$ is positive, $4A^2+8A+1$ is also positive, so the square root will always give two distinct real values. Moreover, \\[\\left(4A+1\\right)^2 = 16A^2+8A+1 > 4A^2+8A+1,\\] so $4A+1-\\sqrt{4A^2+8A+1} > 0$ , meaning that both solutions for $x^2$ are positive. Hence both solutions will give $2$ distinct values of $x$ (the positive and negative square roots), and each of these will correspond to a distinct point of intersection of the graphs, so there are $2 \\cdot 2 = \\boxed{4}$ points of intersection.", "Firstly, note that $y = Ax^2$ is an upward-facing parabola (since $A > 0$ ) whose vertex is at the origin. We now manipulate the equation of the second graph as follows: \\begin{align*}y^2+3 = x^2+4y &\\iff y^2-4y+3-x^2 = 0 \\\\ &\\iff y^2-4y+4-x^2 = 1 \\\\ &\\iff \\frac{(y-2)^2}{1}-\\frac{(x-0)^2}{1} = 1,\\end{align*} showing that it is a vertical (upward- and downward-opening) hyperbola with center $(0,2)$ and asymptotes $y=x+1$ and $y=-x+1$ . It therefore remains to consider graphically where the parabola will intersect the hyperbola.\nOn the lower branch of the hyperbola, the maximum point is $(0,2-1) = (0,1)$ , which is above the vertex of the parabola. Therefore, by continuity and the symmetry of both the parabola and the hyperbola in the $y$ -axis, there are always exactly $2$ intersection points here.\nFor the top branch, as it approaches the asymptote $y = x+1$ , its slope also approaches that of this asymptote, which is $1$ . However, for any upward-opening parabola, the slope approaches infinity as $x$ does, so no matter how small $A$ is (i.e. how 'flat' the parabola is), the parabola will eventually overtake the hyperbola, giving a point of intersection with positive $x$ -coordinate. As above, symmetry gives another point of intersection with negative $x$ -coordinate, so that there are $2$ intersection points with this branch too.\nThus there are a total of $2+2 = \\boxed{4}$ intersection points." ]

[ "Let's express the number in terms of $10^n$ . We can obtain $(10-1)+(10^2-1)+(10^3-1)+\\cdots+(10^{321}-1)$ . By the commutative and associative property, we can group it into $(10+10^2+10^3+\\cdots+10^{321})-321$ . We know the former will yield $1111....10$ , so we only have to figure out what the last few digits are. There are currently $321$ 1's. We know the last four digits are $1110$ , and that the others will not be affected if we subtract $321$ . If we do so, we get that $1110-321=789$ . This method will remove three $1$ 's, and add a $7$ $8$ and $9$ . Therefore, the sum of the digits is $(321-3)+7+8+9=\\boxed{342}$", "Observe how adding results in the last term but with a $1$ concatenated in front and also a $1$ subtracted ( $09$ $108$ $1107$ $11106$ ). Then for any index of terms, $n$ , the sum is $11...10-n$ , where the first term is of length $n+1$ . Here, that is $\\boxed{342}$" ]

[ "Using the chart, $(2*4)=3$ and $(1*3)=3$ . Therefore, $(2*4)*(1*3)=3*3=\\boxed{4}$", "By the chart, we can see that the \" $*$ \" operation is actually multiplication modulo $5$ . Thus, we can do $(2*4)*(1*3)\\rightarrow(2\\cdot4)\\cdot(1\\cdot3)=8\\cdot3=24\\rightarrow\\boxed{4}$" ]

[ "As shown in the image above, let $D$ $E$ , and $F$ be the midpoints of $\\overline{BC}$ $\\overline{CA}$ , and $\\overline{AB}$ , respectively. Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $\\triangle ABC$ . The crux of this problem is the following lemma.\nLemma: The point $O$ is the orthocenter of $\\triangle ABC$\nProof. Observe that \\[OF^2 - OE^2 = PF^2 - PE^2 = AF^2 - AE^2;\\] the first equality follows by the Pythagorean Theorem, while the second follows from $AF = FP$ and $AE = EP$ . Thus, by the Perpendicularity Lemma, $AO$ is perpendicular to $FE$ and hence $BC$ . Analogously, $O$ lies on the $B$ -altitude and $C$ -altitude of $\\triangle ABC$ , and so $O$ is, indeed, the orthocenter of $\\triangle ABC$\nTo find the coordinates of $O$ , we need to find the intersection point of altitudes $BE$ and $AD$ . The equation of $BE$ is simply $x=16$ $AD$ is perpendicular to line $BC$ , so the slope of $AD$ is equal to the negative reciprocal of the slope of $BC$ $BC$ has slope $\\frac{24-0}{16-34}=-\\frac{4}{3}$ , therefore $y=\\frac{3}{4} x$ . These two lines intersect at $(16,12)$ , so that's the base of the height of the tetrahedron.\nLet $S$ be the foot of altitude $BS$ in $\\triangle BPQ$ . From the Pythagorean Theorem $h=\\sqrt{BS^2-SO^2}$ . However, since $S$ and $O$ are, by coincidence, the same point, $SO=0$ and $h=12$\nThe area of the base is $102$ , so the volume is $\\frac{102*12}{3}=\\boxed{408}$", "Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates $(17, 0, 0)$ $(8, 12, 0)$ , and $(25, 12, 0)$ . We can compute the area of this triangle as $102$ . Suppose $(x, y, z)$ are the coordinates of the vertex of the resulting pyramid. Call this point $V$ . Clearly, the height of the pyramid is $z$ . The desired volume is thus $\\frac{102z}{3} = 34z$\nWe note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, $VR = RA$ $VP = PB$ , and $VQ = QC$ . We then use distance formula to find the distances from $V$ to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding $z = 12$ . The desired volume is thus $34 \\times 12 = \\boxed{408}$", "The formed tetrahedron has pairwise parallel planar and oppositely equal length ( $4\\sqrt{13},15,17$ ) edges and can be inscribed in a parallelepiped (rectangular box) with the six tetrahedral edges as non-intersecting diagonals of the box faces. Let the edge lengths of the parallelepiped be $p,q,r$ and solve (by Pythagoras)\n$p^2+q^2=4^2\\cdot{13}$\n$q^2+r^2=15^2$\n$r^2+p^2=17^2$\nto find that $(p^2,q^2,r^2)=(153,136,72)=(3^2\\cdot{17},2^3\\cdot{17},2^3\\cdot{3^2}).$\nUse the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is $\\tfrac{1}{3}$ and then the volume is\n$\\tfrac{1}{3}pqr=\\tfrac{1}{3}\\sqrt{2^6\\cdot{3^4}\\cdot{17^2}}=\\boxed{408}$", "Let $A = (0,0), B = (16, 24), C = (34,0).$ Then define $D,E,F$ as the midpoints of $BC, AC, AB$ . By Pythagorean theorem, $EF = \\frac{1}{2} BC = 15, DE = \\frac{1}{2}AB = 4 \\sqrt{13}, DF = \\frac{1}{2} AC = 17.$ Then let $P$ be the point in space which is the vertex of the tetrahedron with base $DEF$\nNote that $\\triangle DEP \\cong \\triangle EDF$ . Create point $F'$ on the plane of $DEF$ such that $\\triangle DEP \\cong \\triangle DEF'$ (i.e by reflecting $F$ over the perpendicular bisector of $DE$ ). Project $F, P$ onto $DE$ as $X, Y$ . Note by the definition of $F'$ then $\\angle PYF'$ is the dihedral angle between planes $DEP, DEF$\nNow see that by Heron's, \\[[DEP] = [DEF] = \\sqrt{(16 + 2 \\sqrt{13})(16 - 2 \\sqrt{13})(1 + 2 \\sqrt{13})(-1 + 2 \\sqrt{13})} = 102.\\] So $PY$ , the hypotenuse $DEP$ has length $\\frac{102 \\cdot 2}{4 \\sqrt{13}} = \\frac{51}{\\sqrt{13}}$ . Similarly $F'Y = \\frac{51}{\\sqrt{13}}.$ Further from Pythagoras $DY = \\sqrt{DP^2 - PY^2} = \\frac{18}{\\sqrt{13}}.$ Symmetrically $EX = \\frac{18}{\\sqrt{13}}.$ Therefore $XY = DE - DY - EX = \\frac{16}{\\sqrt{13}}.$\nBy Law of Cosines on $\\triangle PYF'$ \\begin{align*} PF'^2 &= PY^2 + F'Y^2 - 2 \\cdot PY \\cdot F'Y \\cos{\\angle PYF'} \\\\ PF^2 - XY^2 &= 2 (\\frac{51}{\\sqrt{13}})^2 \\cos{\\angle PYF'} \\\\ (4\\sqrt{13})^2 - (\\frac{16}{\\sqrt{13}})^2 &= 2 (\\frac{51}{\\sqrt{13}})^2 \\cos{\\angle PYF'} \\\\ \\cos{\\angle PYF'} &= \\frac{9}{17} \\\\ \\sin{\\angle PYF'} &= \\frac{4 \\sqrt{13}}{17}. \\end{align*}.\nTherefore the altitude of the tetrahedron from vertex $P$ to base $DEF$ is $PY \\sin{\\angle PYF'} = \\frac{51}{\\sqrt{13}} \\frac{4 \\sqrt{13}}{17} = 12.$ So the area is $\\frac{1}{3}bh = \\frac{1}{3} 12 \\cdot 102 = \\boxed{408}.$" ]

[ "Let the first point on the line $x=10$ be $(10,45+a)$ where a is the height above $(10,45)$ . Let the second point on the line $x=28$ be $(28, 153-a)$ . For two given points, the line will pass the origin if the coordinates are proportional (such that $\\frac{y_1}{x_1} = \\frac{y_2}{x_2}$ ). Then, we can write that $\\frac{45 + a}{10} = \\frac{153 - a}{28}$ . Solving for $a$ yields that $1530 - 10a = 1260 + 28a$ , so $a=\\frac{270}{38}=\\frac{135}{19}$ . The slope of the line (since it passes through the origin) is $\\frac{45 + \\frac{135}{19}}{10} = \\frac{99}{19}$ , and the solution is $m + n = \\boxed{118}$", "You can clearly see that a line that cuts a parallelogram into two congruent pieces must go through the center of the parallelogram. Taking the midpoint of $(10,45)$ , and $(28,153)$ gives $(19,99)$ , which is the center of the parallelogram. Thus the slope of the line must be $\\frac{99}{19}$ , and the solution is $\\boxed{118}$", "Note that the area of the parallelogram is equivalent to $69 \\cdot 18 = 1242,$ so the area of each of the two trapezoids with congruent area is $621.$ Therefore, since the height is $18,$ the sum of the bases of each trapezoid must be $69.$\nThe points where the line in question intersects the long side of the parallelogram can be denoted as $(10, \\frac{10m}{n})$ and $(28, \\frac{28m}{n}),$ respectively. We see that $\\frac{10m}{n} - 45 + \\frac{28m}{n} - 84 = 69,$ so $\\frac{38m}{n} = 198 \\implies \\frac{m}{n} = \\frac{99}{19} \\implies \\boxed{118}.$", "$(\\Sigma x_i /4, \\Sigma y_i /4)$ is the centroid, which generates $(19,99)$ , so the answer is $\\boxed{118}$ . This is the fastest way because you do not need to find the opposite vertices by drawing." ]

[ "We divide the path into eight “ $R$ ” movements and eight “ $U$ ” movements. Five sections of alternative $RURUR$ or $URURU$ are necessary in order to make four “turns.” We use the first case and multiply by $2$\nFor $U$ , we have seven ordered pairs of positive integers $(a,b)$ such that $a+b=8$\nFor $R$ , we subtract $1$ from each section (to make the minimum stars of each section $0$ ) and we use Stars and Bars to get ${7 \\choose 5}=21$\nThus our answer is $7\\cdot21\\cdot2=\\boxed{294}$", "Notice that the $RURUR$ case and the $URURU$ case is symmetrical. WLOG, let's consider the RURUR case.\nNow notice that there is a one-to-one correspondence between this problem and the number of ways to distribute 8 balls into 3 boxes and also 8 other balls into 2 other boxes, such that each box has a nonzero amount of balls.\nThere are ${8+2-3 \\choose 2}$ ways for the first part, and ${8+1-2 \\choose 1}$ ways for the second part, by stars and bars.\nThe answer is $2\\cdot {7 \\choose 2} \\cdot {7 \\choose 1} = \\boxed{294}$", "Starting at the origin, you can either first go up or to the right. If you go up first, you will end on the side opposite to it (the right side) and if you go right first, you will end up on the top. It can then be observed that if you choose the turning points in the middle $7 \\times 7$ grid, that will automatically determine your start and ending points. For example, in the diagram if you choose the point $(3,2)$ and $(5,3)$ , you must first move three up or two right, determining your first point, and move 5 up or 3 right, determining your final point. Knowing this is helpful because if we first move anywhere horizontally, we have $7$ points on each column to choose from and starting from left to right, we have $6,5,4,3,2,1$ points on that row to choose from. This gives us $7(6)+7(5)+7(4)+7(3)+7(2)+7(1)$ which simplifies to $7\\cdot21$ . The vertical case is symmetrical so we have $7\\cdot21\\cdot2 = \\boxed{294}$", "As in Solution 1, there are two cases: $RURUR$ or $URURU$ . We will work with the first case and multiply by $2$ at the end. We use stars and bars; we can treat the $R$ s as the stars and the $U$ s as the bars. However, we must also use stars and bars on the $U$ s to see how many different patterns of bars we can create for the reds. We must have $1$ bar in $8$ blacks, so we use stars and bars on the equation \\[x + y = 8\\] . However, each divider must have at least one black in it, so we do the change of variable $x' = x-1$ and $y' = x-1$ . Our equation becomes \\[x' + y' = 6\\] . By stars and bars, this equation has $\\binom{6 + 2 - 1}{1} = 7$ valid solutions. Now, we use stars and bars on the reds. We must distribute two bars amongst the reds, so we apply stars and bars to \\[x + y + z = 8\\] . Since each group must have one red, we again do a change of variables with $x' = x-1$ $y' = y-1$ , and $z' = z-1$ . We are now working on the equation \\[x' + y' + z' = 5\\] . By stars and bars, this has $\\binom{5 + 3 - 1}{2} = 21$ solutions. The number of valid paths in this case is the number of ways to create the bars times the number of valid arrangements of the stars given fixed bars, which equals $21 \\cdot 7 = 147$ . We must multiply by two to account for both cases, so our final answer is $147 \\cdot 2 = \\boxed{294}$" ]

[ "Let $(a,b)$ denote a normal vector of the side containing $A$ . Note that $\\overline{AC}, \\overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$ $ax+by=8a$ $bx-ay=10b-9a$ , and $bx-ay=-4b-7a$ . The lines form a square, so the distance between $C$ and the line through $A$ equals the distance between $D$ and the line through $B$ , hence $8a+0b-12b=-4b-7a-10b+9a$ , or $-3a=b$ . We can take $a=-1$ and $b=3$ . So the side of the square is $\\frac{44}{\\sqrt{10}}$ , the area is $K=\\frac{1936}{10}$ , and the answer to the problem is $\\boxed{936}$" ]

[ "The degree of $P(x)$ is $1024+512+256+\\cdots+1=2047$ . We want to find the coefficient of $x^{2012}$ , so we need to omit the powers of $2$ that add up to $2047-2012=35$ . We find that $35=2^0+2^1+2^5$ . From here, we know that the answer is $2^0\\cdot2^1\\cdot2^5=2^6$ . Therefore, the answer is $\\boxed{6.}$" ]

[ "When we use long division to divide $P(x)$ by $Q(x)$ , the remainder is $x^2-x+1$\nSo, since $z_1$ is a root, $P(z_1)=(z_1)^2-z_1+1$\nNow this also follows for all roots of $Q(x)$ Now \\[P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1\\]\nNow by Vieta's we know that $-z_4-z_3-z_2-z_1=-1$ ,\nso by Newton's Sums we can find $z_1^2+z_2^2+z_3^2+z_4^2$\n$a_ns_2+a_{n-1}s_1+2a_{n-2}=0$\n$(1)(s_2)+(-1)(1)+2(-1)=0$\n$s_2-1-2=0$\n$s_2=3$\nSo finally $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\\boxed{006}.$", "Let $S_k=z_1^k+z_2^k+z_3^k+z_4^k$ then by Vieta's Formula we have \\[S_{-1}=\\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0\\] \\[S_0=4\\] \\[S_1=1\\] \\[S_2=3\\] By Newton's Sums we have \\[a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0\\]\nApplying the formula couples of times yields $P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\\boxed{6}$" ]

[ "Let $z=a+bi \\implies \\frac{z}{40}=\\frac{a}{40}+\\frac{b}{40}i$ . Since $0\\leq \\frac{a}{40},\\frac{b}{40}\\leq 1$ we have the inequality \\[0\\leq a,b \\leq 40\\] which is a square of side length $40$\nAlso, $\\frac{40}{\\overline{z}}=\\frac{40}{a-bi}=\\frac{40a}{a^2+b^2}+\\frac{40b}{a^2+b^2}i$ so we have $0\\leq a,b \\leq \\frac{a^2+b^2}{40}$ , which leads to: \\[(a-20)^2+b^2\\geq 20^2\\] \\[a^2+(b-20)^2\\geq 20^2\\]\nWe graph them:\nTo find the area outside the two circles but inside the square, we want to find the unique area of the two circles. We can do this by adding the area of the two circles and then subtracting out their overlap. There are two methods of finding the area of overlap:\n1. Consider that the area is just the quarter-circle with radius $20$ minus an isosceles right triangle with base length $20$ , and then doubled (to consider the entire overlapped area)\n2. Consider that the circles can be converted into polar coordinates, and their equations are $r = 40sin\\theta$ and $r = 40cos\\theta$ . Using calculus with the appropriate bounds, we can compute the overlapped area.\nUsing either method, we compute the overlapped area to be $200\\pi + 400$ , and so the area of the intersection of those three graphs is $40^2-(200\\pi + 400) \\Rightarrow 1200 - 200\\pi \\approx 571.68$\n$\\boxed{572}$" ]

[ "The average of a list is the sum of all numbers divided by the size of the list.\nThe sum of the list can be found by adding the numbers in pairs: $(1 + -2) + (3 + -4) + ... + (199 + -200)$\nThe sum of each pair is $-1$ , and there are $100$ pairs, so the total sum is $-100$\nThere are $200$ numbers on the list, so the average is $\\frac{-100}{200} = -0.5$ , and the answer is $\\boxed{0.5}$" ]

[ "We know that $a_{1}=\\tfrac{t}{t+1}$ when $t=2020$ so $1$ is a possible value of $j$ . Note also that $a_{2}=\\tfrac{2038}{2040}=\\tfrac{1019}{1020}=\\tfrac{t}{t+1}$ for $t=1019$ . Then $a_{2+q}=\\tfrac{1019+18q}{1020+19q}$ unless $1019+18q$ and $1020+19q$ are not relatively prime which happens when $q+1$ divides $18q+1019$ (by the Euclidean Algorithm), or $q+1$ divides $1001$ . Thus, the least value of $q$ is $6$ and $j=2+6=8$ . We know $a_{8}=\\tfrac{1019+108}{1020+114}=\\tfrac{1127}{1134}=\\tfrac{161}{162}$ . Now $a_{8+q}=\\tfrac{161+18q}{162+19q}$ unless $18q+161$ and $19q+162$ are not relatively prime which happens the first time $q+1$ divides $18q+161$ or $q+1$ divides $143$ or $q=10$ , and $j=8+10=18$ . We have $a_{18}=\\tfrac{161+180}{162+190}=\\tfrac{341}{352}=\\tfrac{31}{32}$ . Now $a_{18+q}=\\tfrac{31+18q}{32+19q}$ unless $18q+31$ and $19q+32$ are not relatively prime. This happens the first time $q+1$ divides $18q+31$ implying $q+1$ divides $13$ , which is prime so $q=12$ and $j=18+12=30$ . We have $a_{30}=\\tfrac{31+216}{32+228}=\\tfrac{247}{260}=\\tfrac{19}{20}$ . We have $a_{30+q}=\\tfrac{18q+19}{19q+20}$ , which is always reduced by EA, so the sum of all $j$ is $1+2+8+18+30=\\boxed{059}$", "Let $a_{j_1}, a_{j_2}, a_{j_3}, \\ldots, a_{j_u}$ be all terms in the form $\\frac{t}{t+1},$ where $j_1<j_2<j_3<\\cdots<j_u,$ and $t$ is some positive integer.\nWe wish to find $\\sum_{i=1}^{u}{j_i}.$ Suppose $a_{j_i}=\\frac{m}{m+1}$ for some positive integer $m.$\nTo find $\\boldsymbol{a_{j_{i+1}},}$ we look for the smallest positive integer $\\boldsymbol{k'}$ for which \\[\\boldsymbol{a_{j_{i+1}}=a_{j_i+k'}=\\frac{m+18k'}{m+1+19k'}}\\] is reducible:\nIf $\\frac{m+18k'}{m+1+19k'}$ is reducible, then there exists a common factor $d>1$ for $m+18k'$ and $m+1+19k'.$ By the Euclidean Algorithm , we have \\begin{align*} d\\mid m+18k' \\text{ and } d\\mid m+1+19k' &\\implies d\\mid m+18k' \\text{ and } d\\mid k'+1 \\\\ &\\implies d\\mid m-18 \\text{ and } d\\mid k'+1. \\end{align*} Since $m-18$ and $k'+1$ are not relatively prime, and $m$ is fixed, the smallest value of $k'$ such that $\\frac{m+18k'}{m+1+19k'}$ is reducible occurs when $k'+1$ is the smallest prime factor of $m-18.$\nWe will prove that for such value of $\\boldsymbol{k',}$ the number $\\boldsymbol{a_{j_{i+1}}}$ can be written in the form $\\boldsymbol{\\frac{t}{t+1}:}$ \\[a_{j_{i+1}}=a_{j_i+k'}=\\frac{m+18k'}{m+1+19k'}=\\frac{(m-18)+18(k'+1)}{(m-18)+19(k'+1)}=\\frac{\\frac{m-18}{k'+1}+18}{\\frac{m-18}{k'+1}+19}, \\hspace{10mm} (*)\\] where $t=\\frac{m-18}{k'+1}+18$ must be a positive integer.\nWe start with $m=2020$ and $a_{j_1}=a_1=\\frac{2020}{2021},$ then find $a_{j_2}, a_{j_3}, \\ldots, a_{j_u}$ by filling out the table below recursively: \\[\\begin{array}{c|c|c|c|c|c} & & & & & \\\\ [-2ex] \\boldsymbol{i} & \\boldsymbol{m} & \\boldsymbol{m-18} & \\boldsymbol{k'+1} & \\boldsymbol{k'} & \\boldsymbol{a_{j_{i+1}} \\left(\\textbf{by } (*)\\right)} \\\\ [0.5ex] \\hline & & & & & \\\\ [-1.5ex] 1 & 2020 & 2002 & 2 & 1 & \\hspace{4.25mm} a_2 = \\frac{1019}{1020} \\\\ [1ex] 2 & 1019 & 1001 & 7 & 6 & \\hspace{2.75mm} a_8 = \\frac{161}{162} \\\\ [1ex] 3 & 161 & 143 & 11 & 10 & a_{18} = \\frac{31}{32} \\\\ [1ex] 4 & 31 & 13 & 13 & 12 & a_{30} = \\frac{19}{20} \\\\ [1ex] 5 & 19 & 1 & \\text{N/A} & \\text{N/A} & \\text{N/A} \\\\ [1ex] \\end{array}\\] As $\\left(j_1,j_2,j_3,j_4,j_5\\right)=(1,2,8,18,30),$ the answer is $\\sum_{i=1}^{5}{j_i}=\\boxed{059}.$" ]

[ "Repeatedly applying the function, and simplifying, we get \\[a,\\quad-\\frac1{a+1},\\quad-\\frac{a+1}a,\\] and then $a$ again. So $a$ must appear at every third term after $u_1$ . The only option given of the form $1+3k$ is $\\boxed{16}$" ]

[ "If $n$ is even, then $t_{(n/2)}$ would be negative, which is not possible. Therefore, $n$ is odd. With this function, backwards thinking is the key. If $t_x < 1$ , then $x$ is odd, and $t_{(x-1)} = \\frac{1}{t_{x}}$ . Otherwise, you keep on subtracting 1 and halving x until $t_\\frac{x}{2^{n}} < 1$ .\nWe can use this logic to go backwards until we reach $t_1 = 1$ , like so:\n$t_n=\\frac{19}{87}\\\\\\\\t_{n-1} = \\frac{87}{19}\\\\\\\\t_{\\frac{n-1}{2}} = \\frac{68}{19}\\\\\\\\t_{\\frac{n-1}{4}} = \\frac{49}{19}\\\\\\\\t_{\\frac{n-1}{8}} = \\frac{30}{19}\\\\\\\\t_{\\frac{n-1}{16}} = \\frac{11}{19}\\\\\\\\t_{\\frac{n-1}{16} - 1} = \\frac{19}{11}\\\\\\\\t_{\\frac{\\frac{n-1}{16} - 1}{2}} = \\frac{8}{11}\\\\\\\\t_{\\frac{\\frac{n-1}{16} - 1}{2} - 1} = \\frac{11}{8}\\\\\\\\t_{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2}} = \\frac{3}{8}\\\\\\\\t_{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1} = \\frac{8}{3}\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{2}} = \\frac{5}{3}\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4}} = \\frac{2}{3}\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1} = \\frac{3}{2}\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2}} = \\frac{1}{2}\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1} = 2\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2}} = t_1 = 1 \\Rightarrow \\frac{\\frac{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2} = 1 \\Rightarrow n = 1905$ , so the answer is $\\boxed{15}$" ]

[ "The sequence is infinite. As there are only $100$ pairs of digits, sooner or later a pair of consecutive digits will occur for the second time. As each next digit only depends on the previous two, from this point on the sequence will be periodic.\n(Additionally, as every two consecutive digits uniquely determine the previous one as well, the first pair of digits that will occur twice must be the first pair $4,7$ .)\nHence it is a good idea to find the period. Writing down more terms of the sequence, we get:\n\\[4,7,1,8,9,7,6,3,9,2,1,3,4,7,\\dots\\]\nand we found the period. The length of the period is $12$ , and its sum is $4+7+\\cdots+1+3 = 60$ . Hence for each $k$ we have $S_{12k} = 60k$\nWe have $\\lfloor 10000/60 \\rfloor = 166$ and $166\\cdot 12 = 1992$ , therefore $S_{1992} = 60\\cdot 166 = 9960$ .\nThe rest can now be computed by hand, we get $S_{1998} = 9960+4+7+1+8+9+7= 9996$ , and $S_{1999}=9996 + 6 = 10002$ , thus the answer is $\\boxed{1999}$" ]

[ "Being on two parabolas means having the same distance from the common focus and both directrices. In particular, you have to be on an angle bisector of the directrices, and clearly on the same \"side\" of the directrices as the focus. So it's easy to see there are at most two solutions per pair of parabolae. Convexity and continuity imply exactly two solutions unless the directrices are parallel and on the same side of the focus.\nSo out of $2\\dbinom{30}{2}$ possible intersection points, only $2*5*2*\\dbinom{3}{2}$ fail to exist. This leaves $870-60=810=\\boxed{810}$ solutions." ]

[ "Through similar reasoning as above in Solution I, determine that two parabolas that have a common focus intersect zero times if there directrixes are parallel and the focus lies on the same side of both directrixes, and intersect twice otherwise. Thereby, as each parabola will intersect $30-3 = 27$ other parabolas twice, we see that the answer is \\[2 \\times \\frac{30 \\times 27}{2} = \\boxed{810}.\\]" ]

[ "You can create the equation $\\frac{x+1}{y+1}=\\frac{11x}{10y}$\nCross multiplying and combining like terms gives $xy + 11x - 10y = 0$\nThis can be factored into $(x - 10)(y + 11) = -110$\n$x$ and $y$ must be positive, so $x > 0$ and $y > 0$ , so $x - 10> -10$ and $y + 11 > 11$\nUsing the factors of 110, we can get the factor pairs: $(-1, 110),$ $(-2, 55),$ and $(-5, 22).$\nBut we can't stop here because $x$ and $y$ must be relatively prime.\n$(-1, 110)$ gives $x = 9$ and $y = 99$ $9$ and $99$ are not relatively prime, so this doesn't work.\n$(-2, 55)$ gives $x = 8$ and $y = 44$ . This doesn't work.\n$(-5, 22)$ gives $x = 5$ and $y = 11$ . This does work.\nWe found one valid solution so the answer is $\\boxed{1}$", "The condition required is $\\frac{x+1}{y+1}=\\frac{11}{10}\\cdot\\frac{x}{y}$\nObserve that $x+1 > \\frac{11}{10}\\cdot x$ so $x$ is at most $9.$\nBy multiplying by $\\frac{y+1}{x}$ and simplifying we can rewrite the condition as $y=\\frac{11x}{10-x}$ . Since $x$ and $y$ are integer, this only has solutions for $x\\in\\{5,8,9\\}$ . However, only the first yields a $y$ that is relative prime to $x$\nThere is only one valid solution so the answer is $\\boxed{1}$" ]

[ "Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$\nWe can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).\n$\\det \\left(\\begin{array}{c} P \\\\ Q\\end{array}\\right)=\\det \\left(\\begin{array}{cc}x_1 &y_1\\\\x_2&y_2\\end{array}\\right).$\nSince the triangle has half the area of the parallelogram, we just need the determinant to be even.\nThe determinant is\n\\[(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))\\]\nSince $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$ 's must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$ . There are $25$ even and $25$ odd numbers available for use as coordinates and thus there are $(_{25}C_2)+(_{25}C_2)=\\boxed{600}$ such triangles.", "As in the solution above, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$\nIf the coordinates of $P$ and $Q$ have nonnegative integer coordinates, $P$ and $Q$ must be lattice points either\nWe can calculate the y-intercept of the line $41x+y=2009$ to be $(0,2009)$ and the x-intercept to be $(49,0)$\nUsing the point-to-line distance formula, we can calculate the height of $\\triangle OPQ$ from vertex $O$ (the origin) to be:\n$\\dfrac{|41(0) + 1(0) - 2009|}{\\sqrt{41^2 + 1^2}} = \\dfrac{2009}{\\sqrt{1682}} = \\dfrac{2009}{29\\sqrt2}$\nLet $b$ be the base of the triangle that is part of the line $41x+y=2009$\nThe area is calculated as: $\\dfrac{1}{2}\\times b \\times \\dfrac{2009}{29\\sqrt2} = \\dfrac{2009}{58\\sqrt2}\\times b$\nLet the numerical area of the triangle be $k$\nSo, $k = \\dfrac{2009}{58\\sqrt2}\\times b$\nWe know that $k$ is an integer. So, $b = 58\\sqrt2 \\times z$ , where $z$ is also an integer.\nWe defined the points $P$ and $Q$ as $P=(x_1,y_1)$ and $Q=(x_2,y_2)$\nChanging the y-coordinates to be in terms of x, we get:\n$P=(x_1,2009-41x_1)$ and $Q=(x_2,2009-41x_2)$\nThe distance between them equals $b$\nUsing the distance formula, we get\n$PQ = b = \\sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\\sqrt2 \\times |x_2 - x_1| = 58\\sqrt2\\times z$ $(*)$\nWLOG, we can assume that $x_2 > x_1$\nTaking the last two equalities from the $(*)$ string of equalities and putting in our assumption that $x_2>x_1$ , we get\n$29\\sqrt2\\times (x_2-x_1) = 58\\sqrt2\\times z$\nDividing both sides by $29\\sqrt2$ , we get\n$x_2-x_1 = 2z$\nAs we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well.\n...\nSumming them up, we get that there are $2+4+\\dots + 48 = \\boxed{600}$ triangles." ]

[ "First, substitute in $z=a+bi$\n\\[|1+(a+bi)+(a+bi)^2|=4\\] \\[|(1+a+a^2-b^2)+ (b+2ab)i|=4\\] \\[(1+a+a^2-b^2)^2+ (b+2ab)^2=16\\] \\[(1+a+a^2-b^2)^2+ b^2(1+4a+4a^2)=16\\]\nLet $p=b^2$ and $q=1+a+a^2$\n\\[(q-p)^2+ p(4q-3)=16\\] \\[p^2-2pq+q^2 + 4pq -3p=16\\]\nWe are trying to maximize $b=\\sqrt p$ , so we'll turn the equation into a quadratic to solve for $p$ in terms of $q$\n\\[p^2+(2q-3)p+(q^2-16)=0\\] \\[p=\\frac{(-2q+3)\\pm \\sqrt{-12q+73}}{2}\\]\nWe want to maximize $p$ . Since $q$ is always negatively contributing to $p$ 's value, we want to minimize $q$\nDue to the trivial inequality: $q=1+a+a^2=(a+\\frac 12)^2+\\frac{3}4 \\geq \\frac{3}4$\nIf we plug $q$ 's minimum value in, we get that $p$ 's maximum value is \\[p=\\frac{(-2(\\frac 34)+3)+ \\sqrt{-12(\\frac 34)+73}}{2}=\\frac{\\frac 32+ 8}{2}=\\frac{19}{4}\\]\nThen \\[b=\\frac{\\sqrt{19}}{2}\\] and \\[m+n=\\boxed{21}\\]", "We are given that $1+z+z^2=c$ where $c$ is some complex number with magnitude $4$ . Rearranging the quadratic to standard form and applying the quadratic formula, we have \\[z=\\frac{-1\\pm \\sqrt{1^2-4(1)(1-c)}}{2}=\\frac{-1\\pm\\sqrt{4c-3}}{2}.\\]\nThe imaginary part of $z$ is maximized when $c=-4$ . (Why? See note below.)\nThus $z=i\\sqrt{19}/2$ , and so the answer is $\\boxed{21}$", " We can write the given condition as \\[\\left|\\left(z+\\frac{1}{2}\\right)^2 + \\frac{3}{4}\\right| = 4.\\] Letting $u = \\left(z+\\frac{1}{2}\\right)^2$ , the equation $\\left|u + \\frac{3}{4}\\right| = 4$ equates to the circle centered at $-\\frac{3}{4}$ with radius $4$ in the complex plane, call it $\\omega$ . Thus the locus of $\\left(z+\\frac{1}{2}\\right)^2$ is $\\omega$ . Let $v = z+\\frac{1}{2}$ , and since the $+\\frac{1}{2}$ does not change $z$ 's imaginary part, we now need to find $v$ with the largest imaginary part such that $v^2$ lies on $\\omega$\nNote that the point on $\\omega$ with largest magnitude is $19/4$ and has argument $\\pi$ , call it $\\phi$ (The leftmost point on $\\omega$ ). The value $v'$ with positive imaginary part such that $(v')^2 = \\phi$ has an argument of $\\frac{\\pi}{2}$ and a magnitude of $\\frac{\\sqrt{19}}{2}$\nSince across all values of $v$ the imaginary part is given by $r\\sin{\\theta}$ and $v'$ has the largest possible $r$ and the largest possible value of $\\sin{\\theta},$ it must have the largest imaginary part.\nThis can non-rigorously be seen by sketching the oval which is the locus of $v$\nThis gives $19 + 2 \\implies \\boxed{21}$", "To start, we factor $1+z+z^2$ to get:\n\\[|(z-\\frac{-1+\\sqrt{3}i}{2})(z-\\frac{-1-\\sqrt{3}i}{2})|=4\\]\nNote that since the magnitude of a product of complex numbers is equal to the product of the magnitudes:\n\\[|(z+\\frac{1-\\sqrt{3}i}{2})||(z+\\frac{1+\\sqrt{3}i}{2})|=4\\]\nNow, we substitute $z=a+bi$ (Note that we are trying to maximize b):\n\\[|a+\\frac{1}{2}+(b+\\frac{\\sqrt{3}}{2})i||a+\\frac{1}{2}+(b-\\frac{\\sqrt{3}}{2})i|=4\\]\nSince we are trying to maximize $b$ , we want the real parts of the components to be as small as possible, which we can do by setting $a=-\\frac{1}{2}$ . This leaves us with:\n\\[|(b+\\frac{\\sqrt{3}}{2})i||(b-\\frac{\\sqrt{3}}{2})i|=4\\] \\[(b+\\frac{\\sqrt{3}}{2})(b-\\frac{\\sqrt{3}}{2})=4\\] \\[b^2-\\frac{3}{4}=4\\] \\[b^2=\\frac{19}{4}\\] \\[b=\\frac{\\sqrt{19}}{2}\\]\nThis gives us $19 + 2 \\implies \\boxed{21}$" ]

[ "The requested ratio is \\[\\dfrac{10^{10}}{10^9 + 10^8 + \\ldots + 10 + 1}.\\] Using the formula for a geometric series, we have \\[10^9 + 10^8 + \\ldots + 10 + 1 = \\dfrac{10^{10} - 1}{10 - 1} = \\dfrac{10^{10} - 1}{9},\\] which is very close to $\\dfrac{10^{10}}{9},$ so the ratio is very close to $\\boxed{9}.$", "The problem asks for the value of \\[\\dfrac{10^{10}}{10^9 + 10^8 + \\ldots + 10 + 1}.\\] Written in base 10, we can find the value of $10^9 + 10^8 + \\ldots + 10 + 1$ to be $1111111111.$ Long division gives us the answer to be $\\boxed{9}.$", "Let $f(n)=\\dfrac{10^n}{1+10+10^2+10^3+\\cdots+10^{n-1}}$ . We are approximating $f(10)$ . Trying several small values of $n$ gives answers very close to $9$ , so our answer is $\\boxed{9}\\approx9.09.$ ~Technodoggo" ]

[ "The set can be broken into several parts: the big $3\\times 4 \\times 5$ parallelepiped, $6$ external parallelepipeds that each share a face with the large parallelepiped and have a height of $1$ , the $1/8$ spheres (one centered at each vertex of the large parallelepiped), and the $1/4$ cylinders connecting each adjacent pair of spheres.\nThe combined volume of these parts is $60+94+\\frac{4}{3}\\pi+12\\pi = \\frac{462+40\\pi}{3}$ . Thus, the answer is $m+n+p = 462+40+3 = \\boxed{505}$" ]

[ "Since a counterexample must be a value of $n$ which is not prime, $n$ must be composite, so we eliminate $\\text{A}$ and $\\text{C}$ . Now we subtract $2$ from the remaining answer choices, and we see that the only time $n-2$ is not prime is when $n = \\boxed{27}$" ]

[ "Since a counterexample must be a value of $n$ which is not prime, $n$ must be composite, so we eliminate $\\text{A}$ and $\\text{C}$ . Now we subtract $2$ from the remaining answer choices, and we see that the only time $n-2$ is not prime is when $n = \\boxed{27}$" ]

[ "Statement $1$ states that $p$ is true and $q$ is false. Therefore, $p \\rightarrow q$ is false, because a premise being true and a conclusion being false is, itself, false. This means that $(p \\rightarrow q) \\rightarrow X$ , where $X$ is any logical statement (or series of logical statements) must be true - if your premise is false, then the implication is automatically true. So statement $1$ implies the truth of the given statement.\nStatement $3$ similarly has $p$ as true and $q$ is false, so it also implies the truth of the given statement.\nStatement $2$ states that $p$ and $q$ are both false. This in turn means that $p \\rightarrow q$ is true. Since $r$ is also true from statement $2$ , this means that $(p \\rightarrow q) \\rightarrow r$ is true, since $T \\rightarrow T$ is $T$ . Thus statement $2$ implies the truth of the given statement.\nStatement $4$ states that $p$ is false and $q$ is true. In this case, $p \\rightarrow q$ is true - your conclusion can be true even if your premise is false. And, since $r$ is also true from statement $4$ , this means $(p \\rightarrow q) \\rightarrow r$ is true. Thus, statement $4$ implies the truth of the given statement.\nAll four statements imply the truth of the given statement, so the answer is $\\boxed{4}$" ]

[ "We can substitute $5$ $3$ , and $6$ into the functions' definitions: \\begin{align*} (5 \\, \\blacklozenge \\, 3) \\, \\bigstar \\, 6 &= \\left(5^2-3^2\\right) \\, \\bigstar \\, 6 \\\\ (5 \\, \\blacklozenge \\, 3) \\, \\bigstar \\, 6 &= \\left(25-9\\right) \\, \\bigstar \\, 6 \\\\ &= 16 \\, \\bigstar \\, 6 \\\\ &= (16-6)^2 \\\\ &= \\boxed{100} ~pog ~MathFun1000 (Minor Edits)" ]

[ "Since $81$ students took the test, the median is the score of the $41^{st}$ student. The five rightmost intervals include $2+3+6+12+16=39$ students, so the $41^{st}$ one must lie in the next interval, which is $\\boxed{70}$" ]

[ "Recall that $p$ is the fundamental period of function $f$ iff $p$ is the smallest positive $p$ such that $f(x) = f(x + p)$ for all $x$\nIn this case, we know that $f(x+ 4) + f(x - 4) = f(x)$ . Plugging in $x+4$ in for $x$ to get the next equation in the recursion, we also get $f(x + 8) + f(x) = f(x + 4)$ . Adding those two equations gives $f(x + 8) + f(x - 4) = 0$ after cancelling out common terms.\nAgain plugging in $x + 4$ in for $x$ in that last equation (in order to get $f(x)$ ), we find that $f(x) = -f(x + 12)$ . Now, plugging in $x+12$ for $x$ , we get $f(x + 12) = -f(x + 24)$ . This proves that $f(x) = f(x + 24)$ , so there is a period of $24$ , which gives answer $\\boxed{24}$ . We now eliminate answers $A$ through $C$" ]

[ "The two equations of the balls are\n\\[x^2 + y^2 + \\left(z - \\frac{21}{2}\\right)^2 \\le 36\\]\n\\[x^2 + y^2 + (z - 1)^2 \\le \\frac{81}{4}\\]\nNote that along the $z$ axis, the first ball goes from $10.5 \\pm 6$ , and the second ball goes from $1 \\pm 4.5$ . The only integer value that $z$ can be is $z=5$\nPlugging that in to both equations, we get:\n\\[x^2 + y^2 \\le \\frac{23}{4}\\]\n\\[x^2 + y^2 \\le \\frac{17}{4}\\]\nThe second inequality implies the first inequality, so the only condition that matters is the second inequality.\nFrom here, we do casework, noting that $|x|, |y| \\le 3$\nFor $x=\\pm 2$ , we must have $y=0$ . This gives $2$ points.\nFor $x = \\pm 1$ , we can have $y\\in \\{-1, 0, 1\\}$ . This gives $2\\cdot 3 = 6$ points.\nFor $x = 0$ , we can have $y \\in \\{-2, -1, 0, 1, 2\\}$ . This gives $5$ points.\nThus, there are $\\boxed{13}$", "Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a 2-dimensional plane (the previous z-axis is the new x-axis while the y-axis remains the same).\nThe spheres now become circles with centers at $(1,0)$ and $(\\frac{21}{2},0)$ . They have radii $\\frac{9}{2}$ and $6$ , respectively.\nLet circle $A$ be the circle centered on $(1,0)$ and circle $B$ be the one centered on $(\\frac{21}{2},0)$\nThe point on circle $A$ closest to the center of circle $B$ is $(\\frac{11}{2},0)$ . The point on circle B closest to the center of circle $A$ is $(\\frac{11}{2},0)$\nTaking a look back at the 3-dimensional coordinate grid with the spheres, we can see that their intersection appears to be a circle with congruent \"dome\" shapes on either end. Because the tops of the \"domes\" are at $(0,0,\\frac{9}{2})$ and $(0,0,\\frac{11}{2})$ , respectively, the lattice points inside the area of intersection must have z-value $5$ (because $5$ is the only integer between $\\frac{9}{2}$ and $\\frac{11}{2}$ ). Thus, the lattice points in the area of intersection must all be on the 2-dimensional circle. The radius of the circle will be the distance from the z-axis.\nNow, looking at the 2-dimensional coordinate plane, we see that the radius of the circle (now the distance from the x-axis, because there is no more z-axis) is the altitude of a triangle with two points on centers of circles $A$ and $B$ and third point at the first quadrant intersection of the circles. Let's call that altitude $h$\nWe know all three side lengths of this triangle: $\\frac{9}{2}$ (the radius of circle $A$ ), $6$ (the radius of circle $B$ ), and $\\frac{19}{2}$ (the distance between the centers of circles $A$ and $B$ ). We can now find the area of the triangle using Heron's formula:\n\\[s=\\frac{\\frac{9}{2}+6+\\frac{19}{2}}{2}=20\\]\n\\[A=\\sqrt{(20)(20-\\frac{9}{2})(20-6)(20-\\frac{19}{2})}=\\sqrt{110}\\]\nUsing the area of the triangle, we can find that altitude $h$ from the x-axis:\n\\[\\sqrt{110}=\\frac{h*\\frac{19}{2}}{2}\\]\n\\[h=\\frac{4*\\sqrt{110}}{19}\\]\nRemember, the altitude $h$ is also the radius of the circle containing all the solutions to the problem.\nGoing back to the 3-dimensional grid and looking at the circle, we can again make the figure 2-dimensional.\nThe radius $h$ of the circle in a 2-dimensional plane is $\\frac{4\\sqrt{110}}{19}$ , a little greater than $2$ . We know that $h>2$ because $4\\sqrt{110}>4*10$ , and $\\frac{40}{19}>2$\nFinally, looking at a circle with radius slightly larger than $2$ , we see that there are $\\boxed{13}$", "Note that the spheres are on the same $x$ and $y$ axis. Therefore, we can draw the spheres so that only the $x$ and $z$ axis are featured.\n$A = (1, 0, 5)$ $\\sqrt{1^2+4^2} = \\sqrt{17}<\\frac92$ $A$ is inside the smaller sphere. $\\sqrt{1^2+(\\frac{11}{2})^2} = \\frac{\\sqrt{126}}{2}<6$ $A$ is inside the larger sphere. Point $A$ is inside the intersection.\n$B = (2, 0, 5)$ $\\sqrt{2^2+4^2} = \\sqrt{20}<\\frac92$ $B$ is inside the smaller sphere. $\\sqrt{2^2+(\\frac{11}{2})^2} = \\frac{\\sqrt{137}}{2}<6$ $B$ is inside the larger sphere. Point $B$ is inside the intersection.\n$C = (3, 0, 5)$ $\\sqrt{3^2+4^2} = \\sqrt{25}>\\frac92$ $C$ is outside the smaller sphere. $\\sqrt{3^2+(\\frac{11}{2})^2} = \\frac{\\sqrt{157}}{2}>6$ $C$ is outside the larger sphere. Point $C$ is outside the intersection.\nThe intersection of $2$ spheres is a circle, by drawing the circle flat we can see that there are $4$ more points within the intersection.\nWe can see that points with the same $x$ -coordinates with $A$ are still inside the circle. $B$ is very close to the circle, the points with the same $x$ -coordinates with $B$ are not necessarily inside the circle.\nFor example, $D = (2,1,5)$ $\\sqrt{2^2+1^2+4^2}=\\sqrt{21}>\\frac92$ $D$ is outside the smaller sphere. $\\sqrt{2^2 + 1^2 + (\\frac{11}{2})^2} = \\frac{\\sqrt{141}}{2}<6$ $D$ is inside the larger sphere. Point $D$ is outside the intersection.\nTherefore, the answer is $9+4 = \\boxed{13}$" ]

[ "Of the six shapes used to create the polygon, the triangle and octagon are adjacent to the others on one side, and the others are adjacent on two sides. In the triangle and octagon $3+8-2(1)=9$ sides are on the outside of the final polygon. In the other shapes $4+5+6+7-4(2) = 14$ sides are on the outside. The resulting polygon has $9+14 = \\boxed{23}$ sides.", "We can quickly see a pattern if we draw out the other shapes. Every shape will have two of its sides taken out except the triangle and octagon. We can then make the expression $2+2+3+4+5+7$ which is $\\boxed{23}$" ]

[ "Assume the incircle touches $AB$ $BC$ $CD$ $DE$ $EA$ at $P,Q,R,S,T$ respectively. Then let $PB=x=BQ=RD=SD$ $ET=y=ES=CR=CQ$ $AP=AT=z$ . So we have $x+y=6$ $x+z=5$ and $y+z$ =7, solve it we have $x=2$ $z=3$ $y=4$ . Let the center of the incircle be $I$ , by SAS we can proof triangle $BIQ$ is congruent to triangle $DIS$ , and triangle $CIR$ is congruent to triangle $SIE$ . Then we have $\\angle AED=\\angle BCD$ $\\angle ABC=\\angle CDE$ . Extend $CD$ , cross ray $AB$ at $M$ , ray $AE$ at $N$ , then by AAS we have triangle $END$ is congruent to triangle $BMC$ . Thus $\\angle M=\\angle N$ . Let $EN=MC=a$ , then $BM=DN=a+2$ . So by law of cosine in triangle $END$ and triangle $ANM$ we can obtain \\[\\frac{2a+8}{2(a+7)}=\\cos N=\\frac{a^2+(a+2)^2-36}{2a(a+2)}\\] , solved it gives us $a=8$ , which yield triangle $ANM$ to be a triangle with side length 15, 15, 24, draw a height from $A$ to $NM$ divides it into two triangles with side lengths 9, 12, 15, so the area of triangle $ANM$ is 108. Triangle $END$ is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is $108-48=\\boxed{60}$", "Suppose that the circle intersects $\\overline{AB}$ $\\overline{BC}$ $\\overline{CD}$ $\\overline{DE}$ , and $\\overline{EA}$ at $P$ $Q$ $R$ $S$ , and $T$ respectively. Then $AT = AP = a$ $BP = BQ = b$ $CQ = CR = c$ $DR = DS = d$ , and $ES = ET = e$ . So $a + b = 5$ $b + c = 6$ $c + d = 6$ $d + e = 6$ , and $e + a = 7$ . Then $2a + 2b + 2c + 2d + 2e = 30$ , so $a + b + c + d + e= 15$ . Then we can solve for each individually. $a = 3$ $b = 2$ $c = 4$ $d = 2$ , and $e = 4$ . To find the radius, we notice that $4 \\arctan(\\frac{2}{r}) + 4 \\arctan(\\frac{4}{r}) + 2 \\arctan (\\frac{3}{r}) = 360 ^ \\circ$ , or $2 \\arctan(\\frac{2}{r}) + 2 \\arctan(\\frac{4}{r}) + \\arctan (\\frac{3}{r}) = 180 ^ \\circ$ . Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, $(r + 2i)^2 \\cdot (r + 4i)^2 \\cdot (r + 3i)$ is real. Expanding, we get:\n\\[(r^2 + 4ir - 4)(r^2 + 8ir -16)(r + 3i)\\]\n\\[(r^4 + 12ir^3 - 52r^2 - 96ir + 64)(r + 3i)\\]\nOn the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0.\n\\[15ir^4 - 252ir^2 + 192i = 0\\]\n\\[5r^4 - 84r^2 + 64 = 0\\]\n\\[(5r^2 - 4)(r^2 - 16) = 0\\]\n$r$ must equal 4, as r cannot be negative or be approximately equal to 1. \nThus, the area of $ABCDE$ is $4 \\cdot (a + b + c + d + e) = 4 \\cdot 15 = \\boxed{60}$", "This pentagon is very close to a regular pentagon with side lengths $6$ . The area of a regular pentagon with side lengths $s$ is $\\frac{5s^2}{4\\sqrt{5-2\\sqrt{5}}}$ $5-2\\sqrt{5}$ is slightly greater than $\\frac{1}{2}$ given that $2\\sqrt{5}$ is slightly less than $\\frac{9}{2}$ $4\\sqrt{5-2\\sqrt{5}}$ is then slightly greater than $2\\sqrt{2}$ . We will approximate that to be $2.9$ . The area is now roughly $\\frac{180}{2.9}$ , but because the actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the $2.9$ into $3$ thus turning the area into $\\frac{180}{3}$ which is $60$ and since $60$ is a multiple of the semiperimeter $15$ , we can safely say that the answer is most likely $\\boxed{60}$", "Let $\\omega$ be the inscribed circle, $I$ be its center, and $r$ be its radius. The area of $ABCDE$ is equal to its semiperimeter, $15,$ times $r$ , so the problem is reduced to finding $r$ . Let $a$ be the length of the tangent segment from $A$ to $\\omega$ , and analogously define $b$ $c$ $d$ , and $e$ . Then $a+b=5$ $b+c= c+d=d+e=6$ , and $e+a=7$ , with a total of $a+b+c+d+e=15$ . Hence $a=3$ $b=d=2$ , and $c=e=4$ . It follows that $\\angle B= \\angle D$ and $\\angle C= \\angle E$ . Let $Q$ be the point where $\\omega$ is tangent to $\\overline{CD}$ . Then $\\angle IAE = \\angle IAB =\\frac{1}{2}\\angle A$ . Now we claim that points $A, I, Q$ are collinear, which can be proved if $\\angle{AIQ}=\\angle{QIA}=180^{\\circ}$ . The sum of the internal angles in polygons $ABCQI$ and $AIQDE$ are equal, so $\\angle IAE + \\angle AIQ + \\angle IQD + \\angle D + \\angle E = \\angle IAB + \\angle B + \\angle C + \\angle CQI + \\angle QIA$ , which implies that $\\angle AIQ$ must be $180^\\circ$ . Therefore points $A$ $I$ , and $Q$ are collinear. Because $\\overline{AQ} \\perp \\overline{CD}$ , it follows that \\[AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.\\] Another expression for $AC^2-AD^2$ can be found as follows. Note that $\\tan \\left(\\frac{\\angle B}{2}\\right) = \\frac{r}{2}$ and $\\tan \\left(\\frac{\\angle E}{2}\\right) = \\frac{r}{4}$ , so \\[\\cos (\\angle B) =\\frac{1-\\tan^2 \\left(\\frac{\\angle B}{2}\\right)}{1+\\tan^2 \\left(\\frac{\\angle B}{2}\\right)} = \\frac{4-r^2}{4+r^2}\\] and \\[\\cos (\\angle E) = \\frac{1-\\tan^2 \\left(\\frac{\\angle E}{2}\\right)}{1+\\tan^2 \\left(\\frac{\\angle E}{2}\\right)}= \\frac{16-r^2}{16+r^2}.\\] Applying the Law of Cosines to $\\triangle ABC$ and $\\triangle AED$ gives \\[AC^2=AB^2+BC^2-2\\cdot AB\\cdot BC\\cdot \\cos (\\angle B) = 5^2+6^2-2 \\cdot 5 \\cdot 6 \\cdot \\frac{4-r^2}{4+r^2}\\] and \\[AD^2=AE^2+DE^2-2 \\cdot AE \\cdot DE \\cdot \\cos(\\angle E) = 7^2+6^2-2 \\cdot 7 \\cdot 6 \\cdot \\frac{16-r^2}{16+r^2}.\\] Hence\n\\[12=AC^2- AD^2= 5^2-2\\cdot 5 \\cdot 6\\cdot \\frac{4-r^2}{4+r^2} -7^2+2\\cdot 7 \\cdot 6 \\cdot \\frac{16-r^2}{16+r^2},\\] yielding \\[2\\cdot 7 \\cdot 6 \\cdot \\frac{16-r^2}{16+r^2}- 2\\cdot 5 \\cdot 6\\cdot \\frac{4-r^2}{4+r^2}= 36;\\] equivalently \\[7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).\\] Substituting $x=r^2$ gives the quadratic equation $5x^2-84x+64=0$ , with solutions $\\frac{42 - 38}{5}=\\frac45$ , and $\\frac{42 + 38}{5}= 16$ . The solution $r^2=\\frac45$ corresponds to a five-pointed star, which is not convex. Indeed, if $r<3$ , then $\\tan \\left(\\frac{\\angle A}{2}\\right)$ $\\tan \\left(\\frac{\\angle C}{2}\\right)$ , and $\\tan \\left(\\frac{\\angle E}{2}\\right)$ are less than $1,$ implying that $\\angle A$ $\\angle C$ , and $\\angle E$ are acute, which cannot happen in a convex pentagon. Thus $r^2=16$ and $r=4$ . The requested area is $15\\cdot4 = \\boxed{60}$" ]

[ "Let $E$ be a point on $\\overline{AB}$ such that $BCDE$ is a parallelogram. Suppose that $BC=ED=b, CD=BE=c,$ and $DA=d,$ so $AE=18-c,$ as shown below. We apply the Law of Cosines to $\\triangle ADE:$ \\begin{align*} AD^2 + AE^2 - 2\\cdot AD\\cdot AE\\cdot\\cos 60^\\circ &= DE^2 \\\\ d^2 + (18-c)^2 - d(18-c) &= b^2 \\\\ (18-c)^2 - d(18-c) &= b^2 - d^2 \\\\ (18-c)(18-c-d) &= (b+d)(b-d). \\hspace{15mm}(\\bigstar) \\end{align*} Let $k$ be the common difference of the arithmetic progression of the side-lengths. It follows that $b,c,$ and $d$ are $18-k, 18-2k,$ and $18-3k,$ in some order. It is clear that $0\\leq k<6.$\nIf $k=0,$ then $ABCD$ is a rhombus with side-length $18,$ which is valid.\nIf $k\\neq0,$ then we have six cases:\nTogether, the sum of all possible values of $a$ is $18+(13+3+8)+(14+12+16)=\\boxed{84}.$", "Let $b, c$ , and $d$ denote the sides $BC, CD$ , and $AD$ respectively. Since $AB\\parallel CD$ , we get \\[\\tfrac{\\sqrt 3}{2}\\ d = b\\sin\\theta \\quad \\textrm{and}\\quad \\tfrac 12 d + c + b\\cos\\theta = 18.\\] Using $b^2\\sin^2\\theta + b^2\\cos^2\\theta = b^2$ , we eliminate $\\theta$ from above to get $(36-2c-d)^2+3d^2=4b^2$ , which rearranges to $(36-2c-d)^2-d^2=4(b^2-d^2)$ , and, upon factoring, yields \\begin{align} (18-c)(18-c-d)=(b+d)(b-d). \\end{align} We divide into two cases, depending on whether $c$ is the smallest side.\nIf $c$ is not the smallest side then $18-c=\\pm (b-d)$ . If $c=18$ , we get a rhombus of side $18$ , so one possible value is $a=18$ . Otherwise, we can cancel the common factor from $(1)$ . After rearranging we get \\[18-c=-b \\quad \\textrm{or}\\quad 18-c=b+2d.\\] The first condition is false because $-b< 0 <18-c$ ; the second condition is false because $b+2d > |b-d| = 18-c$\nIf $c$ is the smallest side, then $18-c = \\pm 3(b-d)$ . Assuming $c<18$ we can cancel common factors in $(1)$ to get \\[8b=13d \\quad \\textrm{or}\\quad 8b=7d.\\] The first condition yields the solution $(c,d,b)=(3,8,13)$ and the second condition yields the solution $(c,b,d)=(12,14,16)$\nTogether, the sum of all possible values of $a$ is $18+(3+8+13)+(12+14+16)=\\boxed{84}.$", "Denote $x = AD$ $\\theta = \\angle B$ .\nHence, $BC = \\frac{\\sqrt{3}}{2} \\cdot \\frac{x}{\\sin \\theta}$ $DC = 18 - \\frac{x}{2} - \\frac{\\sqrt{3}}{2} x \\cot \\theta$\n$\\textbf{Case 1}$ $DC = AD = BC = AB$\nThis is a rhombus. So each side has length $18$\nFor the following cases, we consider four sides that have distinct lengths.\nTo make their lengths an arithmetic sequence, we must have $\\theta \\neq 120^\\circ$\nTherefore, in the subsequent analysis, we exclude the solution $\\theta = 120^\\circ$\n$\\textbf{Case 2}$ $DC < AD < BC < AB$\nBecause the lengths of these sides form an arithmetic sequence, we have the following system of equations: \\[ AB - BC = BC - AD = AD - DC . \\]\nHence, \\begin{eqnarray*} & 18 - \\frac{\\sqrt{3}}{2}\\cdot\\frac{x}{\\sin \\theta} = \\frac{\\sqrt{3}}{2}\\cdot\\frac{x}{\\sin \\theta} - x = x - \\left( 18 - \\frac{x}{2} - \\frac{\\sqrt{3}}{2} x \\cot \\theta \\right) . & \\end{eqnarray*}\nBy solving this system of equations, we get $\\left( \\cos \\theta , \\sin \\theta , x\\right) = \\left( \\frac{11}{13} , \\frac{4 \\sqrt{3}}{13} , 8 \\right)$\nThus, in this case, $DC = 3$ $AD = 8$ $BC = 13$\n$\\textbf{Case 3}$ $DC < BC < AD < AB$\nBecause the lengths of these sides form an arithmetic sequence, we have the following system of equations: \\[ AB - AD = AD - BC = BC - DC . \\]\nHence, \\begin{eqnarray*} & 18 - x = x - \\frac{\\sqrt{3}}{2}\\cdot\\frac{x}{\\sin \\theta} = \\frac{\\sqrt{3}}{2}\\cdot\\frac{x}{\\sin \\theta} - \\left( 18 - \\frac{x}{2} - \\frac{\\sqrt{3}}{2} x \\cot \\theta \\right) . & \\end{eqnarray*}\nBy solving this system of equations, we get $\\left( \\cos \\theta , \\sin \\theta , x\\right) = \\left( - \\frac{1}{7} , \\frac{4 \\sqrt{3}}{7} , 16 \\right)$\nThus, in this case, $DC = 12$ $AD = 16$ $BC = 14$\n$\\textbf{Case 4}$ $BC < CD < AD < AB$\nBy doing the similar analysis, we can show there is no solution in this case.\n$\\textbf{Case 5}$ $BC < AD < CD < AB$\nBy doing the similar analysis, we can show there is no solution in this case.\n$\\textbf{Case 6}$ $AD < CD < BC < AB$\nBy doing the similar analysis, we can show there is no solution in this case.\n$\\textbf{Case 7}$ $AD < BC < CD < AB$\nBy doing the similar analysis, we can show there is no solution in this case.\nTherefore, the sum of all possible values of $a$ is \\begin{align*} 18 + \\left( 3 + 8 + 13 \\right) + \\left( 12 + 14 + 16 \\right) & = 84 . \\end{align*}\nTherefore, the answer is $\\boxed{84}$" ]

[ "The easiest way for the areas of the triangles to be equal would be if they were congruent [1] . A way for that to work would be if $ABCD$ were simply an isosceles trapezoid! Since $AC = 14$ and $AE:EC = 3:4$ (look at the side lengths and you'll know why!), $\\boxed{6}$" ]

[ "Using the fact that $[AED] = [BEC]$ and the fact that $\\triangle AEB \\sim \\triangle EDC$ (which should be trivial given the two equal triangles) we have that\n\\[\\frac{AE}{DC} = \\frac{BE}{EC} = \\frac{9}{12}\\]\nWe know that $DC=EC,$ so we have\n\\[\\frac{AE}{EC} = \\frac{BE}{EC} = \\frac{3}{4}\\]\nThus\n\\[\\frac{AE}{EC} = \\frac{3}{4}\\]\nBut $EC = 14 - AE$ so we have\n\\[\\frac{AE}{14 - AE} = \\frac{3}{4}\\]\nSimplifying gives $AE = \\boxed{6}.$" ]

[ "The easiest way for the areas of the triangles to be equal would be if they were congruent [1] . A way for that to work would be if $ABCD$ were simply an isosceles trapezoid! Since $AC = 14$ and $AE:EC = 3:4$ (look at the side lengths and you'll know why!), $\\boxed{6}$" ]

[ "Using the fact that $[AED] = [BEC]$ and the fact that $\\triangle AEB \\sim \\triangle EDC$ (which should be trivial given the two equal triangles) we have that\n\\[\\frac{AE}{DC} = \\frac{BE}{EC} = \\frac{9}{12}\\]\nWe know that $DC=EC,$ so we have\n\\[\\frac{AE}{EC} = \\frac{BE}{EC} = \\frac{3}{4}\\]\nThus\n\\[\\frac{AE}{EC} = \\frac{3}{4}\\]\nBut $EC = 14 - AE$ so we have\n\\[\\frac{AE}{14 - AE} = \\frac{3}{4}\\]\nSimplifying gives $AE = \\boxed{6}.$" ]

[ "Note that by the pythagorean theorem, $AC=5$ . Also note that $\\angle CAD$ is a right angle because $\\triangle CAD$ is a right triangle. The area of the quadrilateral is the sum of the areas of $\\triangle ABC$ and $\\triangle CAD$ which is equal to \\[\\frac{3\\times4}{2} + \\frac{5\\times12}{2} = 6 + 30 = \\boxed{36}\\]" ]

[ "Let $n$ be the number of steps. We have\n\\[\\left\\lceil \\frac{n}{2} \\right\\rceil - 19 = \\left\\lceil \\frac{n}{5} \\right\\rceil\\]\nWe will proceed to solve this equation via casework.\nCase $1$ $\\left\\lceil \\frac{n}{2} \\right\\rceil = \\frac{n}{2}$\nOur equation becomes $\\frac{n}{2} - 19 = \\frac{n}{5} + \\frac{j}{5}$ , where $j \\in \\{0,1,2,3,4\\}$ Using the fact that $n$ is an integer, we quickly find that $j=1$ and $j=4$ yield $n=64$ and $n=66$ , respectively.\nCase $2$ $\\left\\lceil \\frac{n}{2} \\right\\rceil = \\frac{n}{2}+\\frac{1}{2}$\nOur equation becomes $\\frac{n}{2} +\\frac{1}{2} - 19 = \\frac{n}{5} + \\frac{j}{5}$ , where $j \\in \\{0,1,2,3,4\\}$ Using the fact that $n$ is an integer, we quickly find that $j=2$ yields $n=63$ . Summing up we get $63+64+66=193$ . The sum of the digits is $\\boxed{13}$", "We know from the problem that Dash goes $3$ steps further than Cozy per jump (assuming they aren't within $4$ steps from the top). That means that if Dash takes $19$ fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least $3 \\cdot 19=57$ steps high. We then start using guess-and-check:\n$57$ steps: $\\left \\lceil {57/2} \\right \\rceil = 29$ jumps for Cozy, and $\\left \\lceil {57/5} \\right \\rceil = 12$ jumps for Dash, giving a difference of $17$ jumps.\n$58$ steps: $\\left \\lceil {58/2} \\right \\rceil = 29$ jumps for Cozy, and $\\left \\lceil {58/5} \\right \\rceil = 12$ jumps for Dash, giving a difference of $17$ jumps.\n$59$ steps: $\\left \\lceil {59/2} \\right \\rceil = 30$ jumps for Cozy, and $\\left \\lceil {59/5} \\right \\rceil = 12$ jumps for Dash, giving a difference of $18$ jumps.\n$60$ steps: $\\left \\lceil {60/2} \\right \\rceil = 30$ jumps for Cozy, and $\\left \\lceil {60/5} \\right \\rceil = 12$ jumps for Dash, giving a difference of $18$ jumps.\n$\\vdots$\nBy the time we test $61$ steps, we notice that when the number of steps exceeds a multiple of $2$ , the difference in jumps increases. So, we have to find the next number that will increase the difference. $62$ doesn't because both both Cozy's and Dash's number of jumps increases, but $63$ does, and $64$ $65$ actually gives a difference of $20$ jumps, but $66$ goes back down to $19$ (because Dash had to take another jump when Cozy didn't). We don't need to go any further because the difference will stay above $19$ onward.\nTherefore, the possible numbers of steps in the staircase are $63$ $64$ , and $66$ , giving a sum of $193$ . The sum of those digits is $13$ , so the answer is $\\boxed{13}$" ]

[ "We're looking for natural numbers $x$ such that $\\left \\lceil{\\frac{x}{5}}\\right \\rceil + 19 = \\left \\lceil{\\frac{x}{2}}\\right \\rceil$\nLet's call $x = 10a + b$ . We now have $2a + \\left \\lceil{\\frac{b}{5}}\\right \\rceil + 19 = 5a + \\left \\lceil{\\frac{b}{2}}\\right \\rceil$ , or\n$19 - 3a = \\left \\lceil{\\frac{b}{2}}\\right \\rceil - \\left \\lceil{\\frac{b}{5}}\\right \\rceil$\nObviously, since $b \\le 10$ , this will not work for any value of $a$ under $6$ . In addition, since obviously $\\frac{b}{2} \\ge \\frac{b}{5}$ , this will not work for any value over six, so we have $a = 6$ and $\\left \\lceil{\\frac{b}{2}}\\right \\rceil - \\left \\lceil{\\frac{b}{5}}\\right \\rceil = 1.$\nThis can be achieved when $\\left \\lceil{\\frac{b}{5}}\\right \\rceil = 1$ and $\\left \\lceil{\\frac{b}{2}}\\right \\rceil = 2$ , or when $\\left \\lceil{\\frac{b}{5}}\\right \\rceil = 2$ and $\\left \\lceil{\\frac{b}{2}}\\right \\rceil = 3$\nCase One:\nWe have $b \\le 5$ and $3 \\le b \\le 4$ , so $b = 3, 4$\nCase Two:\nWe have $6 \\le b \\le 9$ and $5 \\le b \\le 6$ , so $b = 6$\nWe then have $63 + 64 + 66 = 193$ , which has a digit sum of $\\boxed{13}$", "Translate the problem into following equation:\n$n = 5D - \\{0 \\sim 4\\} = 2C - \\{0 \\sim 1\\}$\nSince $C = D + 19$ , we have\n$5D - \\{0 \\sim 4\\} = 2D + 38 - \\{0 \\sim 1\\}$\ni.e.,\n$3D = 38 + \\{0 \\sim 4\\} - \\{0 \\sim 1\\}$\nWe then have $D = 13$ when $\\{1\\} - \\{0\\}$ or $\\{2\\} - \\{1\\}$ (the dog's last jump has $2$ steps and the cat's last jump has $1$ step), which yields $n = 64$ and $n = 63$ respectively.\nAnother solution is $D = 14$ when $\\{4\\} - \\{0\\}$ , which yields $n = 66$\nTherefore, with $63 + 64 + 66 = 193$ , the digit sum is $\\boxed{13}$" ]

[ "We can translate this wordy problem into this simple equation:\n\\[\\left\\lceil \\frac{s}{2} \\right\\rceil - 19 = \\left\\lceil \\frac{s}{5} \\right\\rceil\\]\nWe will proceed to solve this equation via casework.\nCase 1: $\\left\\lceil \\frac{s}{2} \\right\\rceil = \\frac{s}{2}$\nOur equation becomes $\\frac{s}{2} - 19 = \\frac{s}{5} + \\frac{j}{5}$ , where $j \\in \\{0,1,2,3,4\\}$ Using the fact that $s$ is an integer, we quickly find that $j=1$ and $j=4$ yield $s=64$ and $s=66$ , respectively.\nCase 2: $\\left\\lceil \\frac{s}{2} \\right\\rceil = \\frac{s}{2}+\\frac{1}{2}$\nOur equation becomes $\\frac{s}{2} +\\frac{1}{2} - 19 = \\frac{s}{5} + \\frac{j}{5}$ , where $j \\in \\{0,1,2,3,4\\}$ Using the fact that $s$ is an integer, we quickly find that $j=2$ yields $s=63$\nSumming up we get $63+64+66=193$ . The sum of the digits is $\\boxed{13}$", "It can easily be seen that the problem can be expressed by the equation: \\[\\left\\lceil \\frac{s}{2} \\right\\rceil - \\left\\lceil \\frac{s}{5} \\right\\rceil = 19\\]\nHowever, because the ceiling function is difficult to work with, we can rewrite the previous equation as:\n\\[\\frac{s+a}{2} - \\frac{s+b}{5} = 19\\] Where $a \\in \\{0,1\\}$ and $b \\in \\{0,1,2,3,4\\}$ Multiplying both sides by ten and simplifying, we get: \\[5s+5a-2s-2b=190\\] \\[3s = 190+2b-5a\\] \\[s = 63 + \\frac{1+2b-5a}{3}\\]\nBecause s must be an integer, we need to find the values of $a$ and $b$ such that $2b-5a \\equiv 2 \\mod 3$ . We solve using casework.\nCase 1: $a = 0$\nIf $a = 0$ , we have $2b \\equiv 2 \\mod 3$ . We can easily see that $b = 1$ or $b = 4$ , which when plugged into our original equation lead to $s = 64$ and $s=66$ respectively.\nCase 2: $a = 1$\nIf $a = 1$ , we have $2b-5 \\equiv 2 \\mod 3$ , which can be rewritten as $2b \\equiv 1 \\mod 3$ . We can again easily see that $b = 2$ is the only solution, which when plugged into our original equation lead to $s = 63$\nAdding these together we get $64+66+63=193$ . The sum of the digits is $\\boxed{13}$", "As before, we write the equation:\n\\[\\left\\lceil \\frac{s}{2} \\right\\rceil - 19 = \\left\\lceil \\frac{s}{5} \\right\\rceil.\\]\nTo get a ballpark estimate of where $s$ might lie, we remove the ceiling functions to find:\n\\[\\frac{s}{2} - 19 = \\frac{s}{5}.\\]\nThis gives $\\frac{3s}{10} = 19$ , and thus values for $s$ will be around $\\frac{190}{3} = 63.\\overline3$\nNow, to establish some bounds around this estimated working value, we note that if $s=60$ , Cozy takes 30 steps while Dash takes 12, a difference of 18. If $s=70$ , Cozy takes 35 steps while Dash takes 14, a difference of 21. When $s$ increases from a multiple of ten, the difference will never decrease beyond what it is at the multiple of ten, and likewise, when it decreases, it never becomes greater than at the multiple of ten, so any working values of $s$ will be between $60$ and $70$\nThen, by inspection, $s=63, 64,$ or $66$ , so $\\sum s = 193 \\implies \\boxed{13}.$", "Notice that the possible number of steps in the staircase is around 60 to 70. By testing all of the values between 60 and 70, we see that 63, 64 and 66 work. Adding those up gives 193, so the answer is $1+9+3=\\boxed{13}.$", "We represent C's steps with $2c + a = n$ and D's steps with $5d + b = n$ , where $a \\in \\{1,2\\}$ and $b \\in \\{1,2,3,4,5\\}$ , where $n$ is the number of steps, $c$ is the number of jumps C takes bar the last one, and $d$ is the number of jumps D takes bar the last.\nThe reason for starting at 1 and ending at 5 instead of 0 through 4 is that the last step can be quite problematic to deal with, especially if it is possible to make it in one go, so we treat it as a different jump that can take all possible jump values. We know that D makes it in 19 fewer than C, so $c+1-(d+1) = 19 \\implies c = 19+d \\implies 5d + b = 2(19+d) + a \\implies 3d = 38 + a - b$\nNow that we have this nice equivalence, we can do the thing and it works.\nIf we take both sides mod 3 and rearrange, we get $b \\equiv 2+a \\pmod{3}$ This gives us the following satisfactory $(a,b)$ relational pairs: $\\{(1,3),(2,1),(2,4)\\}$ . We can now just find the corresponding $d$ value for each pair, sum it all up using $5d + b = n$ and sum the digits to reveal $\\boxed{13}$ as the answer." ]

[ "Crystal runs north one mile, then her next two moves can be broken up into four individual moves: for her northeast section, it forms a $45-45-90$ triangle whose legs are each $\\frac{\\sqrt{2}}{2}$ . For her southeast section, it is also a $45-45-90$ triangle whose legs are each $\\frac{\\sqrt{2}}{2}$ . Notice that the two of the legs cancel each other out; she moves north $\\frac{\\sqrt{2}}{2}$ units and also south $\\frac{\\sqrt{2}}{2}$ units. So her net movement these two moves is $\\sqrt{2}$ to the right. Finally, after the third move, she is at the corner of a right triangle with legs $1$ and $\\sqrt{2}$ . Using the Pythagorean theorem, $d^{2}=1^{2}+\\left(\\sqrt{2}\\right)^{2}=1+2=3$ and $d=\\sqrt{3} \\Longrightarrow \\boxed{3}$" ]

[ "Define a coordinate system with $D$ at the origin and $C,$ $A,$ and $H$ on the $x$ $y$ , and $z$ axes respectively. Then $D=(0,0,0),$ $M=(.5,1,0),$ and $N=(1,0,.5).$ It follows that the plane going through $D,$ $M,$ and $N$ has equation $2x-y-4z=0.$ Let $Q = (1,1,.25)$ be the intersection of this plane and edge $BF$ and let $P = (1,2,0).$ Now since $2(1) - 1(2) - 4(0) = 0,$ $P$ is on the plane. Also, $P$ lies on the extensions of segments $DM,$ $NQ,$ and $CB$ so the solid $DPCN = DMBCQN + MPBQ$ is a right triangular pyramid. Note also that pyramid $MPBQ$ is similar to $DPCN$ with scale factor $.5$ and thus the volume of solid $DMBCQN,$ which is one of the solids bounded by the cube and the plane, is $[DPCN] - [MPBQ] = [DPCN] - \\left(\\frac{1}{2}\\right)^3[DPCN] = \\frac{7}{8}[DPCN].$ But the volume of $DPCN$ is simply the volume of a pyramid with base $1$ and height $.5$ which is $\\frac{1}{3} \\cdot 1 \\cdot .5 = \\frac{1}{6}.$ So $[DMBCQN] = \\frac{7}{8} \\cdot \\frac{1}{6} = \\frac{7}{48}.$ Note, however, that this volume is less than $.5$ and thus this solid is the smaller of the two solids. The desired volume is then $[ABCDEFGH] - [DMBCQN] = 1 - \\frac{7}{48} = \\frac{41}{48} \\rightarrow p+q = \\boxed{089.}$", "Define a coordinate system with $D = (0,0,0)$ $M = (1, \\frac{1}{2}, 0)$ $N = (0,1,\\frac{1}{2})$ . The plane formed by $D$ $M$ , and $N$ is $z = \\frac{y}{2} - \\frac{x}{4}$ . It intersects the base of the unit cube at $y = \\frac{x}{2}$ . The z-coordinate of the plane never exceeds the height of the unit cube for $0 \\leq x \\leq 1, 0 \\leq y \\leq 1$ . Therefore, the volume of one of the two regions formed by the plane is \\[\\int_0^1 \\int_{\\frac{x}{2}}^1 \\int_0^{\\frac{y}{2}-\\frac{x}{4}}dz\\,dy\\,dx = \\frac{7}{48}\\] Since $\\frac{7}{48} < \\frac{1}{2}$ , our answer is $1-\\frac{7}{48} = \\frac{41}{48} \\rightarrow p+q = \\boxed{089}$", "Same coordinate system, but we will use domains instead of using triple integrals. By the way, the method to obtain the equation of the plane is the cross product ( $\\vec{[1,.5,0]}\\times\\vec{[0,1,.5]}=\\vec{[.25,-.5,1]}$ ). We can multiply this vector by $-4$ to make things look cleaner and get $\\vec{[-1,2,-4]}$ . We then get the desired plane, $-x+2y-4z=0$ , or $z=\\frac{2y-x}{4}$ . We use a double integral with a Type I domain. Observing the diagram, the domain is where $0 \\leq x \\leq 1, .5x \\leq y \\leq 1$ . The integral is then \\[.25\\int_0^1\\int_{.5x}^1 2y - x\\] which becomes \\[.25\\int_0^1 (y^2-xy)|_{.5x}^1\\] which becomes \\[.25\\int_0^1 1-x+.25x^2\\] which then becomes \\[.25(1-\\frac{1}{2}+\\frac{1}{12})\\] and finally \\[\\frac{7}{48}\\] So our answer is $1^3-\\frac{7}{48} = \\frac{41}{48}, 41+48 = \\boxed{089}$", "Let $Q$ be the intersection of the plane with edge $FB,$ then $\\triangle MQB$ is similar to $\\triangle DNC$ and the volume $[DNCMQB]$ is a sum of areas of cross sections of similar triangles running parallel to face $ABFE.$ Let $x$ be the distance from face $ABFE,$ let $h$ be the height parallel to $AB$ of the cross-sectional triangle at that distance, and $a$ be the area of the cross-sectional triangle at that distance. $A(x)=\\frac{h^2}{4},$ and $h=\\frac{x+1}{2},$ then $A=\\frac{(x+1)^2}{16}$ , and the volume $[DNCMQB]$ is $\\int^1_0{A(x)}\\,\\mathrm{d}x=\\int^1_0{\\frac{(x+1)^2}{16}}\\,\\mathrm{d}x=\\frac{7}{48}.$ Thus the volume of the larger solid is $1-\\frac{7}{48}=\\frac{41}{48} \\rightarrow p+q = \\boxed{089}$", "The volume of a frustum is $\\frac{h_2b_2 -h_1b_1}3$ where $b_i$ is the area of the base and $h_i$ is the height from the chopped off apex to the base.\nWe can easily see that from symmetry, the area of the smaller front base is $\\frac{1}{16}$ and the area of the larger back base is $\\frac{1}4$\nNow to find the height of the apex.\nExtend the $DM$ and (call the intersection of the plane with $FB$ G) $NG$ to meet at $x$ . Now from similar triangles $XMG$ and $XDN$ we can easily find the total height of the triangle $XDN$ to be $2$\nNow straight from our formula, the volume is $\\frac{7}{48}$ Thus the answer is:\n$1-\\text{Volume} \\Longrightarrow \\boxed{089}$", "We will solve for the area of the smaller region, and then subtract it from 1.\nLet $X$ be the point where plane $DMN$ intersects $FB$ . Then $DMBCNX$ can be split into triangular pyramid $DMBX$ and quadrilateral pyramid $BCNXD$\nPyramid $DMBX$ has base $DMB$ with area $\\frac{1}2 \\cdot 1 \\div 2 = \\frac{1}4$ . The height is $BX = \\frac{1}4$ , so the volume of $DMBX$ is $\\frac{1}4 \\cdot \\frac{1}4 \\div 3 = \\frac{1}{48}$\nSimilarly, pyramid $BCNXD$ has base $BCNX$ with area $(\\frac{1}4 + \\frac{1}2) \\cdot 1 = \\frac{3}8$ . The height is $CD = 1$ , so the volume of $BCNXD$ is $\\frac{3}8 \\cdot 1 \\div 3 = \\frac{1}8$\nAdding up the volumes of $DMBX$ and $BCNXD$ , we find that the volume of $DMBCNX$ is $\\frac{1}{48} + \\frac{6}{48} = \\frac{7}{48}$ . Therefore the volume of the larger solid is $1 - \\frac{7}{48} = \\frac{41}{48} \\rightarrow p+q = \\boxed{089}$", "We use a coordinate system with $C = (0,0,0)$ $D = (1,0,0)$ $M = (\\frac{1}{2},1,0)$ $N = (0,0,\\frac{1}{2})$ . Then the plane going through $D$ $M$ , and $N$ has equation $z = \\frac{1}{2} - \\frac{1}{2}x - \\frac{1}{4}y$ . We set up a double integral in this coordinate system. Consider the region to be integrated over in the $xy$ -plane. From $x=0$ to $x=\\frac{1}{2}$ , the upper bound of the region is $y = 1$ . From $x= \\frac{1}{2}$ to $x=1$ , the upper bound of the region is $y = 2 - 2x$ . In both cases, the lower bound of the region is $y = 0$ . Thus, we have the double integral $\\int_{0}^{\\frac{1}{2}} \\int_{0}^{1} \\frac{1}{2} - \\frac{1}{2}x - \\frac{1}{4}y dydx + \\int_{\\frac{1}{2}}^{1} \\int_{0}^{2-2x} \\frac{1}{2} - \\frac{1}{2}x - \\frac{1}{4}y dydx$ . We find that this sum evaluates to $\\frac{7}{48}$ . However, this is the volume of the smaller region, so the larger region's volume is that of the cube minus that of the smaller region. Since the cube has side length $1$ , its volume is $1$ , so the volume of the larger region is $1 - \\frac{7}{48} = \\frac{41}{48}$ . Thus, our answer is $41 + 48 = \\boxed{089}$" ]

[ "We can set up an algebraic equation for this problem.\nFrom what's given, we have that $100c+10b+a=55x+100a+10b+c$\nThis simplifies to be $0=55x+99a-99c\\implies -55x=99a-99c$\nFactoring, we get that $-55x=99(a-c)\\implies x=-\\frac{9(a-c)}{5}$\nHence, notice that we want $a-c=-5$ so that $x=9$\nThe only pair that works for this problem that satisfies the original requirements is $(1,6)$\nHence, $a=1, b=0, c=6$\nChecking, we have that $106+55(9)=601\\implies 601=601$\nHence, the answer is $1^2+0^2+6^2=37\\implies\\boxed{37}$" ]

[ "We know that the number of miles she drove is divisible by $5$ , so $a$ and $c$ must either be the equal or differ by $5$ . We can quickly conclude that the former is impossible, so $a$ and $c$ must be $5$ apart. Because we know that $c > a$ and $a + c \\le 7$ and $a \\ge 1$ , we find that the only possible values for $a$ and $c$ are $1$ and $6$ , respectively. Because $a + b + c \\le 7$ $b = 0$ . Therefore, we have \\[a^2 + b^2 + c^2 = 36 + 0 + 1 = \\boxed{37}\\]", "Let the number of hours Danica drove be $k$ . Then we know that $100a + 10b + c + 55k$ $100c + 10b + a$ . Simplifying, we have $99c - 99a = 55k$ , or $9c - 9a = 5k$ . Thus, k is divisible by $9$ . Because $55 * 18 = 990$ $k$ must be $9$ , and therefore $c - a = 5$ . Because $a + b + c \\leq{7}$ and $a \\geq{1}$ $a = 1$ $c = 6$ and $b = 0$ , and our answer is $a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37$ , or $\\boxed{37}$" ]

[ "The least multiple of 6 greater than 23 is 24. So she will need to add $24-23=\\boxed{1}$ more model car.\n~avamarora", "6 x 4 = 24, which is 1 more than 23. So, the answer is $\\boxed{1}$" ]

[ " Label the bottom left corner of the larger rectangle (without the square cut out) as $A$ and the top right as $D$ $w$ is the width of the rectangle and $\\ell$ is the length. Now we have vertices $E, F, G, H$ as vertices of the irregular octagon created by cutting out the squares. Let $I, J$ be the two closest vertices formed by the squares.\nThe distance between the two closest vertices of the squares is thus $IJ=\\left(4\\sqrt{2}\\right).$ Substituting, we get\n\\[(IJ)^2 = (w-2)^2 + (\\ell-2)^2 = \\left(4\\sqrt{2}\\right)^2 = 32 \\implies w^2+\\ell^2-4w-4\\ell = 24.\\] Using the fact that the diagonal of the rectangle is $8,$ we get \\[w^2+\\ell^2 = 64.\\] Subtracting the first equation from the second equation, we get \\[4w+4\\ell=40 \\implies w+\\ell = 10.\\] Squaring yields \\[w^2 + 2w\\ell + \\ell^2 = 100.\\] Subtracting the second equation from this, we get $2w\\ell = 36,$ and thus area of the original rectangle is $w\\ell = \\boxed{18}.$" ]

[ "The $365$ -day time period can be split up into $6$ $60$ -day time periods, because after $60$ days, all three of them visit again (Least common multiple of $3$ $4$ , and $5$ ).\nYou can find how many times each pair of visitors can meet by finding the LCM of their visiting days and dividing that number by 60.\nRemember to subtract $1$ , because you do not wish to count the $60$ th day, when all three visit.\nA and B visit $\\frac{60}{3 \\cdot 4}-1 = 4$ times.\nA and C visit $\\frac{60}{3 \\cdot 5}-1 = 3$ times.\nB and C visit $\\frac{60}{4 \\cdot 5}-1 = 2$ times.\nThis is a total of $9$ visits per $60$ day period.\nTherefore, the total number of $2$ -person visits is $9 \\cdot 6 = \\boxed{54}$", "From the information above, we get that $A=3x$ $B=4x$ $C=5x$\nNow, we want the days in which exactly two of these people meet up\nThe three pairs are $(A,B)$ $(B,C)$ $(A,C)$\nNotice that we are trying to find the LCM of each pair.\nHence, $LCM(A,B)=12x$ $LCM(B,C)=20x$ $LCM(A,C)=15x$\nNotice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.\nHence, $LCM(A,B,C)=60x$\nNow, we add all of the days up(including overcount).\nWe get $30+18+24=72$ .\nNow, because $60(6)=360$ , we have to subtract $6$ days from every pair.\nHence, our answer is $72-18=\\boxed{54}$" ]

[ "There are $12 \\cdot 11 = 132$ possible situations ( $12$ choices for the initially assigned gate, and $11$ choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most $400$ feet apart.\nIf we number the gates $1$ through $12$ , then gates $1$ and $12$ have four other gates within $400$ feet, gates $2$ and $11$ have five, gates $3$ and $10$ have six, gates $4$ and $9$ have have seven, and gates $5$ $6$ $7$ $8$ have eight. Therefore, the number of valid gate assignments is \\[2\\cdot(4+5+6+7)+4\\cdot8 = 2 \\cdot 22 + 4 \\cdot 8 = 76\\] so the probability is $\\frac{76}{132} = \\frac{19}{33}$ . The answer is $19 + 33 = \\boxed{052}$", "As before, derive that there are $132$ possibilities for Dave's original and replacement gates.\nNow suppose that Dave has to walk $100k$ feet to get to his new gate. This means that Dave's old and new gates must be $k$ gates apart. (For example, a $100$ foot walk would consist of the two gates being adjacent to each other.) There are $12-k$ ways to pick two gates which are $k$ gates apart, and $2$ possibilities for gate assignments, for a total of $2(12-k)$ possible assignments for each $k$\nAs a result, the total number of valid gate arrangements is \\[2\\cdot 11 + 2\\cdot 10 + 2\\cdot 9 + 2\\cdot 8 = 76\\] and so the requested probability is $\\tfrac{19}{33}$ for a final answer of $\\boxed{052}$" ]

[ "There are many almost equivalent approaches that lead to summing a geometric series. For example, we can compute the probability of the opposite event. Let $p$ be the probability that Dave will make at least two more throws than Linda. Obviously, $p$ is then also the probability that Linda will make at least two more throws than Dave, and our answer will therefore be $1-2p$\nHow to compute $p$\nSuppose that Linda made exactly $t$ throws. The probability that this happens is $(5/6)^{t-1}\\cdot (1/6)$ , as she must make $t-1$ unsuccessful throws followed by a successful one. In this case, we need Dave to make at least $t+2$ throws. This happens if his first $t+1$ throws are unsuccessful, hence the probability is $(5/6)^{t+1}$\nThus for a fixed $t$ the probability that Linda makes $t$ throws and Dave at least $t+2$ throws is $(5/6)^{2t} \\cdot (1/6)$\nThen, as the events for different $t$ are disjoint, $p$ is simply the sum of these probabilities over all $t$ . Hence:\n\\begin{align*} p & = \\sum_{t=1}^\\infty \\left(\\frac 56\\right)^{2t} \\cdot \\frac 16 \\\\ & = \\frac 16 \\cdot \\left(\\frac 56\\right)^2 \\cdot \\sum_{x=0}^\\infty \\left(\\frac{25}{36}\\right)^x \\\\ & = \\frac 16 \\cdot \\frac{25}{36} \\cdot \\frac 1{1 - \\dfrac{25}{36}} \\\\ & = \\frac 16 \\cdot \\frac{25}{36} \\cdot \\frac{36}{11} \\\\ & = \\frac {25}{66} \\end{align*}\nHence the probability we were supposed to compute is $1 - 2p = 1 - 2\\cdot \\frac{25}{66} = 1 - \\frac{25}{33} = \\frac 8{33}$ , and the answer is $8+33 = \\boxed{041}$", "Let $p$ be the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. (We will call this event \"a win\", and the opposite event will be \"a loss\".)\nLet both players roll their first die.\nWith probability $\\frac 1{36}$ , both throw a six and we win.\nWith probability $\\frac{10}{36}$ exactly one of them throws a six. In this case, we win if the remaining player throws a six in their next throw, which happens with probability $\\frac 16$\nFinally, with probability $\\frac{25}{36}$ none of them throws a six. Now comes the crucial observation: At this moment, we are in exactly the same situation as in the beginning. Hence in this case we will win with probability $p$\nWe just derived the following linear equation: \\[p = \\frac 1{36} + \\frac{10}{36} \\cdot \\frac 16 + \\frac{25}{36} \\cdot p\\]\nSolving for $p$ , we get $p=\\frac 8{33}$ , hence the answer is $8+33 = \\boxed{041}$", "Let's write out the probabilities with a set number of throws that Linda rolls before getting a 6. The probability of Linda rolling once and gets 6 right away is $\\frac{1}{6}$ . The probability that Dave will get a six in the same, one less, or one more throw is $\\frac{1}{6} + \\frac{5}{6} * \\frac{5}{6}$ . Thus the combined probability is $\\frac{11}{216}$\nLet's do the same with the probability that Linda rolls twice and getting a six. This time it is $\\frac{5}{6} * \\frac{1}{6}$ . The probability that Dave meets the requirements set is $\\frac{1}{6} + \\frac{5}{6} * \\frac{1}{6} + \\frac{5}{6} * \\frac{5}{6} * \\frac{1}{6}$ . Combine the probabilities again to get $\\frac{455}{7776}$ . (or not, because you can simplify without calculating later)\nIt's clear that as the number of rolls before getting a six increases, the probability that Dave meets the requirements is multiplied by $\\frac{5}{6} * \\frac{5}{6}$ . We can use this pattern to solve for the sum of an infinite geometric series.\nFirst, set the case where Linda rolls only once aside. It doesn't fit the same pattern as the rest, so we'll add it separately at the end. Next, let $a = (\\frac{5}{6} * \\frac{1}{6}) * (\\frac{1}{6} + \\frac{5}{6} * \\frac{1}{6} + \\frac{5}{6} * \\frac{5}{6} * \\frac{1}{6}) = \\frac{455}{7776}$ as written above. Each probability where the number of tosses Linda makes increases by one will be $a * (\\frac{25}{36})^{n+1}$ . Let $S$ be the sum of all these probabilities.\n$S = a + a * \\frac{25}{36} + a * (\\frac{25}{36})^2...$\n$S * \\frac{25}{36} = a * \\frac{25}{36} + a * (\\frac{25}{36})^2 + a * (\\frac{25}{36})^3...$\nSubtract the second equation from the first to get\n$S * \\frac{11}{36} = a$\n$S = a * \\frac{36}{11}$\n$S = \\frac{455}{2376}$\nDon't forget to add the first case where Linda rolls once.\n$\\frac{455}{2376} + \\frac{11}{216} = \\frac{8}{33}$\n$8 + 33 = \\boxed{41}$" ]

[ "Suppose our four sides lengths cut out arc lengths of $2a$ $2b$ $2c$ , and $2d$ , where $a+b+c+d=180^\\circ$ . Then, we only have to consider which arc is opposite $2a$ . These are our three cases, so \\[\\varphi_A=a+c\\] \\[\\varphi_B=a+b\\] \\[\\varphi_C=a+d\\] Our first case involves quadrilateral $ABCD$ with $\\overarc{AB}=2a$ $\\overarc{BC}=2b$ $\\overarc{CD}=2c$ , and $\\overarc{DA}=2d$\nThen, by Law of Sines, $AC=2\\sin\\left(\\frac{\\overarc{ABC}}{2}\\right)=2\\sin(a+b)$ and $BD=2\\sin\\left(\\frac{\\overarc{BCD}}{2}\\right)=2\\sin(a+d)$ . Therefore,\n\\[K=\\frac{1}{2}\\cdot AC\\cdot BD\\cdot \\sin(\\varphi_A)=2\\sin\\varphi_A\\sin\\varphi_B\\sin\\varphi_C=\\frac{24}{35},\\] so our answer is $24+35=\\boxed{059}$", "Suppose the four side lengths of the quadrilateral cut out arc lengths of $2a$ $2b$ $2c$ , and $2d$ $a+b+c+d=180^\\circ$ . \nTherefore, without losing generality,\n\\[\\varphi_A=a+b\\] \\[\\varphi_B=b+c\\] \\[\\varphi_C=a+c\\]\n$(1)+(3)-(2)$ $(1)+(2)-(3)$ , and $(2)+(3)-(1)$ yields\n\\[2a=\\varphi_A+\\varphi_C-\\varphi_B\\] \\[2b=\\varphi_A+\\varphi_B-\\varphi_C\\] \\[2c=\\varphi_B+\\varphi_C-\\varphi_A\\]\nBecause $2d=360^\\circ-2a-2b-2c,$ Therefore,\n\\[2d=360^\\circ-\\varphi_A-\\varphi_B-\\varphi_C\\]\nUsing the sum-to-product identities , our area of the quadrilateral $K$ then would be\n\\begin{align*} K&=\\frac{1}{2}(\\sin(2a)+\\sin(2b)+\\sin(2c)+\\sin(2d))\\\\ &=\\frac{1}{2}(\\sin(\\varphi_A+\\varphi_B-\\varphi_C)+\\sin(\\varphi_B+\\varphi_C-\\varphi_A)+\\sin(\\varphi_C+\\varphi_A-\\varphi_B)-\\sin(\\varphi_A+\\varphi_B+\\varphi_C))\\\\ &=\\frac{1}{2}(2\\sin\\varphi_B\\cos(\\varphi_A-\\varphi_C)-2\\sin\\varphi_B\\cos(\\varphi_A+\\varphi_C))\\\\ &=\\frac{1}{2}\\cdot2\\cdot2\\sin\\varphi_A\\sin\\varphi_B\\sin\\varphi_C\\\\ &=2\\sin\\varphi_A\\sin\\varphi_B\\sin\\varphi_C\\\\ &=\\frac{24}{35}\\\\ \\end{align*}\nTherefore, our answer is $24+35=\\boxed{059}$", "Let the four stick lengths be $a$ $b$ $c$ , and $d$ . WLOG, let’s say that quadrilateral $A$ has sides $a$ and $d$ opposite each other, quadrilateral $B$ has sides $b$ and $d$ opposite each other, and quadrilateral $C$ has sides $c$ and $d$ opposite each other. The area of a convex quadrilateral can be written as $\\frac{1}{2} d_1 d_2 \\sin{\\theta}$ , where $d_1$ and $d_2$ are the lengths of the diagonals of the quadrilateral and $\\theta$ is the angle formed by the intersection of $d_1$ and $d_2$ . By Ptolemy's theorem $d_1 d_2 = ad+bc$ for quadrilateral $A$ , so, defining $K_A$ as the area of $A$ \\[K_A = \\frac{1}{2} (ad+bc)\\sin{\\varphi_A}\\] Similarly, for quadrilaterals $B$ and $C$ \\[K_B = \\frac{1}{2} (bd+ac)\\sin{\\varphi_B}\\] and \\[K_C = \\frac{1}{2} (cd+ab)\\sin{\\varphi_C}\\] Multiplying the three equations and rearranging, we see that \\[K_A K_B K_C = \\frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\\sin{\\varphi_A}\\sin{\\varphi_B}\\sin{\\varphi_B}\\] \\[K^3 = \\frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\\left(\\frac{2}{3}\\right) \\left(\\frac{3}{5}\\right) \\left(\\frac{6}{7}\\right)\\] \\[\\frac{70}{3}K^3 = (ab+cd)(ac+bd)(ad+bc)\\] The circumradius $R$ of a cyclic quadrilateral with side lengths $a$ $b$ $c$ , and $d$ and area $K$ can be computed as $R = \\frac{\\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}$ .\nInserting what we know, \\[1 = \\frac{\\sqrt{\\frac{70}{3}K^3}}{4K}\\quad \\Rightarrow \\quad 16K^2 = \\frac{70}{3}K^3\\quad \\Rightarrow \\quad \\frac{24}{35} = K\\] So our answer is $24 + 35 = \\boxed{059}$", "Let the sides of the quadrilaterals be $a,b,c,$ and $d$ in some order such that $A$ has $a$ opposite of $c$ $B$ has $a$ opposite of $b$ , and $C$ has $a$ opposite of $d$ . Then, let the diagonals of $A$ be $e$ and $f$ . Similarly to solution $2$ , we get that $\\tfrac{2}{3}(ac+bd)=\\tfrac{3}{5}(ab+cd)=\\tfrac{6}{7}(ad+bc)=2K$ , but this is also equal to $2\\cdot\\tfrac{eab+ecd}{4(1)}=2\\cdot\\tfrac{fad+fbc}{4(1)}$ using the area formula for a triangle using the circumradius and the sides, so $\\tfrac{e(ab+cd)}{2}=\\tfrac{3}{5}(ab+cd)$ and $\\tfrac{f(ad+bc)}{2}=\\tfrac{6}{7}(ad+bc)$ . Solving for $e$ and $f$ , we get that $e=\\tfrac{6}{5}$ and $f=\\tfrac{12}{7}$ , but $K=\\tfrac{1}{2}\\cdot\\tfrac{2}{3}\\cdot{}ef$ , similarly to solution $2$ , so $K=\\tfrac{24}{35}$ and the answer is $24+35=\\boxed{059}$" ]

[ "Notice that $L_{17}$ contains the highest power of every prime below $17$ since higher primes cannot divide $L_{17}$ . Thus, $L_{17}=16\\cdot 9 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17$\nWhen writing the sum under a common fraction, we multiply the denominators by $L_{17}$ divided by each denominator. However, since $L_{17}$ is a multiple of $17$ , all terms will be a multiple of $17$ until we divide out $17$ , and the only term that will do this is $\\frac{1}{17}$ . Thus, the remainder of all other terms when divided by $17$ will be $0$ , so the problem is essentially asking us what the remainder of $\\frac{L_{17}}{17} = L_{16}$ divided by $17$ is. This is equivalent to finding the remainder of $16 \\cdot 9 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13$ divided by $17$\nWe use modular arithmetic to simplify our answer:\nThis is congruent to $-1 \\cdot 9 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 \\pmod{17}$\nEvaluating, we get: \\begin{align*} (-1) \\cdot 9 \\cdot 35 \\cdot 11 \\cdot 13 &\\equiv (-1) \\cdot 9 \\cdot 1 \\cdot 11 \\cdot 13 \\pmod{17} \\\\ &\\equiv 9 \\cdot 11 \\cdot (-13) \\pmod{17} \\\\ &\\equiv 9 \\cdot 11 \\cdot 4\\pmod{17} \\\\ &\\equiv 2 \\cdot 11 \\pmod{17} \\\\ &\\equiv 5\\pmod{17} \\end{align*} Therefore the remainder is $\\boxed{5}$", "As in solution 1, we express the LHS as a sum under one common denominator. We note that \\[\\frac{1}{1} + \\frac{1}{2} + \\dots + \\frac{1}{17} = \\frac{\\frac{17!}{1}}{17!} + \\frac{\\frac{17!}{2}}{17!} + \\frac{\\frac{17!}{3}}{17!} + \\dots + \\frac{\\frac{17!}{17}}{17!}\\]\nNow, we have $h = L_{17}\\left(\\frac{\\frac{17!}{1} + \\frac{17!}{2} + \\frac{17!}{3} + \\dots + \\frac{17!}{17}}{17!}\\right)$ . We'd like to find $h \\pmod{17},$ so we can evaluate our expression $\\pmod{17}.$ Since $\\frac{\\frac{17!}{1}}{17!}, \\frac{\\frac{17!}{2}}{17!}, \\dots, \\frac{\\frac{17!}{16}}{17!}$ don't have a factor of $17$ in their denominators, and since $L_{17}$ is a multiple of $17,$ multiplying each of those terms and adding them will get a multiple of $17.$ $\\pmod{17}$ , that result is $0.$ Thus, we only need to consider $L_{17}\\cdot \\frac{\\frac{17!}{17}}{17!} = \\frac{L_{17}}{17} \\pmod{17}.$ Proceed with solution $1$ to get $\\boxed{5}$", "Using Wolstenholmes' Theorem, we can rewrite $1 + \\frac{1}{2} \\dots + \\frac{1}{16}$ as $\\frac{17^2 n}{(17 - 1)!} = \\frac{17^2 n}{16!}$ (for some $n \\in \\mathbb{Z}$ ). Adding the $\\frac{1}{17}$ to $\\frac{17^2 n}{16!}$ , we get $\\frac{17^3 n + 16!}{17!}$\nNow we have $\\frac{17^3 n + 16!}{17!} = \\frac{h}{L_{17}}$ and we want $h \\pmod{17}$ . We find that $\\frac{L_{17}(17^3 n + 16!)}{17!} = \\frac{L_{16}(17^3 n + 16!)}{16!} = h$ . Taking $\\pmod{17}$ and multiplying, we get $L_{16}(17^3 n + 16!) \\equiv 16! \\cdot h \\pmod{17}$\nApplying Wilson's Theorem on $16!$ and reducing, we simplify the congruence to $L_{16}(0 - 1) \\equiv -L_{16} \\equiv -h \\pmod{17}$ . Now we proceed with Solution 1 and find that $L_{16} \\equiv 5 \\pmod{17}$ , so our answer is $\\boxed{5}$" ]

[ "If we graph $4g(f(x))$ , we see it forms a sawtooth graph that oscillates between $0$ and $1$ (for values of $x$ between $-1$ and $1$ , which is true because the arguments are between $-1$ and $1$ ). Thus by precariously drawing the graph of the two functions in the square bounded by $(0,0)$ $(0,1)$ $(1,1)$ , and $(1,0)$ , and hand-counting each of the intersections, we get $\\boxed{385}$", "We will denote $h(x)=4g(f(x))$ for simplicity. Denote $p(x)$ as the first equation and $q(y)$ as the graph of the second. We notice that both $f(x)$ and $g(x)$ oscillate between 0 and 1. The intersections are thus all in the square $(0,0)$ $(0,1)$ $(1,1)$ , and $(1,0)$ . Every $p(x)$ wave going up and down crosses every $q(y)$ wave. Now, we need to find the number of times each wave touches 0 and 1.\nWe notice that $h(x)=0$ occurs at $x=-\\frac{3}{4}, -\\frac{1}{4}, \\frac{1}{4}, \\frac{3}{4}$ , and $h(x)=1$ occurs at $x=-1, -\\frac{1}{2}, 0,\\frac{1}{2},1$ . A sinusoid passes through each point twice during each period, but it only passes through the extrema once. $p(x)$ has 1 period between 0 and 1, giving 8 solutions for $p(x)=0$ and 9 solutions for $p(x)=1$ , or 16 up and down waves. $q(y)$ has 1.5 periods, giving 12 solutions for $q(y)=0$ and 13 solutions for $q(y)=1$ , or 24 up and down waves. This amounts to $16\\cdot24=384$ intersections.\nHowever, we have to be very careful when counting around $(1, 1)$ . At this point, $q(y)$ has an infinite downwards slope and $p(x)$ is slanted, giving us an extra intersection; thus, we need to add 1 to our answer to get $\\boxed{385}$", "We can easily see that only $x, y \\in \\left[0,1 \\right]$ may satisfy both functions.\nWe call function $y = 4g \\left( f \\left( \\sin \\left( 2 \\pi x \\right) \\right) \\right)$ as Function 1 and function $x = 4g \\left( f \\left( \\cos \\left( 3 \\pi y \\right) \\right) \\right)$ as Function 2.\nFor Function 1, in each interval $\\left[ \\frac{i}{4} , \\frac{i+1}{4} \\right]$ with $i \\in \\left\\{ 0, 1, \\cdots, 3 \\right\\}$ , Function 1's value oscillates between 0 and 1. It attains 1 at $x = \\frac{i}{4}$ $\\frac{i+1}{4}$ and another point between these two.\nBetween two consecutive points whose functional values are 1, the function first decreases from 1 to 0 and then increases from 0 to 1. \nSo the graph of this function in this interval consists of 4 monotonic pieces.\nFor Function 2, in each interval $\\left[ \\frac{i}{6} , \\frac{i+1}{6} \\right]$ with $i \\in \\left\\{ 0, 1, \\cdots, 5 \\right\\}$ , Function 2's value oscillates between 0 and 1. It attains 1 at $y = \\frac{i}{6}$ $\\frac{i+1}{6}$ and another point between these two.\nBetween two consecutive points whose functional values are 1, the function first decreases from 1 to 0 and then increases from 0 to 1.\nSo the graph of this function in this interval consists of 4 monotonic curves.\nConsider any region $\\left[ \\frac{i}{4} , \\frac{i+1}{4} \\right] \\times \\left[ \\frac{j}{6} , \\frac{j+1}{6} \\right]$ with $i \\in \\left\\{ 0, 1, \\cdots, 3 \\right\\}$ and $j \\in \\left\\{0, 1, \\cdots , 5 \\right\\}$ but $\\left( i, j \\right) \\neq \\left( 3, 5 \\right)$ .\nBoth functions have four monotonic pieces.\nBecause Function 1's each monotonic piece can take any value in $\\left[ \\frac{j}{6} , \\frac{j+1}{6} \\right]$ and Function 2' each monotonic piece can take any value in $\\left[ \\frac{i}{4} , \\frac{i+1}{4} \\right]$ , Function 1's each monotonic piece intersects with Function 2's each monotonic piece. \nTherefore, in the interval $\\left[ \\frac{i}{4} , \\frac{i+1}{4} \\right] \\times \\left[ \\frac{j}{6} , \\frac{j+1}{6} \\right]$ , the number of intersecting points is $4 \\cdot 4 = 16$\nNext, we prove that if an intersecting point is on a line $x = \\frac{i}{4}$ for $i \\in \\left\\{ 0, 1, \\cdots, 4 \\right\\}$ , then this point must be $\\left( 1, 1 \\right)$\nFor $x = \\frac{i}{4}$ , Function 1 attains value 1. \nFor Function 2, if $y = 1$ , then $x = 1$ . \nTherefore, the intersecting point is $\\left( 1, 1 \\right)$\nSimilarly, we can prove that if an intersecting point is on a line $y = \\frac{i}{6}$ for $i \\in \\left\\{ 0, 1, \\cdots, 6 \\right\\}$ , then this point must be $\\left( 1, 1 \\right)$\nTherefore, in each region $\\left[ \\frac{i}{4} , \\frac{i+1}{4} \\right] \\times \\left[ \\frac{j}{6} , \\frac{j+1}{6} \\right]$ with $i \\in \\left\\{ 0, 1, \\cdots, 3 \\right\\}$ and $j \\in \\left\\{0, 1, \\cdots , 5 \\right\\}$ but $\\left( i, j \\right) \\neq \\left( 3, 5 \\right)$ , all 16 intersecting points are interior.\nThat is, no two regions share any common intersecting point.\nNext, we study region $\\left[ \\frac{3}{4} , 1 \\right] \\times \\left[ \\frac{5}{6} , 1 \\right]$ .\nConsider any pair of monotonic pieces, where one is from Function 1 and one is from Function 2, except the pair of two monotonic pieces from two functions that attain $\\left( 1 , 1 \\right)$ .\nTwo pieces in each pair intersects at an interior point on the region.\nSo the number of intersecting points is $4 \\cdot 4 - 1 = 15$\nFinally, we compute the number intersection points of two functions' monotonic pieces that both attain $\\left( 1, 1 \\right)$\nOne trivial intersection point is $\\left( 1, 1 \\right)$ .\nNow, we study whether they intersect at another point.\nDefine $x = 1 - x'$ and $y = 1 - y'$ .\nThus, for positive and sufficiently small $x'$ and $y'$ , Function 1 is reduced to \\[ y' = 4 \\sin 2 \\pi x' \\hspace{1cm} (1) \\] and Function 2 is reduced to \\[ x' = 4 \\left( 1 - \\cos 3 \\pi y' \\right) . \\hspace{1cm} (2) \\]\nNow, we study whether there is a non-zero solution.\nBecause we consider sufficiently small $x'$ and $y'$ , to get an intuition and quick estimate, we do approximations of the above equations.\nEquation (1) is approximated as \\[ y' = 4 \\cdot 2 \\pi x' \\] and Equation (2) is approximated as \\[ x' = 2 \\left( 3 \\pi y' \\right)^2 \\]\nTo solve these equations, we get $x' = \\frac{1}{8^2 \\cdot 18 \\pi^4}$ and $y' = \\frac{1}{8 \\cdot 18 \\pi^3}$ .\nTherefore, two functions' two monotonic pieces that attain $\\left( 1, 1 \\right)$ have two intersecting points.\nPutting all analysis above, the total number of intersecting points is $16 \\cdot 4 \\cdot 6 + 1 = \\boxed{385}$" ]

[ "First, note that $(2n)!! = 2^n \\cdot n!$ , and that $(2n)!! \\cdot (2n-1)!! = (2n)!$\nWe can now take the fraction $\\dfrac{(2i-1)!!}{(2i)!!}$ and multiply both the numerator and the denominator by $(2i)!!$ . We get that this fraction is equal to $\\dfrac{(2i)!}{(2i)!!^2} = \\dfrac{(2i)!}{2^{2i}(i!)^2}$\nNow we can recognize that $\\dfrac{(2i)!}{(i!)^2}$ is simply ${2i \\choose i}$ , hence this fraction is $\\dfrac{{2i\\choose i}}{2^{2i}}$ , and our sum turns into $S=\\sum_{i=1}^{2009} \\dfrac{{2i\\choose i}}{2^{2i}}$\nLet $c = \\sum_{i=1}^{2009} {2i\\choose i} \\cdot 2^{2\\cdot 2009 - 2i}$ .\nObviously $c$ is an integer, and $S$ can be written as $\\dfrac{c}{2^{2\\cdot 2009}}$ .\nHence if $S$ is expressed as a fraction in lowest terms, its denominator will be of the form $2^a$ for some $a\\leq 2\\cdot 2009$\nIn other words, we just showed that $b=1$ .\nTo determine $a$ , we need to determine the largest power of $2$ that divides $c$\nLet $p(i)$ be the largest $x$ such that $2^x$ that divides $i$\nWe can now return to the observation that $(2i)! = (2i)!! \\cdot (2i-1)!! = 2^i \\cdot i! \\cdot (2i-1)!!$ . Together with the obvious fact that $(2i-1)!!$ is odd, we get that $p((2i)!)=p(i!)+i$\nIt immediately follows that $p\\left( {2i\\choose i} \\right) = p((2i)!) - 2p(i!) = i - p(i!)$ ,\nand hence $p\\left( {2i\\choose i} \\cdot 2^{2\\cdot 2009 - 2i} \\right) = 2\\cdot 2009 - i - p(i!)$\nObviously, for $i\\in\\{1,2,\\dots,2009\\}$ the function $f(i)=2\\cdot 2009 - i - p(i!)$ is is a strictly decreasing function. \nTherefore $p(c) = p\\left( {2\\cdot 2009\\choose 2009} \\right) = 2009 - p(2009!)$\nWe can now compute $p(2009!) = \\sum_{k=1}^{\\infty} \\left\\lfloor \\dfrac{2009}{2^k} \\right\\rfloor = 1004 + 502 + \\cdots + 3 + 1 = 2001$ .\nHence $p(c)=2009-2001=8$\nAnd thus we have $a=2\\cdot 2009 - p(c) = 4010$ , and the answer is $\\dfrac{ab}{10} = \\dfrac{4010\\cdot 1}{10} = \\boxed{401}$", "Using the steps of the previous solution we get $c = \\sum_{i=1}^{2009} {2i\\choose i} \\cdot 2^{2\\cdot 2009 - 2i}$ and if you do the small cases(like $1, 2, 3, 4, 5, 6$ ) you realize that you can \"thin-slice\" the problem and simply look at the cases where $i=2009, 2008$ (they're nearly identical in nature but one has $4$ with it) since $\\dbinom{2i}{I}$ hardly contains any powers of $2$ or in other words it's very inefficient and the inefficient cases hold all the power so you can just look at the highest powers of $2$ in $\\dbinom{4018}{2009}$ and $\\dbinom{4016}{2008}$ and you get the minimum power of $2$ in either expression is $8$ so the answer is $\\frac{4010}{10} \\implies \\boxed{401}$ since it would violate the rules of the AIME and the small cases if $b>1$", "We can logically deduce that the $b$ value will be 1. Listing out the first few values of odd and even integers, we have: $1, 3, 5, 7, 9, 11, 13...$ and $2, 4, 6, 8, 10, 12, 14, 16...$ . Obviously, none of the factors of $2$ in the denominator will cancel out, since the numerator is odd. Starting on the second term of the numerator, a factor of $3$ occurs every $3$ terms, and starting out on the third term of the denominator, a factor of $3$ appears also every $3$ terms. Thus, the factors of $3$ on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator of the final expression will be in the form $1 \\cdot 2^a$ . Since there will be no odd factors in the denominators, all the denominators will be in the form $2^a$ where $a$ is the number of factors of $2$ in $(2009 \\cdot 2)! = 4018!$ . This is simply $\\sum_{n=1}^{11} \\left \\lfloor{\\frac{4018}{2^n}}\\right \\rfloor = 4010$ . Therefore, our answer is $\\boxed{401}$", "Using the initial steps from Solution 1, $S=\\sum_{i=1}^{2009} \\dfrac{{2i\\choose i}}{2^{2i}}$ . Clearly $b = 1$ as in all the summands there are no non-power of 2 factors in the denominator. So we seek to find $a$ . Note that $2^a$ would be the largest denominator in all the summands, so when they are summed it is the common denominator.\nTaking the p-adic valuations of each term, the powers of 2 in the denominator for $i$ is $(2i) - v_2(\\binom{2i}{i}).$ We can use Kummers theorem to see that $v_2(\\binom{2i}{i})$ is the number of digits carried over when $i$ is added to $i$ in base $2$ . This is simply the number of $1$ 's in the binary representation of $i$\nLooking at the binary representations of some of the larger $i$ we see $2009 = 11111011001_2$ having eight $1$ 's. So the power of two is $2 \\cdot 2009 - 8 = 4010$ . Experimenting with $2008, 2007, 2006$ we see that the power of two are all $< 4010$ , and under $2005$ the power of two $2i - v_2(\\binom{2i}{i}) < 2i < 4010$ . Therefore $a = 4010$ and $\\frac{ab}{10} = \\boxed{401}.$" ]

[ "We start by looking at the answers. Examining statement C, we notice:\n$x \\heartsuit 0 = |x-0| = |x|$\n$|x| \\neq x$ when $x<0$ , but statement C says that it does for all $x$\nTherefore the statement that is not true is $\\boxed{0}$" ]

[ "We start by looking at the answers. Examining statement C, we notice:\n$x \\heartsuit 0 = |x-0| = |x|$\n$|x| \\neq x$ when $x<0$ , but statement C says that it does for all $x$\nTherefore the statement that is not true is $\\boxed{0}$" ]

[ "We have \\begin{align*} (1\\diamond(2\\diamond3))-((1\\diamond2)\\diamond3) &= |1-|2-3|| - ||1-2|-3| \\\\ &= |1-1| - |1-3| \\\\ &= 0-2 \\\\ &= \\boxed{2} ~MRENTHUSIASM", "Observe that the $\\diamond$ function is simply the positive difference between two numbers. Thus, we evaluate: the difference between $2$ and $3$ is $1;$ the difference between $1$ and $1$ is $0;$ the difference between $1$ and $2$ is $1;$ the difference between $1$ and $3$ is $2;$ and finally, $0-2=\\boxed{2}.$" ]

[ "We have \\begin{align*} (1\\diamond(2\\diamond3))-((1\\diamond2)\\diamond3) &= |1-|2-3|| - ||1-2|-3| \\\\ &= |1-1| - |1-3| \\\\ &= 0-2 \\\\ &= \\boxed{2} ~MRENTHUSIASM", "Observe that the $\\diamond$ function is simply the positive difference between two numbers. Thus, we evaluate: the difference between $2$ and $3$ is $1;$ the difference between $1$ and $1$ is $0;$ the difference between $1$ and $2$ is $1;$ the difference between $1$ and $3$ is $2;$ and finally, $0-2=\\boxed{2}.$" ]

[ "We perform casework on $P(n)\\leq0:$\nTogether, the answer is $100+5000=\\boxed{5100}.$", "Notice that $P(x)$ is nonpositive when $x$ is between $100^2$ and $99^2, 98^2$ and $97^2, \\cdots$ $2^2$ and $1^2$ (inclusive), because there are an odd number of negatives, which means that the number of values equals \\[((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \\cdots + ((2+1)(2-1)+1).\\] This reduces to \\[200 + 196 + 192 + \\cdots + 4 = 4(1+2+\\cdots + 50) = 4 \\cdot\\frac{50 \\cdot 51}{2} = \\boxed{5100}.\\] ~Zeric", "We know that $P(x)$ is a $100$ -degree function with a positive leading coefficient. That is, $P(x)=x^{100}+ax^{99}+bx^{98}+...+\\text{(constant)}$\nSince the degree of $P(x)$ is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach $\\infty$ as $x$ goes in either direction, from which \\[\\lim_{x\\to-\\infty} P(x)=\\lim_{x\\to\\infty} P(x)=\\infty.\\] So the first time $P(x)$ is going to be negative is when it intersects the $x$ -axis at an $x$ -intercept and it's going to dip below. This happens at $1^2$ , which is the smallest intercept.\nHowever, when it hits the next intercept, it's going to go back up again into positive territory, we know this happens at $2^2$ . And when it hits $3^2$ , it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until $100^2$\nTo get the amount of integers below and/or on the $x$ -axis, we simply need to count the integers. For example, the amount of integers in between the $[1^2,2^2]$ interval we got earlier, we subtract and add one. $(2^2-1^2+1)=4$ integers, so there are four integers in this interval that produce a negative result.\nDoing this with all of the other intervals, we have \\[(2^2-1^2+1)+(4^2-3^2+1)+\\cdots+(100^2-99^2+1)=\\boxed{5100}\\] from Solution 2's result.", "We know $P(x) \\leq 0$ when an odd number of its factors are positive and negative. For example, to make the first factor positive, $x \\in [1^2, 2^2]$ . then there will be a even number of positive factors. We would do $2^2 - 1^2 + 1 (inclusive)$ to find all integers that work. In short we can generalize too: \\begin{align*} x^2 - (x-1)^2 + 1 &= 2x \\\\ x^2 - (x^2 - 2x + 1) + 1 &= 2x \\\\ x^2 - x^2 + 2x - 1 + 1 &= 2x. \\\\ \\end{align*} But remember this only works when $x \\in \\{2, 4, 6, 8 \\cdots 98, 100\\}$ because only then will there be a odd amount of positive and negative factors. So we can set $x = 2k$ , for $k \\in \\{1, 2, 3, 4, \\cdots 49, 50\\}$ Now we only have to solve: \\[\\sum_{k=1}^{k=50}2(2k) = 2\\sum_{k = 1}^{k = 50}2k = 4\\sum_{k = 1}^{k = 50}k = 4 \\cdot \\dfrac{(50)(51)}{2} = 2 \\cdot (50)(51) = \\boxed{5100}.\\] ~Wiselion" ]

[ "There are three letters to make the first letter in the sequence. However, after the first letter (whatever it is), only two letters can follow it, since one of the letters is restricted. Therefore, the number of seven-letter good words is $3*2^6=192$\nTherefore, there are $\\boxed{192}$ seven-letter good words.", "There are three choices for the first letter and two choices for each subsequent letter, so there are $3\\cdot2^{n-1}\\ n$ -letter good words. Substitute $n=7$ to find there are $3\\cdot2^6=\\boxed{192}$ seven-letter good words. ~ aopsav (Credit to AoPS Alcumus)", "We solve this problem using recursion. Let $f(x)$ be the number of $x$ -letter good words. Thus $f(1) = 3$ (A, B or C) and the answer is just $f(7)$ . The recurrence relation can be found by considering the last letter of one of the valid strings of length $x - 1$ . There are $2$ possibilities for the next letter and thus $f(x) = 2 \\cdot f(x-1)$ . Now we can find a closed form as $f(x) = 3 \\cdot 2 ^{x-1}$ (easy to prove by induction) and thus $f(7) = 64 * 3 = \\boxed{192}$ seven-letter good words. ~AK2006" ]

[ "The T-grid can be considered as a tic-tac-toe board: five $1$ 's (or X's) and four $0$ 's (or O's).\nThere are only $\\dbinom{9}{5} = 126$ ways to fill the board with five $1$ 's and four $0$ 's. Now we just need to subtract the number of bad grids. Bad grids are ones with more than one person winning, or where someone has won twice.\nLet three-in-a-row/column/diagonal be a \"win\" and let player $0$ be the one that fills in $0$ and player $1$ fills in $1$\nCase $1$ Each player wins once.\nIf player X takes a diagonal, player Y cannot win. If either takes a row, all the columns are blocked, and visa versa. Therefore, they either both take a row or they both take a column.\nCase $1$ $36$ cases\nCase $2$ Player $1$ wins twice. (Player $0$ cannot win twice because he only has 4 moves.)\nCase $2$ total: $22$\nThus, the answer is $126-22-36=\\boxed{68}$", "We can use generating functions to compute the case that no row or column is completely filled with $1$ 's. Given a row, let $a$ $b$ $c$ be the events that the first, second, third respective squares are $1$ 's. Then the generating function representing the possible events that exclude a row of $1,1,1$ or $0,0,0$ from occuring is \\[ab+bc+ca+a+b+c.\\] Therefore, the generating function representing the possible grids where no row is filled with $0$ 's and $1$ 's is \\[P(a,b,c)=((ab+bc+ca)+(a+b+c))^3.\\] We expand this by the Binomial Theorem to find \\[P(a,b,c)=(ab+bc+ca)^3+3(ab+bc+ca)^2(a+b+c)+3(ab+bc+ca)(a+b+c)^2+(a+b+c)^3.\\] Recall that our grid has five $1$ 's, hence we only want terms where the sum of the exponents is $5$ . This is given by \\[3(ab+bc+ca)^2(a+b+c).\\] When we expand this, we find \\[3(2abc(a+b+c)+a^2b^2+b^2c^2+c^2a^2)(a+b+c).\\] We also want to make sure that each of $a$ $b$ $c$ appears at least once (so there is no column filled with $0$ 's) and the power of each of $a$ $b$ $c$ is not greater than or equal to $3$ (so there is no column filled with $1$ 's). The sum of the coefficients of the above polynomial is clearly $81$ (using $a,b,c=1$ ), and the sum of the coefficients of the terms $a^3bc$ $ab^3c$ , and $abc^3$ is $6+6+6+3+3+3+3+3+3=36$ , hence the sum of the coefficients of the desired terms is $81-36=45$ . This counts the number of grids where no column or row is filled with $0$ 's or $1$ 's. However, we could potentially have both diagonals filled with $1$ 's, but this is the only exception to our $45$ possibilities, hence the number of $T$ -grids with no row or column filled with the same digit is $44$\nOn the other hand, if a row (column) is filled with $0$ 's, then by the Pigeonhole Principle, another row (column) must be filled with $1$ 's. Hence this is impossible, so all other\tpossible $T$ -grids have a row (column) filled with $1$ 's. If the top row is filled with $1$ 's, then we have two $1$ 's left to place. Clearly they cannot go in the same row, because then the other row is filled with $0$ 's. They also cannot appear in the same column. This leaves $3\\cdot 2$ arrangements--3 choices for the location of the $1$ in the second row, and 2 choices for the location of the $1$ in the last row. However, two of these arrangements will fill a diagonal with $1$ 's. Hence there are only $4$ $T$ -grids where the top row is filled with $1$ 's. The same argument applies if any other row or column is filled with $1$ 's. Hence there are $4\\cdot 6=24$ such $T$ -grids.\nThus the answer is $44+24=\\boxed{68}$" ]

[ "We can draw a comparison between the domino a set of 40 points (labeled 1 through 40) in which every point is connected with every other point. The connections represent the dominoes.\nYou need to have all even number of segments coming from each point except 0 or 2 which have an odd number of segments coming from the point. (Reasoning for this: Everytime you go to a vertex, you have to leave the vertex, so every vertex reached is equivalent to adding 2 more segments. So the degree of each vertex must be even, with the exception of endpoints) Since there are 39 segments coming from each point it is impossible to touch every segment.\nBut you can get up to 38 on each segment because you go in to the point then out on a different segment. Counting going out from the starting and ending at the ending point we have:\n$\\frac{38\\cdot 38 + 2\\cdot 39}2 = \\boxed{761}$" ]

[ "This is a recursive function, which means the function refers back to itself to calculate subsequent terms. To solve this, we must identify the base case, $f(1)=2$ . We also know that when $n$ is odd, $f(n)=f(n-2)+2$ . Thus we know that $f(2017)=f(2015)+2$ . Thus we know that n will always be odd in the recursion of $f(2017)$ , and we add $2$ each recursive cycle, which there are $1008$ of. Thus the answer is $1008*2+2=2018$ , which is answer $\\boxed{2018}$ .\nNote that when you write out a few numbers, you find that $f(n)=n+1$ for any $n$ , so $f(2017)=2018$" ]

[ "Let the number of zeros at the end of $m!$ be $f(m)$ . We have $f(m) = \\left\\lfloor \\frac{m}{5} \\right\\rfloor + \\left\\lfloor \\frac{m}{25} \\right\\rfloor + \\left\\lfloor \\frac{m}{125} \\right\\rfloor + \\left\\lfloor \\frac{m}{625} \\right\\rfloor + \\left\\lfloor \\frac{m}{3125} \\right\\rfloor + \\cdots$\nNote that if $m$ is a multiple of $5$ $f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)$\nSince $f(m) \\le \\frac{m}{5} + \\frac{m}{25} + \\frac{m}{125} + \\cdots = \\frac{m}{4}$ , a value of $m$ such that $f(m) = 1991$ is greater than $7964$ . Testing values greater than this yields $f(7975)=1991$\nThere are $\\frac{7975}{5} = 1595$ distinct positive integers, $f(m)$ , less than $1992$ . Thus, there are $1991-1595 = \\boxed{396}$ positive integers less than $1992$ that are not factorial tails.", "After testing various values of $m$ in $f(m)$ of solution 1 to determine $m$ for which $f(m) = 1992$ , we find that $m \\in \\{7980, 7981, 7982, 7983, 7984\\}$ . WLOG, we select $7980$ . Furthermore, note that every time $k$ reaches a multiple of $25$ $k!$ will gain two or more additional factors of $5$ and will thus skip one or more numbers.\nWith this logic, we realize that the desired quantity is simply $\\left \\lfloor \\frac{7980}{25} \\right \\rfloor + \\left \\lfloor \\frac{7980}{125} \\right \\rfloor \\cdots$ , where the first term accounts for every time $1$ number is skipped, the second term accounts for each time $2$ numbers are skipped, and so on. Evaluating this gives us $319 + 63 + 12 + 2 = \\boxed{396}$ . - Spacesam(edited by srisainandan6)" ]

[ "We use the Principle of Inclusion-Exclusion (PIE).\nIf we join the adjacent vertices of the regular $n$ -star, we get a regular $n$ -gon. We number the vertices of this $n$ -gon in a counterclockwise direction: $0, 1, 2, 3, \\ldots, n-1.$\nA regular $n$ -star will be formed if we choose a vertex number $m$ , where $0 \\le m \\le n-1$ , and then form the line segments by joining the following pairs of vertex numbers: $(0 \\mod{n}, m \\mod{n}),$ $(m \\mod{n}, 2m \\mod{n}),$ $(2m \\mod{n}, 3m \\mod{n}),$ $\\cdots,$ $((n-2)m \\mod{n}, (n-1)m \\mod{n}),$ $((n-1)m \\mod{n}, 0 \\mod{n}).$\nIf $\\gcd(m,n) > 1$ , then the star degenerates into a regular $\\frac{n}{\\gcd(m,n)}$ -gon or a (2-vertex) line segment if $\\frac{n}{\\gcd(m,n)}= 2$ . Therefore, we need to find all $m$ such that $\\gcd(m,n) = 1$\nNote that $n = 1000 = 2^{3}5^{3}.$\nLet $S = \\{1,2,3,\\ldots, 1000\\}$ , and $A_{i}= \\{i \\in S \\mid i\\, \\textrm{ divides }\\,1000\\}$ . The number of $m$ 's that are not relatively prime to $1000$ is: $\\mid A_{2}\\cup A_{5}\\mid = \\mid A_{2}\\mid+\\mid A_{5}\\mid-\\mid A_{2}\\cap A_{5}\\mid$ $= \\left\\lfloor \\frac{1000}{2}\\right\\rfloor+\\left\\lfloor \\frac{1000}{5}\\right\\rfloor-\\left\\lfloor \\frac{1000}{2 \\cdot 5}\\right\\rfloor$ $= 500+200-100 = 600.$\nVertex numbers $1$ and $n-1=999$ must be excluded as values for $m$ since otherwise a regular n-gon, instead of an n-star, is formed.\nThe cases of a 1st line segment of (0, m) and (0, n-m) give the same star. Therefore we should halve the count to get non-similar stars.\nTherefore, the number of non-similar 1000-pointed stars is $\\frac{1000-600-2}{2}= \\boxed{199}.$" ]

[ "A pattern starts to emerge as the function is continued. The repeating pattern is $0,1,1,2,0,2,2,1\\ldots$ The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating pattern, we just need to find the sum of the numbers in the sequence, which is $\\boxed{9}$" ]

[ "First it will be shown by induction that the zeros of $f_n$ are the integers $a, {a+2,} {a+4,} \\dots, {a + n(n-1)}$ , where $a = n - \\frac{n(n-1)}2.$\nThis is certainly true for $n=1$ . Suppose that it is true for $n = m-1 \\ge 1$ , and note that the zeros of $f_m$ are the solutions of $|x - m| = k$ , where $k$ is a nonnegative zero of $f_{m-1}$ . Because the zeros of $f_{m-1}$ form an arithmetic sequence with common difference $2,$ so do the zeros of $f_m$ . The greatest zero of $f_{m-1}$ is \\[m-1+\\frac{(m-1)(m-2)}2 =\\frac{m(m-1)}2,\\] so the greatest zero of $f_m$ is $m+\\frac{m(m-1)}2$ and the least is $m-\\frac{m(m-1)}2$\nIt follows that the number of zeros of $f_n$ is $\\frac{n(n-1)}2+1=\\frac{n^2-n+2}2$ , and their average value is $n$ . The sum of the zeros of $f_n$ is \\[\\frac{n^3-n^2+2n}2.\\] Let $S(n)=n^3-n^2+2n$ , so the sum of the zeros exceeds $500000$ if and only if $S(n) > 1000000 = 100^3\\!.$ Because $S(n)$ is increasing for $n > 2$ , the values $S(100) = 1000000 - 10000 + 200 = 990200$ and $S(101)=1030301 - 10201 + 202 = 1020302$ show that the requested value of $n$ is $\\boxed{101}$", "Starting from $f_1(x)=|x-1|$ , we can track the solutions, the number of solutions, and their sum.\n\\[\\begin{array}{c|c|c|c} n&Solutions&number&sum\\\\ 1&1&1&1\\\\ 2&1,3&2&4\\\\ 3&0,2,4,6&4&12\\\\ 4&-2,0,2...10&7&28\\\\ 5&-5,-3,-1...15&11&55\\\\ \\end{array}\\]\nIt is clear that the solutions form arithmetic sequences with a difference of 2, and the negative solutions cancel out all but $n$ of the $1+\\frac{n(n-1)}{2}$ solutions. Thus, the sum of the solutions is $n \\cdot [1+\\frac{n(n-1)}{2}]$ , which is a cubic function.\n$n \\cdot [1+\\frac{n(n-1)}{2}]>500,000$\nMultiplying both sides by $2$\n$n \\cdot [2+n(n-1)]>1,000,000$\n1 million is $10^6=100^3$ , so the solution should be close to $100$\n100 is slightly too small, so $\\boxed{101}$ works." ]

[ "Let $t_n=\\frac{s_n}{5}$ . Then, we have $s_n=\\frac{s_{n-1}+1}{s_{n-2}}$ where $s_1 = 100$ and $s_2 = 105$ . By substitution, we find $s_3 = \\frac{53}{50}$ $s_4=\\frac{103}{105\\cdot50}$ $s_5=\\frac{101}{105}$ $s_6=100$ , and $s_7=105$ . So $s_n$ has a period of $5$ . Thus $s_{2020}=s_5=\\frac{101}{105}$ . So, $\\frac{101}{105\\cdot 5}\\implies 101+525=\\boxed{626}$ .\n~mn28407", "Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given $t_1 = 20$ and $t_2 = 21$ , so now we are able to determine the numerical value of $t_3$ using this information: \\[t_3 = \\frac{5t_{3-1}+1}{25t_{3-2}} = \\frac{5t_{2}+1}{25t_{1}} = \\frac{5(21) + 1}{25(20)} = \\frac{105 + 1}{500}t_3 = \\frac{106}{500} = \\frac{53}{250}\\] \\[t_4 = \\frac{5t_{4-1}+1}{25t_{4-2}} = \\frac{5t_{3}+1}{25t_{2}} = \\frac{5(\\frac{53}{250}) + 1}{25(21)} = \\frac{\\frac{53}{50} + 1}{525} = \\frac{\\frac{103}{50}}{525} = \\frac{103}{26250}\\] \\[t_5 = \\frac{5t_{5-1}+1}{25t_{5-2}} = \\frac{5t_{4}+1}{25t_{3}} = \\frac{5(\\frac{103}{26250}) + 1}{25(\\frac{53}{250})} = \\frac{\\frac{103}{5250} + 1}{\\frac{53}{10}} = \\frac{\\frac{5353}{5250}}{\\frac{53}{10}} = \\frac{101}{525}\\] \\[t_6 = \\frac{5t_{6-1}+1}{25t_{6-2}} = \\frac{5t_{5}+1}{25t_{4}} = \\frac{5(\\frac{101}{525}) + 1}{25(\\frac{103}{26250})} = \\frac{\\frac{101}{105} + 1}{\\frac{103}{1050}} = \\frac{\\frac{206}{105}}{\\frac{103}{1050}} \\implies t_6 = 20\\]\nAlas, we have figured this sequence is period 5! But since $2020 \\equiv 5 \\pmod 5$ , we can state that $t_{2020} = t_5 = \\frac{101}{525}$ . According to the original problem statement, our answer is $\\boxed{626}$ . ~ nikenissan" ]

[ "We first prove that $x_n > 4$ for all $n \\ge 0$ , by induction. Observe that \\[x_{n+1} - 4 = \\frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \\frac{(x_n - 4)(x_n+5)}{x_n+6}.\\] so (since $x_n$ is clearly positive for all $n$ , from the initial definition), $x_{n+1} > 4$ if and only if $x_{n} > 4$\nWe similarly prove that $x_n$ is decreasing: \\[x_{n+1} - x_n = \\frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \\frac{4-x_n}{x_n+6} < 0.\\]\nNow we need to estimate the value of $x_{n+1}-4$ , which we can do using the rearranged equation: \\[x_{n+1} - 4 = (x_n-4)\\cdot\\frac{x_n + 5}{x_n+6}.\\] Since $x_n$ is decreasing, $\\frac{x_n + 5}{x_n+6}$ is also decreasing, so we have \\[\\frac{9}{10} < \\frac{x_n + 5}{x_n+6} \\le \\frac{10}{11}\\] and \\[\\frac{9}{10}(x_n-4) < x_{n+1} - 4 \\le \\frac{10}{11}(x_n-4).\\]\nThis becomes \\[\\left(\\frac{9}{10}\\right)^n = \\left(\\frac{9}{10}\\right)^n \\left(x_0-4\\right) < x_{n} - 4 \\le \\left(\\frac{10}{11}\\right)^n \\left(x_0-4\\right) = \\left(\\frac{10}{11}\\right)^n.\\] The problem thus reduces to finding the least value of $n$ such that \\[\\left(\\frac{9}{10}\\right)^n < x_{n} - 4 \\le \\frac{1}{2^{20}} \\text{ and } \\left(\\frac{10}{11}\\right)^{n-1} > x_{n-1} - 4 > \\frac{1}{2^{20}}.\\]\nTaking logarithms, we get $n \\ln \\frac{9}{10} < -20 \\ln 2$ and $(n-1)\\ln \\frac{10}{11} > -20 \\ln 2$ , i.e.\n\\[n > \\frac{20\\ln 2}{\\ln\\frac{10}{9}} \\text{ and } n-1 < \\frac{20\\ln 2}{\\ln\\frac{11}{10}} .\\]\nAs approximations, we can use $\\ln\\frac{10}{9} \\approx \\frac{1}{9}$ $\\ln\\frac{11}{10} \\approx \\frac{1}{10}$ , and $\\ln 2\\approx 0.7$ . These approximations allow us to estimate \\[126 < n < 141,\\] which gives $\\boxed{81,242}$", "The condition where $x_m\\leq 4+\\frac{1}{2^{20}}$ gives the motivation to make a substitution to change the equilibrium from $4$ to $0$ . We can substitute $x_n = y_n + 4$ to achieve that. Now, we need to find the smallest value of $m$ such that $y_m\\leq \\frac{1}{2^{20}}$ given that $y_0 = 1$\nFactoring the recursion $x_{n+1} = \\frac{x_n^2 + 5x_n+4}{x_n + 6}$ , we get:\n$x_{n+1}=\\dfrac{(x_n + 4)(x_n + 1)}{x_n + 6} \\Rightarrow y_{n+1}+4=\\dfrac{(y_n+8)(y_n+5)}{y_n+10}$\n$y_{n+1}+4=\\dfrac{y_n^2+13y_n+40}{y_n+10} = \\dfrac{y_n^2+9y_n +(4y_n+40)}{y_n+10}$\n$y_{n+1}+4=\\dfrac{y_n^2+9y_n}{y_n+10} + 4$\n$y_{n+1}=\\dfrac{y_n^2+9y_n}{y_n+10}$\nUsing wishful thinking, we can simplify the recursion as follows:\n$y_{n+1} = \\frac{y_n^2 + 9y_n + y_n - y_n}{y_n + 10}$\n$y_{n+1} = \\frac{y_n(y_n + 10) - y_n}{y_n + 10}$\n$y_{n+1} = y_n - \\frac{y_n}{y_n + 10}$\n$y_{n+1} = y_n\\left(1 - \\frac{1}{y_n + 10}\\right)$\nThe recursion looks like a geometric sequence with the ratio changing slightly after each term. Notice from the recursion that the $y_n$ sequence is strictly decreasing, so all the terms after $y_0$ will be less than 1. Also, notice that all the terms in sequence will be positive. Both of these can be proven by induction.\nWith both of those observations in mind, $\\frac{9}{10} < 1 - \\frac{1}{y_n + 10} \\leq \\frac{10}{11}$ . Combining this with the fact that the recursion resembles a geometric sequence, we conclude that $\\left(\\frac{9}{10}\\right)^n < y_n \\leq \\left(\\frac{10}{11}\\right)^n.$\n$\\frac{9}{10}$ is approximately equal to $\\frac{10}{11}$ and the ranges that the answer choices give us are generous, so we should use either $\\frac{9}{10}$ or $\\frac{10}{11}$ to find a rough estimate for $m$\nSince $\\dfrac{1}{2}=0.5$ , that means $\\frac{1}{\\sqrt{2}}=2^{-\\frac{1}{2}} \\approx 0.7$ . Additionally, $\\left(\\frac{9}{10}\\right)^3=0.729$\nTherefore, we can estimate that $2^{-\\frac{1}{2}} < y_3$\nRaising both sides to the 40th power, we get $2^{-20} < (y_3)^{40}$\nBut $y_3 = (y_0)^3$ , so $2^{-20} < (y_0)^{120}$ and therefore, $2^{-20} < y_{120}$\nThis tells us that $m$ is somewhere around 120, so our answer is $\\boxed{81,242}$", "Since the choices are rather wide ranges, we can use approximation to make it easier. Notice that \\[x_{n+1} - x_n = \\frac{4-x_n}{x_n+6}\\] And $x_0 =5$ , we know that $x_n$ is a declining sequence, and as it get close to 4 its decline will slow, never falling below 4. So we'll use 4 to approximate $x_n$ in the denominator so that we have a solvable difference equation: \\[x_{n+1} - x_n = \\frac{4-x_n}{10}\\] \\[x_{n+1} = \\frac{9}{10}x_n + \\frac{2}{5}\\] Solve it with $x_0 = 5$ , we have \\[x_n = 4 + (\\frac{9}{10})^n\\] Now we wish to find $n$ so that \\[(\\frac{9}{10})^n \\approx \\frac{1}{2^{20}}\\] \\[n \\approx \\frac{\\log{2^{20}}}{\\log{10}-\\log9} \\approx \\frac{20*0.3}{0.05} = 120\\] Since 120 is safely within the range of [81,242], we have the answer. $\\boxed{81,242}$", "We start by simplifying the recursion: $x_{n+1} = x_n + \\frac{4-x_n}{x_n+6}$ .\nWhile $x_n > 4$ $x_n$ is a decreasing sequence. Let $x_n = 4 + f_n$ . Then \\[x_{n} - x_{n+1} = \\frac{x_n-4}{x_n+6} = \\frac{f_n}{10 + f_n} \\approx \\frac{f_n}{10}.\\] Now notice that we want to find $m$ , such that $x_m \\leq 4 + \\frac{1}{2^{20}}$ , and $x_0 = 4 + \\frac{1}{2^0}$ . \nWe are going to find an approximate number of steps to go from $4 + \\frac{1}{2^i}$ to $4 + \\frac{1}{2^{i+1}}$ . \nIf $x_j = 4 + \\frac{1}{2^i}$ , then $f_j = \\frac{1}{2^i}$ and if $x_k = 4 + \\frac{1}{2^{i+1}}$ , then $f_k = \\frac{1}{2^{i+1}}$ . Then \\[x_j - x_k = \\frac{1}{2^{i+1}} \\approx \\frac{f_j}{10} + \\frac{f_{j+1}}{10} + ... + \\frac{f_{k-1}}{10}.\\] Furthermore, \\[\\frac{1}{2^i} = f_j > f_{j+1} > f_{j+2} > ... > f_{k-1} > f_k = \\frac{1}{2^{i+1}}.\\] Therefore, \\[(k-j) \\cdot \\frac{\\frac{1}{2^i}}{10} > \\frac{f_j}{10} + \\frac{f_{j+1}}{10} + ... + \\frac{f_{k-1}}{10} > (k-j) \\cdot \\frac{\\frac{1}{2^{i+1}}}{10},\\] so $5 < k-j < 10$ . Therefore, to go from $f_0 = \\frac{1}{2^0}$ to $f_m = \\frac{1}{2^{20}}$ , we will need to do this 20 times, which means $100 < m - 0 < 200$ , which is within the range of [81,242], so we have the answer. $\\boxed{81,242}$" ]

[ "We first prove that $x_n > 4$ for all $n \\ge 0$ , by induction. Observe that \\[x_{n+1} - 4 = \\frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \\frac{(x_n - 4)(x_n+5)}{x_n+6}.\\] so (since $x_n$ is clearly positive for all $n$ , from the initial definition), $x_{n+1} > 4$ if and only if $x_{n} > 4$\nWe similarly prove that $x_n$ is decreasing: \\[x_{n+1} - x_n = \\frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \\frac{4-x_n}{x_n+6} < 0.\\]\nNow we need to estimate the value of $x_{n+1}-4$ , which we can do using the rearranged equation: \\[x_{n+1} - 4 = (x_n-4)\\cdot\\frac{x_n + 5}{x_n+6}.\\] Since $x_n$ is decreasing, $\\frac{x_n + 5}{x_n+6}$ is also decreasing, so we have \\[\\frac{9}{10} < \\frac{x_n + 5}{x_n+6} \\le \\frac{10}{11}\\] and \\[\\frac{9}{10}(x_n-4) < x_{n+1} - 4 \\le \\frac{10}{11}(x_n-4).\\]\nThis becomes \\[\\left(\\frac{9}{10}\\right)^n = \\left(\\frac{9}{10}\\right)^n \\left(x_0-4\\right) < x_{n} - 4 \\le \\left(\\frac{10}{11}\\right)^n \\left(x_0-4\\right) = \\left(\\frac{10}{11}\\right)^n.\\] The problem thus reduces to finding the least value of $n$ such that \\[\\left(\\frac{9}{10}\\right)^n < x_{n} - 4 \\le \\frac{1}{2^{20}} \\text{ and } \\left(\\frac{10}{11}\\right)^{n-1} > x_{n-1} - 4 > \\frac{1}{2^{20}}.\\]\nTaking logarithms, we get $n \\ln \\frac{9}{10} < -20 \\ln 2$ and $(n-1)\\ln \\frac{10}{11} > -20 \\ln 2$ , i.e.\n\\[n > \\frac{20\\ln 2}{\\ln\\frac{10}{9}} \\text{ and } n-1 < \\frac{20\\ln 2}{\\ln\\frac{11}{10}} .\\]\nAs approximations, we can use $\\ln\\frac{10}{9} \\approx \\frac{1}{9}$ $\\ln\\frac{11}{10} \\approx \\frac{1}{10}$ , and $\\ln 2\\approx 0.7$ . These approximations allow us to estimate \\[126 < n < 141,\\] which gives $\\boxed{81,242}$", "The condition where $x_m\\leq 4+\\frac{1}{2^{20}}$ gives the motivation to make a substitution to change the equilibrium from $4$ to $0$ . We can substitute $x_n = y_n + 4$ to achieve that. Now, we need to find the smallest value of $m$ such that $y_m\\leq \\frac{1}{2^{20}}$ given that $y_0 = 1$\nFactoring the recursion $x_{n+1} = \\frac{x_n^2 + 5x_n+4}{x_n + 6}$ , we get:\n$x_{n+1}=\\dfrac{(x_n + 4)(x_n + 1)}{x_n + 6} \\Rightarrow y_{n+1}+4=\\dfrac{(y_n+8)(y_n+5)}{y_n+10}$\n$y_{n+1}+4=\\dfrac{y_n^2+13y_n+40}{y_n+10} = \\dfrac{y_n^2+9y_n +(4y_n+40)}{y_n+10}$\n$y_{n+1}+4=\\dfrac{y_n^2+9y_n}{y_n+10} + 4$\n$y_{n+1}=\\dfrac{y_n^2+9y_n}{y_n+10}$\nUsing wishful thinking, we can simplify the recursion as follows:\n$y_{n+1} = \\frac{y_n^2 + 9y_n + y_n - y_n}{y_n + 10}$\n$y_{n+1} = \\frac{y_n(y_n + 10) - y_n}{y_n + 10}$\n$y_{n+1} = y_n - \\frac{y_n}{y_n + 10}$\n$y_{n+1} = y_n\\left(1 - \\frac{1}{y_n + 10}\\right)$\nThe recursion looks like a geometric sequence with the ratio changing slightly after each term. Notice from the recursion that the $y_n$ sequence is strictly decreasing, so all the terms after $y_0$ will be less than 1. Also, notice that all the terms in sequence will be positive. Both of these can be proven by induction.\nWith both of those observations in mind, $\\frac{9}{10} < 1 - \\frac{1}{y_n + 10} \\leq \\frac{10}{11}$ . Combining this with the fact that the recursion resembles a geometric sequence, we conclude that $\\left(\\frac{9}{10}\\right)^n < y_n \\leq \\left(\\frac{10}{11}\\right)^n.$\n$\\frac{9}{10}$ is approximately equal to $\\frac{10}{11}$ and the ranges that the answer choices give us are generous, so we should use either $\\frac{9}{10}$ or $\\frac{10}{11}$ to find a rough estimate for $m$\nSince $\\dfrac{1}{2}=0.5$ , that means $\\frac{1}{\\sqrt{2}}=2^{-\\frac{1}{2}} \\approx 0.7$ . Additionally, $\\left(\\frac{9}{10}\\right)^3=0.729$\nTherefore, we can estimate that $2^{-\\frac{1}{2}} < y_3$\nRaising both sides to the 40th power, we get $2^{-20} < (y_3)^{40}$\nBut $y_3 = (y_0)^3$ , so $2^{-20} < (y_0)^{120}$ and therefore, $2^{-20} < y_{120}$\nThis tells us that $m$ is somewhere around 120, so our answer is $\\boxed{81,242}$", "Since the choices are rather wide ranges, we can use approximation to make it easier. Notice that \\[x_{n+1} - x_n = \\frac{4-x_n}{x_n+6}\\] And $x_0 =5$ , we know that $x_n$ is a declining sequence, and as it get close to 4 its decline will slow, never falling below 4. So we'll use 4 to approximate $x_n$ in the denominator so that we have a solvable difference equation: \\[x_{n+1} - x_n = \\frac{4-x_n}{10}\\] \\[x_{n+1} = \\frac{9}{10}x_n + \\frac{2}{5}\\] Solve it with $x_0 = 5$ , we have \\[x_n = 4 + (\\frac{9}{10})^n\\] Now we wish to find $n$ so that \\[(\\frac{9}{10})^n \\approx \\frac{1}{2^{20}}\\] \\[n \\approx \\frac{\\log{2^{20}}}{\\log{10}-\\log9} \\approx \\frac{20*0.3}{0.05} = 120\\] Since 120 is safely within the range of [81,242], we have the answer. $\\boxed{81,242}$", "We start by simplifying the recursion: $x_{n+1} = x_n + \\frac{4-x_n}{x_n+6}$ .\nWhile $x_n > 4$ $x_n$ is a decreasing sequence. Let $x_n = 4 + f_n$ . Then \\[x_{n} - x_{n+1} = \\frac{x_n-4}{x_n+6} = \\frac{f_n}{10 + f_n} \\approx \\frac{f_n}{10}.\\] Now notice that we want to find $m$ , such that $x_m \\leq 4 + \\frac{1}{2^{20}}$ , and $x_0 = 4 + \\frac{1}{2^0}$ . \nWe are going to find an approximate number of steps to go from $4 + \\frac{1}{2^i}$ to $4 + \\frac{1}{2^{i+1}}$ . \nIf $x_j = 4 + \\frac{1}{2^i}$ , then $f_j = \\frac{1}{2^i}$ and if $x_k = 4 + \\frac{1}{2^{i+1}}$ , then $f_k = \\frac{1}{2^{i+1}}$ . Then \\[x_j - x_k = \\frac{1}{2^{i+1}} \\approx \\frac{f_j}{10} + \\frac{f_{j+1}}{10} + ... + \\frac{f_{k-1}}{10}.\\] Furthermore, \\[\\frac{1}{2^i} = f_j > f_{j+1} > f_{j+2} > ... > f_{k-1} > f_k = \\frac{1}{2^{i+1}}.\\] Therefore, \\[(k-j) \\cdot \\frac{\\frac{1}{2^i}}{10} > \\frac{f_j}{10} + \\frac{f_{j+1}}{10} + ... + \\frac{f_{k-1}}{10} > (k-j) \\cdot \\frac{\\frac{1}{2^{i+1}}}{10},\\] so $5 < k-j < 10$ . Therefore, to go from $f_0 = \\frac{1}{2^0}$ to $f_m = \\frac{1}{2^{20}}$ , we will need to do this 20 times, which means $100 < m - 0 < 200$ , which is within the range of [81,242], so we have the answer. $\\boxed{81,242}$" ]

[ "We first start out when the value of $a=1$\nDoing casework, we discover that $d=5,6,7,8,9,10$ . We quickly find a pattern.\nNow, doing this for the rest of the values of $a$ and $d$ , we see that the answer is simply:\n$(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)$ $+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=\\boxed{080}$", "Notice that if $a+d>b+c$ , then $(11-a)+(11-d)<(11-b)+(11-c)$ , so there is a bijection between the number of ordered quadruples with $a+d>b+c$ and the number of ordered quadruples with $a+d<b+c$\nQuick counting gives that the number of ordered quadruples with $a+d=b+c$ is 50. To count this, consider our numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ . Notice that if, for example, $a+d=b+c=8$ , that the average of $a,d$ and $b,c$ must both be $4$ . In this way, there is a symmetry for this case, centered at $4$ . If instead, say, $a+d=b+c=7$ , an odd number, then there is symmetry with $(a,d);(b,c)$ about $3.5$ . Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at $2.5,3,8,8.5$ , there is $1$ case for each and so on, until centered at $5.5$ , there are $10$ possible cases. Adding these all, we have $2(1+3+6)+10=50$\nThus the answer is $\\frac{\\binom{10}{4}-50}{2} = \\boxed{080}.$", "Think about a,b,c,and d as distinct objects, that we must place in 4 of 10 spaces. However, in only 1 of 24 of these combinations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is $(10*9*8*7/24)=210$\nNext, intuitively, the number of quadruples where $a+d>b+c$ is equal to the number of quadruples where $a+d<b+c$ . So we need to find the number of quadruples where the two quantities are equal. To do this, all we have to do is consider the cases when $a-d$ ranges from 3 to 9. It would seem natural that a range of 3 would produce 1 option, and a range of 4 would produce 2 options. However, since b and c cannot be equal, a range of 3 or 4 produces 1 option each, a range of 5 or 6 produces 2 options each, a range of 7 or 8 produces 3 options each, and a range of 9 will produce 4 options. In addition, a range of n has 10-n options for combinations of a and d. Multiplying the number of combinations of a and d by the corresponding number of options for b and c gives us 50 total quadruplets where $a+d=b+c$\nSo the answer will be $\\frac{210-50}{2} = \\boxed{080}.$", "Let $b = a+x$ and $c = a+x+y$ and $d = a+x+y+z$ for positive integers $x,y,z.$ In order to satisfy the other condition we need $z > x$ so we let $z = x+k.$ Now the only other condition we need to satisfy so $a+2x+y+k \\le 10.$ This condition can be transformed into $a+2x+y+k+b = 11$ for positive $a,x,y,k,b.$ Now we use generating functions to finish. We find the generating function of the whole expression is $(x + x^2 + \\cdots)^4 \\cdot (x^2+x^4 + \\cdots)$ and we are looking for the $x^{11}$ coefficient. This simplifies to finding the $x^5$ coefficient of $(1+x+\\cdots)^4 \\cdot (1+x^2+\\cdots) =\\frac{1}{(1-x)^4} \\cdot\\frac{1}{1+x}.$ Now this expression simplifies to \\[\\left(\\binom{4}{4}+\\binom{5}{4} + \\cdots +\\binom{4+k}{4}x^k\\right)(1-x+x^2-x^3 \\cdots).\\] The $x^5$ coefficient ends up to be $\\binom{9}{4} -\\binom{8}{4} +\\binom{7}{4} -\\binom{6}{4} +\\binom{5}{4} -\\binom{4}{4} = 126 - 70 + 35 - 15 + 5 - 1 = \\boxed{080}.$", "First, let $a=1$ and $d=10$ . If $b=2$ , then $c$ can be from $3$ to $8$ . If $b=3$ , then $4$ to $7$ . If $b=4$ , then $c$ is between $5$ and $6$ . We find a pattern that whenever $b$ increases by $1$ , when $a$ and $d$ are stationary, then the possible values of $c$ decrease by 2, unless it gets to zero or negative, in which case that case ends. Counting up, we have $6+4+2=12$ different possibilities when $a=1$ and $d=10$ . For $a=1$ and $d=9$ $b=2$ , then $c$ can be from $3$ to $7$ . If $b=3$ , then $c$ can be from $4$ to $6$ , and so on. Notice that the possible values for each case of $b$ gets one less than if $d$ were one greater, unless that number is zero, in which it stays zero. We then use this pattern to find all the values: $12+9+6+4+2+1+9+6+4+2+1+6+4+2+1+4+2+1+2+1+1 \\Rightarrow \\\\ 12\\cdot1+9\\cdot2+6\\cdot3+4\\cdot4+2\\cdot5+1\\cdot6= \\\\ 12+18+18+16+10+6=\\boxed{080}.$", "Rearranging the equation obtains $b-a<d-c$ . Let $a-0=e_0$ $b-a=e_3+e_1$ $c-b=e_2$ $d-c=e_3$ $11-d=e_4$ . Add up all of these newly defined equations to obtain $e_0+e_1+e_2+2e_3+e_4=11$ . Note that since all $e_n$ were defined to be $e_n\\ge1$ , to form our stars and bars argument we can let $d_n+1=e_n$ for all $n$ . Then we obtain $d_0+d_1+d_2+2d_3+d_4=5$ where $d_n$ is nonnegative. Now, we can move the $2d_3$ term to the other side and perform casework.\nIf $2d_3=0$ : 5 objects for 4 variables -> $\\binom{8}{3}$\nIf $2d_3=2$ : 3 objects for 4 variables -> $\\binom{6}{3}$\nIf $2d_3=4$ : 1 object for 4 variables -> $\\binom{4}{3}$\nAdding all of these cases up, we get $56+20+4=\\boxed{080}$ as our requested answer." ]

[ "By definition, the recursion becomes $a_n = \\left(n^{\\frac1{\\log_7(n-1)}}\\right)^{\\log_7(a_{n-1})}=n^{\\frac{\\log_7(a_{n-1})}{\\log_7(n-1)}}$ . By the change of base formula, this reduces to $a_n = n^{\\log_{n-1}(a_{n-1})}$ . Thus, we have $\\log_n(a_n) = \\log_{n-1}(a_{n-1})$ . Thus, for each positive integer $m \\geq 3$ , the value of $\\log_m(a_m)$ must be some constant value $k$\nWe now compute $k$ from $a_3$ . It is given that $a_3 = 3\\,\\heartsuit\\,2 = 3^{\\frac1{\\log_7(2)}}$ , so $k = \\log_3(a_3) = \\log_3\\left(3^{\\frac1{\\log_7(2)}}\\right) = \\frac1{\\log_7(2)} = \\log_2(7)$\nNow, we must have $\\log_{2019}(a_{2019}) = k = \\log_2(7)$ . At this point, we simply switch some bases around. For those who are unfamiliar with logarithms, we can turn the logarithms into fractions which are less intimidating to work with.\n$\\frac{\\log{a_{2019}}}{\\log{2019}} = \\frac{\\log{7}}{\\log{2}}\\implies \\frac{\\log{a_{2019}}}{\\log{7}} = \\frac{\\log{2019}}{\\log{2}}\\implies \\log_7(a_{2019}) =\\log_2(2019)$\nWe conclude that $\\log_7(a_{2019}) = \\log_2(2019) \\approx \\boxed{11}$", "Using the recursive definition, $a_4 = (4 \\, \\heartsuit \\, 3) \\, \\diamondsuit\\, (3 \\, \\heartsuit\\, 2)$ or $a_4 = (4^{m})^{k}$ where $m = \\frac{1}{\\log_{7}(3)}$ and $k = \\log_{7}(3^{\\frac{1}{\\log_{7}(2)}})$ . Using logarithm rules, we can remove the exponent of the 3 so that $k = \\frac{\\log_{7}(3)}{\\log_{7}(2)}$ . Therefore, $a_4 = 4^{\\frac{1}{\\log_{7}(2)}}$ , which is $4 \\, \\heartsuit \\, 2$\nWe claim that $a_n = n \\, \\heartsuit \\, 2$ for all $n \\geq 3$ . We can prove this through induction.\nClearly, the base case where $n = 3$ holds.\n$a_n = (n\\, \\heartsuit\\, (n-1)) \\,\\diamondsuit\\, ((n-1) \\, \\heartsuit \\, 2)$\nThis can be simplified as $a_n = (n^{\\log_{n-1}(7)}) \\, \\diamondsuit \\, ((n-1)^{\\log_{2}(7)})$\nApplying the diamond operation, we can simplify $a_n = n^h$ where $h = \\log_{n-1}(7) \\cdot \\log_{7}(n-1)^{\\log_{2}(7)}$ . By using logarithm rules to remove the exponent of $\\log_{7}(n-1)$ and after cancelling, $h = \\frac{1}{\\log_{7}(2)}$\nTherefore, $a_n = n^{\\frac{1}{\\log_{7}(2)}} = n \\, \\heartsuit \\, 2$ for all $n \\geq 3$ , completing the induction.\nWe have $a_{2019} = 2019^{\\log_{2}(7)}$ . Taking $\\log_{2019}$ of both sides gives us ${\\log_{2019}(a_{2019})} = {\\log_{2}(7)}$ . Then, by changing to base $7$ and after cancellation, we arrive at ${\\log_{7}(a_{2019})} = {\\log_{2}(2019)}$ . Because $2^{11} = 2048$ and $2^{10} = 1024$ , our answer is $\\boxed{11}$", "We are given that \\[a_n=(n\\, \\heartsuit\\, (n-1)) \\,\\diamondsuit\\, a_{n-1}\\] \\[a_n=(n^{\\frac{1}{\\log_7(n-1)}})^{\\log_7(a_{n-1})}\\] Since we are asked to find $\\log_7(a_{2019})$ , we directly apply \\[\\log_7(a_n)=\\log_7(n^{\\frac{1}{\\log_7(n-1)}})^{\\log_7(a_{n-1})}\\] Using the property that $\\log_ab^c=c\\log_ab$ \\[\\log_7(a_n)=(\\log_7a_{n-1})(\\log_7(n^{\\frac{1}{\\log_7(n-1)}}))\\] Now using the property that $\\frac{1}{\\log_ab}=\\log_ba$ \\[\\log_7(a_n)=(\\log_7a_{n-1})(\\log_7n^{\\log_{n-1}7})\\] Once again applying the first property yields \\[\\log_7(a_n)=(\\log_7a_{n-1})(\\log_{n-1}7)(\\log_7n)\\] Rearranging the expression, \\[\\log_7(a_n)=(\\log_7n)(\\log_{n-1}7)(\\log_7a_{n-1})\\]\nNow expressing $\\log_7a_{n-1}$ in a similar expression as $\\log_7a_n$\n\\[\\log_7(a_n)=(\\log_7n)(\\log_{n-1}7)(\\log_7n-1)(\\log_{n-2}7)(\\log_7a_{n-2})\\] \\[\\log_7(a_n)=(\\log_7n)(\\log_{n-1}7)(\\log_7n-1)(\\log_{n-2}7)(\\log_7n-2)(\\log_{n-3}7)...(\\log_74)(\\log_37)(\\log_7a_3)\\]\nBecause of the fact that $(\\log_ab)(\\log_ba)=1$ , we can cancel out the terms to get\n\\[\\log_7(a_n)=(\\log_7n)(\\log_37)(\\log_7a_3)\\] \\[\\log_7(a_n)=(\\log_7n)(\\log_37)(\\log_7(3^{\\frac{1}{\\log_72}}))\\] \\[\\log_7(a_n)=(\\log_7n)(\\log_37)(\\log_27)(\\log_73)\\] \\[\\log_7(a_n)=(\\log_27)(\\log_7n)\\]\nUsing the Chain Rule for Logarithm, $(\\log_ab)(\\log_bc)=\\log_ac$ , yields\n\\[\\log_7(a_n)=(\\log_2n)\\] Finally, substituting in $n=2019$ , we have \\[\\log_7(a_{2019})=(\\log_22019)\\] \\[\\log_7(a_{2019})\\approx11\\boxed{11}\\]" ]

[ "Substitute and simplify. \\[(3+5)5 - (5+3)3 = (3+5)2 = 8\\cdot2 = \\boxed{16}\\]", "Note that $(a \\star b) - (b \\star a) = (a+b)b - (b+a)a= (a+b)(b-a)= b^2 - a^2$ . We can substitute $a=3$ and $b=5$ to get $5^2 - 3^2 = \\boxed{16}$" ]

[ "If $n = 1$ $a_n = \\sin(1) > 0$ . Then if $n$ satisfies $a_n < 0$ $n \\ge 2$ , and \\[a_n = \\sum_{k=1}^n \\sin(k) = \\cfrac{1}{\\sin{1}} \\sum_{k=1}^n\\sin(1)\\sin(k) = \\cfrac{1}{2\\sin{1}} \\sum_{k=1}^n\\cos(k - 1) - \\cos(k + 1) = \\cfrac{1}{2\\sin(1)} [\\cos(0) + \\cos(1) - \\cos(n) - \\cos(n + 1)].\\] Since $2\\sin 1$ is positive, it does not affect the sign of $a_n$ . Let $b_n = \\cos(0) + \\cos(1) - \\cos(n) - \\cos(n + 1)$ . Now since $\\cos(0) + \\cos(1) = 2\\cos\\left(\\cfrac{1}{2}\\right)\\cos\\left(\\cfrac{1}{2}\\right)$ and $\\cos(n) + \\cos(n + 1) = 2\\cos\\left(n + \\cfrac{1}{2}\\right)\\cos\\left(\\cfrac{1}{2}\\right)$ $b_n$ is negative if and only if $\\cos\\left(\\cfrac{1}{2}\\right) < \\cos\\left(n + \\cfrac{1}{2}\\right)$ , or when $n \\in [2k\\pi - 1, 2k\\pi]$ . Since $\\pi$ is irrational, there is always only one integer in the range, so there are values of $n$ such that $a_n < 0$ at $2\\pi, 4\\pi, \\cdots$ . Then the hundredth such value will be when $k = 100$ and $n = \\lfloor 200\\pi \\rfloor = \\lfloor 628.318 \\rfloor = \\boxed{628}$", "Notice that $a_n$ is the imaginary part of $\\sum_{k=1}^n e^{ik}$ , by Euler's formula. Using the geometric series formula, we find that this sum is equal to \\[\\frac{e^{i(n+1)}-e^i}{e^i-1} = \\frac{\\cos (n+1) - \\cos 1 + i (\\sin (n+1) - \\sin 1) }{\\cos 1 - 1 + i \\sin 1}\\] We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have \\[\\frac{(\\cos 1 - 1)(\\cos(n+1)-\\cos 1) + (\\sin 1)(\\sin(n+1)-\\sin 1) + i((\\sin(n+1) - \\sin 1)(\\cos 1 - 1) - (\\sin 1)(\\cos(n+1)-\\cos 1))}{\\cos^2 1 - 2 \\cos 1 + 1 + \\sin^2 1}\\] We only need to look at the imaginary part, which is \\[\\frac{(\\sin(n+1) \\cos 1 - \\cos(n+1) \\sin 1) - \\sin 1 \\cos 1 + \\sin 1 - \\sin (n+1) + \\sin 1 \\cos 1}{2-2 \\cos 1} = \\frac{\\sin n - \\sin(n+1) + \\sin 1}{2-2 \\cos 1}\\] Since $\\cos 1 < 1$ $2-2 \\cos 1 > 0$ , so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have $\\sin (n+1) - \\sin n > \\sin 1 \\implies 2 \\cos \\left(n + \\frac{1}{2} \\right) \\sin \\frac{1}{2} > \\sin 1 \\implies \\cos \\left( n + \\frac{1}{2} \\right) > \\frac{\\sin 1}{2 \\sin{\\frac{1}{2}}} = \\cos \\left(\\frac{1}{2} \\right),$ by sum to product. This only holds when $n$ is between $2\\pi k - 1$ and $2\\pi k$ for integer $k$ [continuity proof here], and since this has exactly one integer solution for every such interval, the $100$ th such $n$ is $\\lfloor 200\\pi \\rfloor = \\boxed{628}$", "Similar to solution 2, we set a complex number $z=\\cos 1+i\\sin 1$ . We start from $z$ instead of $1$ because $k$ starts from $1$ : be careful.\nThe sum of $z+z^2+z^3+z^4+z^5\\dots=\\frac{z-z^{n+1}}{1-z}=\\frac{z}{z-1}\\left(z^n-1\\right)$\nWe are trying to make $n$ so that the imaginary part of this expression is negative.\nThe argument of $z$ is $1$ . The argument of $z-1$ , however, is a little more tricky. $z-1$ is on a circle centered on $(-1,0)$ with radius $1$ . The change in angle due to $z$ is $1$ with respect to the center, but the angle that $z-1$ makes with the $y$ -axis is $half$ the change, due to Circle Theorems (this intercepted arc is the argument of $z$ ), because the $y$ - axis is tangent to the circle at the origin. So $\\text{arg}(z-1)=\\frac{\\pi+1}{2}$ . Dividing $z$ by $z-1$ subtracts the latter argument from the former, so the angle of the quotient with the $x$ -axis is $\\frac{1-\\pi}{2}$\nWe want the argument of the whole expression $-\\pi<\\theta<0$ . This translates into $\\frac{-\\pi-1}{2}<\\text{arg}\\left(z^n-1\\right)<\\frac{\\pi-1}{2}$ $z^n-1$ also consists of points on the circle centered at $(-1,0)$ , so we deal with this argument similarly: the argument of $z^n$ is twice the angle $z^n-1$ makes with the $y$ -axis. Since $z^n-1$ is always negative, $\\frac{-3\\pi}{2}<\\text{arg}\\left(z^n-1\\right)<\\frac{-\\pi}{2}$ , and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a $\\frac{\\pi}{2}-\\frac{\\pi-1}{2}=\\frac{-1}{2}$ angle with the $y$ -axis both ways.\nSo the argument of $z^n$ must be in the bound $-1<\\theta<0$ by doubling, namely the last $z^n$ negative before another rotation. Since there is always one $z^n$ in this category per rotation because $\\pi$ is irrational, $n_{100}\\equiv z^{628}$ and the answer is $\\boxed{628}$" ]

[ "Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$ . Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$ , so our answer is $3\\cdot17 + 2\\cdot65 = \\boxed{181}$", "Subtracting the first equation from every one of the other equations yields \\begin{align*} x_2-x_1&=6\\\\ x_3-x_1&=18\\\\ x_4-x_1&=42\\\\ x_5-x_1&=90 \\end{align*} Thus \\begin{align*} 2x_1+x_2+x_3+x_4+x_5&=6\\\\ 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\\\ 6x_1+156&=6\\\\ x_1&=-25 \\end{align*} Using the previous equations, \\[3x_4+2x_5=3(x_1+42)+2(x_1+90)=\\boxed{181}\\]" ]

[ "Since the given expression is a quadratic, the factored form would be $(x-3)(4x+y)$ , where $y$ is a value such that $-12x+yx=-6x$ and $-3(y)=m$ . The only number that fits the first equation is $y=6$ , so $m=-18$ . The only choice that is a multiple of 18 is $\\boxed{36}$" ]

[ "Rewrite the system of equations as \\[\\frac{x^{2}}{t-1}+\\frac{y^{2}}{t-3^{2}}+\\frac{z^{2}}{t-5^{2}}+\\frac{w^{2}}{t-7^{2}}=1.\\] This equation is satisfied when $t \\in \\{4, 16, 36, 64\\}$ . After clearing fractions, for each of the values $t=4,16,36,64$ , we have the equation \\[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\\] where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$ , for $k=1,3,5,7$\nSince the polynomials on each side are equal at $t=4,16,36,64$ , we can express the difference of the two polynomials by a quartic polynomial that has roots at $t=4,16,36,64$ , so \\begin{align} \\tag{\\dag}x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)-F(t) = -(t-4)(t-16)(t-36)(t-64) \\end{align} The leading coefficient of the RHS is $-1$ because the leading coefficient of the LHS is $-1$\nPlug in $t=1^2, 3^2, 5^2, 7^2$ in succession, into $(\\dag)$ . In each case, most terms drop, and we end up with \\begin{align*} x^2=\\frac{3^2\\cdot 5^2\\cdot 7^2}{2^{10}}, \\quad y^2=\\frac{3^3\\cdot 5\\cdot 7\\cdot 11}{2^{10}},\\quad z^2=\\frac{3^2\\cdot 7\\cdot 11\\cdot 13}{2^{10}},\\quad w^2=\\frac{3^2\\cdot 5\\cdot 11\\cdot 13}{2^{10}} \\end{align*} Adding them up we get the sum as $3^2\\cdot 4=\\boxed{036}$", "As in Solution 1, we have \\[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\\] where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$ , for $k=1,3,5,7$\nNow the coefficient of $t^3$ on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the $t^4$ and $t^3$ terms are, so we can eventually apply Vieta's. We can write the long equation as \\[(x^2 + y^2 + z^2 + w^2)t^3 + \\dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \\dots\\] Rearranging gives us \\[t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2)t^3 \\dots = 0.\\] By Vieta's, we know that the sum of the roots of this equation is \\[1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.\\] (recall that the roots of the original and this manipulated form of it had roots $2^2, 4^2, 6^2,$ and $8^2$ ). Thus, \\[x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \\boxed{36}.\\]", "Notice how on each line, we have equations of the form $\\frac{x^2}{a-1^2}+\\frac{y^2}{a-3^2}+\\frac{z^2}{a-5^2}+\\frac{w^2}{a-7^2}=1$ . We can let this be a polynomial, with respect to $a$ . We can say that $w^2$ $x^2$ $y^2$ , and $z^2$ are coefficients, and not variables. So, we can now expand the fractions to get $(a-1)(a-9)(a-25)(a-49)=x^2(a-9)(a-25)(a-49)$ $+ y^2(a-1)(a-25)(a-49)$ $+ z^2(a-1)(a-3)(a-7)$ $+ w^2(a-1)(a-9)(a-25)$\nNow, we have arrived at this huge expression, but what do we do with it?\nWell, we can look at what we want to find - $x^2+y^2+z^2+w^2$ . So, we want the sum of $x^2$ $y^2$ $z^2$ , and $w^2$ . Looking back to our expression, we can note how on the right hand side, the $a^3$ terms add to $x^2+y^2+z^2+w^2$ . Also, on the left hand side, the $a^3$ coefficient is $84$ (which is achievable by Vieta's formulas rather than expanding if you want to save a few seconds). So, moving all the $a^3$ terms to the left hand side, then we have that by Vieta's formulas, the sum of the roots is $-84-x^2-y^2-z^2-w^2=-(2^2+4^2+6^2+8^2)$ . Then, we can solve to find that $x^2+y^2+z^2+w^2=120-84=\\boxed{036}$" ]

[ "Let $P(t)=t^3-at^2+bt-c$ have roots x, y, and z. Then \\[0=P(x)+P(y)+P(z)=3-3a+3b-3c\\] using our system of equations, so $P(1)=0$ . Thus, at least one of x, y, and z is equal to 1; without loss of generality, let $x=1$ . Then we can use the system of equations to find that $y=z=1$ as well, and so $\\boxed{1,1,1}$ is the only solution to the system of equations.", "Let $a=x-1,$ $b=y-1$ and $c=z-1.$ Then \\[a+b+c=0,\\] \\[a^2+b^2+c^2=0,\\] \\[a^3+b^3+c^3=0.\\] We have \\begin{align*} 0&=(a+b+c)^3\\\\ &=(a^3+b^3+c^3)+3a^2(b+c)+3b^3(a+c)+3c^2(a+b)+6abc\\\\ &=0-3a^3-3b^3-3c^3+6abc\\\\ &=6abc. \\end{align*} Then one of $a, b$ and $c$ has to be 0, and easy to prove the other two are also 0. So $\\boxed{1,1,1}$ is the only solution to the system of equations.", "We are going to use Intermediate Algebra Techniques to solve this equation.\nLet's start with the first one: $x+y+z=3$ . This will be referred as the FIRST equation.\nWe are going to use the first equation to relate to the SECOND one ( $x^2+y^2+z^2=3$ ) and the THIRD one ( $x^3+y^3+z^3=3)$\nSquaring this equation: $x^2+y^2+z^2+2xy+2yz+2xz=9$\nSubtracting this equation from the 2nd equation in the problem, we have $2xy+2yz+2xz=6$ , so $xy+xz+yz=3$\nNow we try the same idea with the cubed terms. Cube the first equation:\n$x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3yz^2+3zx^2+3zy^2=27$ . Plug in $x^3+y^3+z^3=3$ and factor partially:\n$3+3(x^2(y+z)+y^2(x+z)+z^2(x+y))+6xyz=27$\nNow here is the key step. Note that $z=3-x-y, y=3-x-z, x=3-y-z$ . So we are going to substitute $y+z, x+z, x+y$ for each of the expressions and we get: $-x^3+3x^2-y^3+3y^2-z^3+3z^2=8-2xyz$ (I rearranged it a bit).\nResubstituting in the second and third equation: $-3+3(3)=8-2xyz$ . So $xyz=1$\nSo now we have three equations for the elementary symmetric sums of $x,y,z$\nEquation 4: $x+y+z=3$ (this is also equation 1)\nEquation 5: $xy+yz+xz=3$\nEquation 6: $xyz=1$\nIf we call the solutions of $t^3-3t^2+3t-1=0$ (Equation 7) $a,b,c$ , then $x,y,z$ are the three roots $a,b,c$ but in some order. Notice that Equation 7 can be factored as $(t-1)^3=0$ , which means that $t=1$ . Therefore $(x,y,z)$ are permutations of $(1,1,1)$ in some order, which can be only $(1,1,1)$ . (This step uses Vieta's formulas)\nTherefore, the only solution is $\\boxed{1,1,1}$" ]

[ "We can write the two digit number in the form of $10a+b$ ; reverse of $10a+b$ is $10b+a$ . The sum of those numbers is: \\[(10a+b)+(10b+a)=132\\] \\[11a+11b=132\\] \\[a+b=12\\] We can use brute force to find order pairs $(a,b)$ such that $a+b=12$ . Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$ . Thus, our ordered pairs are $(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)$ ; or $\\boxed{7}$ ordered pairs.", "Since the numbers are “mirror images,” their average has to be $\\frac{132}{2}=66$ . The highest possible value for the tens digit is $9$ because it is a two-digit number. $9-6=3$ and $6-3=3$ , so our lowest tens digit is $3$ . The numbers between $9$ and $3$ inclusive is $9-3+1=\\boxed{7}$ total possibilities." ]

[ "We first observe that since there will be more 2s than 5s in $1005!$ , we are looking for the largest $n$ such that $5^n$ divides $1005!$ . We will use the fact that:\n\\[n = \\left \\lfloor {\\frac{1005}{5^1}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^2}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^3}}\\right \\rfloor \\cdots\\]\n(This is an application of Legendre's formula).\nFrom $k=5$ and onwards, $\\left \\lfloor {\\frac{1005}{5^k}}\\right \\rfloor = 0$ . Thus, our calculation becomes\n\\[n = \\left \\lfloor {\\frac{1005}{5^1}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^2}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^3}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{5^4}}\\right \\rfloor\\]\n\\[n = \\left \\lfloor {\\frac{1005}{5}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{25}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{125}}\\right \\rfloor + \\left \\lfloor {\\frac{1005}{625}}\\right \\rfloor\\]\n\\[n = 201 + 40 + 8 + 1 = \\boxed{250}\\]" ]

[ "The equation can be rewritten as $\\frac{\\log b}{\\log a} + 6 \\frac{\\log a}{\\log b} = \\frac{(\\log b)^2+6(\\log a)^2}{\\log a \\log b}=5$ Multiplying through by $\\log a \\log b$ and factoring yields $(\\log b - 3\\log a)(\\log b - 2\\log a)=0$ . Therefore, $\\log b=3\\log a$ or $\\log b=2\\log a$ , so either $b=a^3$ or $b=a^2$\nThere are $44-2+1=43$ possibilities for the square case and $12-2+1=11$ possibilities for the cube case. Thus, the answer is $43+11= \\boxed{054}$", "Let $k=\\log_a b$ . Then our equation becomes $k+\\frac{6}{k}=5$ . Multiplying through by $k$ and solving the quadratic gives us $k=2$ or $k=3$ . Hence $a^2=b$ or $a^3=b$\nFor the first case $a^2=b$ $a$ can range from 2 to 44, a total of 43 values.\nFor the second case $a^3=b$ $a$ can range from 2 to 12, a total of 11 values.\nThus the total number of possible values is $43+11=\\boxed{54}$", "Using the change of base formula on the second equation to change to base $a$ , we get $\\log_a(b) + \\frac{6 \\log_a(a)}{\\log_a(b)}$ . If we substitute $x$ for $\\log_a(b)$ , we get $x + \\frac{6}{x}$ . Multiplying by $x$ on both sides and solving, we get $x=3,2$ . Substituting back in, we get $\\log_a(b) = 3,2$ . That means $a^3 = b$ or $a^2 = b$ . Since $b \\leq 2005$ , we can see that for the cubed case, the maximum $a$ can be without exceeding 2005 is 12(because $13^3 = 2197$ ) and for the squared case it can be a maximum of 44. Since $a \\neq 1$ , the number of values is $(44-1)+(12-1) = \\boxed{54}$" ]

[ "Use the change of base formula to see that $\\frac{\\log a}{\\log 8} + \\frac{2 \\log b}{\\log 4} = 5$ ; combine denominators to find that $\\frac{\\log ab^3}{3\\log 2} = 5$ . Doing the same thing with the second equation yields that $\\frac{\\log a^3b}{3\\log 2} = 7$ . This means that $\\log ab^3 = 15\\log 2 \\Longrightarrow ab^3 = 2^{15}$ and that $\\log a^3 b = 21\\log 2 \\Longrightarrow a^3 b = 2^{21}$ . If we multiply the two equations together, we get that $a^4b^4 = 2^{36}$ , so taking the fourth root of that, $ab = 2^9 = \\boxed{512}$", "We can simplify our expressions by changing everything to a common base and by pulling exponents out of the logarithms. The given equations then become $\\frac{\\ln a}{\\ln 8} + \\frac{2 \\ln b}{\\ln 4} = 5$ and $\\frac{\\ln b}{\\ln 8} + \\frac{2 \\ln a}{\\ln 4} = 7$ . Adding the equations and factoring, we get $(\\frac{1}{\\ln 8}+\\frac{2}{\\ln 4})(\\ln a+ \\ln b)=12$ . Rearranging we see that $\\ln ab = \\frac{12}{\\frac{1}{\\ln 8}+\\frac{2}{\\ln 4}}$ . Again, we pull exponents out of our logarithms to get $\\ln ab = \\frac{12}{\\frac{1}{3 \\ln 2} + \\frac{2}{2 \\ln 2}} = \\frac{12 \\ln 2}{\\frac{1}{3} + 1} = \\frac{12 \\ln 2}{\\frac{4}{3}} = 9 \\ln 2$ . This means that $\\frac{\\ln ab}{\\ln 2} = 9$ . The left-hand side can be interpreted as a base-2 logarithm, giving us $ab = 2^9 = \\boxed{512}$", "This solution is very similar to the above two, but it utilizes the well-known fact that $\\log_{m^k}{n^k}= \\log_m{n}.$ Thus, $\\log_8a+\\log_4b^2=5 \\Rightarrow \\log_{2^3}{(\\sqrt[3]{a})^3} + \\log_{2^2}{b^2} = 5 \\Rightarrow \\log_2{\\sqrt[3]{a}} + \\log_2{b} = 5 \\Rightarrow \\log_2{\\sqrt[3]{a}b} = 5.$ Similarly, $\\log_8b+\\log_4a^2=7 \\Rightarrow \\log_2{\\sqrt[3]{b}a} = 7.$ Adding these two equations, we have $\\log_2{a^{\\frac{4}{3}}b^{\\frac{4}{3}}} = 12 \\Rightarrow ab = 2^{12\\times\\frac{3}{4}} = 2^9 = \\boxed{512}$", "We can change everything to a common base, like so: $\\log_8{a} + \\log_8{b^3} = 5,$ $\\log_8{b} + \\log_8{a^3} = 7.$ We set the value of $\\log_8{a}$ to $x$ , and the value of $\\log_8{b}$ to $y.$ Now we have a system of linear equations: \\[x + 3y = 5,\\] \\[y + 3x = 7.\\] Now add the two equations together then simplify, we'll get $x+y=3$ . So $\\log_8{ab} = \\log_8{a} + \\log_8{b} = 3$ $ab = 8^3 = \\boxed{512}$", "Add the two equations to get $\\log_8 {a}+ \\log_8 {b}+ \\log_{a^2}+\\log_{b^2}=12$ . This can be simplified with the log property $\\log_n {x}+\\log_n {y}=log_n {xy}$ . Using this, we get $\\log_8 {ab}+ \\log_4 {a^2b^2}=12$ . Now let $\\log_8 {ab}=c$ and $\\log_4 {a^2b^2}=k$ . Converting to exponents, we get $8^c=ab$ and $4^k=(ab)^2$ . Sub in the $8^c$ to get $k=3c$ . So now we have that $k+c=12$ and $k=3c$ which gives $c=3$ $k=9$ . This means $\\log_4 {a^2b^2}=9$ so $4^9=(ab)^2 \\implies ab=(2^2)^9 \\implies 2^9 \\implies \\boxed{512}$", "By properties of logarithms, we know that $\\log_8 {a}+ \\log_4 {b ^ 2} = \\log_2 {a ^ {1/3}}+ \\log_2 {b} = 5$\nUsing the fact that $\\log_a {b} + \\log_a {c} = log_a{b*c}$ , we get $\\log_2 {a^{1/3} * b} = 5$\nSimilarly, we know that $\\log_2 {a * b^{1/3}} = 7$\nFrom these two equations, we get $a^{1/3} * b = 2^5$ and $a * b^{1/3} = 2^7$\nMultiply the two equations to get $a^{4/3} * b^{4/3} = 2^{12}$ . Solving, we get that $a*b = 2^{12*3/4} = 2^9 =$ $\\boxed{512}$", "Adding both of the equations, we get \\[\\log_8{ab} +2\\log_4{ab}=12\\] Furthermore, we see that $\\log_4 {ab}$ is $\\frac{3}{2}$ times $\\log_8 {ab}.$ Substituting $\\log_8 {ab}$ as $x,$ we get $x+3x=12,$ so $x=3.$ Therefore, we have $\\log_8 {ab} = 3,$ so $ab= 8^3=\\boxed{512}$ ~ math_comb01" ]

[ "Let $E$ be the point of intersection of $L$ and $\\overline{BD}$ . Then, because $AE$ is the altitude to the hypotenuse of right triangle $ABD$ , triangles $ADE$ and $BAE$ are similar, giving \\[\\frac{AE}{BE} = \\frac{ED}{EA},\\] and so \\begin{align*}AE &= \\sqrt{BE \\cdot ED} \\\\ &= \\sqrt{(1+1)(1)} \\\\ &= \\sqrt{2}.\\end{align*} Thus, taking $BD$ and $AE$ as the base and perpendicular height, respectively, of triangle $ABD$ , we may compute its area as $\\frac{1}{2}(3)\\left(\\sqrt{2}\\right) = \\frac{3\\sqrt{2}}{2}$ . By symmetry, the area of the entire rectangle $ABCD$ is \\[2\\left(\\frac{3\\sqrt{2}}{2}\\right) = 3\\sqrt{2} \\approx (3)(1.4) = \\boxed{4.2}.\\]" ]

[ "Consider the classical formula for triangle area: $\\frac 12 \\cdot b \\cdot h$ . \nEach of the triangles that we can make has exactly one side lying on one of the two parallel lines. If we pick this side to be the base, the height will always be the same - it will be the distance between the two lines.\nHence each area is uniquely determined by the length of the base. And it can easily be seen, that the only possible base lengths are $1$ $2$ , and $3$ . Therefore there are only $\\boxed{3}$ possible values for the area.", "No matter what how we draw a triangle by selecting three non-linear points, its height will always remain the same. Therefore, we will only get different areas with different base-lengths. The possibilities are $1$ $2$ , and $3$ units for a total of $\\boxed{3}$" ]

[ "Since $\\angle CAP = \\angle CBP = 10^\\circ$ , quadrilateral $ABPC$ is cyclic (as shown below) by the converse of the theorem \"angles inscribed in the same arc are equal\".\n\nSince $\\angle ACM = 40^\\circ$ $\\angle ACP = 140^\\circ$ , so, using the fact that opposite angles in a cyclic quadrilateral sum to $180^{\\circ}$ , we have $\\angle ABP = 40^\\circ$ . Hence $\\angle ABC = \\angle ABP - \\angle CBP = 40^ \\circ - 10^\\circ = 30^\\circ$\nSince $CA = CB$ , triangle $ABC$ is isosceles, with $\\angle BAC = \\angle ABC = 30^\\circ$ . Now, $\\angle BAP = \\angle BAC - \\angle CAP = 30^\\circ - 10^\\circ = 20^\\circ$ . Finally, again using the fact that angles inscribed in the same arc are equal, we have $\\angle BCP = \\angle BAP = \\boxed{20}$" ]

[ "Because $P$ $Q$ $R$ , and $S$ are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are $(\\pm 3,\\pm 4), (\\pm 4, \\pm 3), (0,\\pm 5),$ and $(\\pm 5,0).$ We want to maximize $PQ$ and minimize $RS.$ They also have to be non perfect squares, because they are both irrational. The greatest value of $PQ$ happens when $P$ and $Q$ are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be $(-4,3)$ and $(3,-4)$ because the two points are almost across from each other. Another possible pair could be $(-4,3)$ and $(5,0)$ . To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from $(-4,3)$ to $(4,-3)$ . The distance between $(3,-4)$ and $(-4,3)$ is greater than the distance between $(5,0)$ and $(4,-3)$ . Therefore, the segment from $(3,-4)$ to $(-4,3)$ is the longest attainable (the other possible coordinates for $P$ and $Q$ are $(4,3)$ and $(-3, -4)$ $(3, 4)$ and $(-4, -3)$ $(-3, 4)$ and $(4, -3)$ . The least value of $RS$ is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, $R$ is $(3,4)$ and $S$ is $(4,3).$ They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point $(3,4)$ than $(4,3)$ and vice versa. Using the distance formula, we get that $PQ$ is $\\sqrt{98}$ and that $RS$ is $\\sqrt{2}.$ $\\frac{\\sqrt{98}}{\\sqrt{2}}=\\sqrt{49}=\\boxed{7}$", "We can look at the option choices. Since we are aiming for the highest possible ratio, let's try using $7$ (though $5 \\sqrt{2}$ actually is the highest ratio.) Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let $RS$ be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of $RS$ is $\\sqrt{1^{2}+1^{2}} = \\sqrt{2}$ . Assuming that $\\frac{PQ}{RS} = 7$ , we plug in $RS = \\sqrt{2}$ and solve for $PQ$ $PQ=7\\sqrt{2}$ . Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of $x$ and $y$ does $\\sqrt{x^{2}+y^{2}}=7\\sqrt2$ ? We see that this can easily be made into a $45-45-90$ triangle. But, instead of substituting $y=x$ into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a $45-45-90$ triangle, the two $45$ degree sides have side length $s$ , then the hypotenuse is $s\\sqrt2$ . Using this, we can see that $s=7$ , and since our equation does in fact yield a sensible solution, we can be assured that our answer is $\\boxed{7}$", "By inspection, when $R$ is at $(3, 4)$ and $S$ is at $(4, 3),$ it makes $RS$ as small as possible with a distance of $\\sqrt{2}$ . The greatest possible length of $PQ$ arises when $P$ is at $(-3, 4)$ and $Q$ is at $(4, -3).$ Using the distance formula, we find that $PQ$ has a length of $7\\sqrt{2}.$ The requested fraction is then $\\dfrac{PQ}{RS} = \\dfrac{7\\sqrt{2}}{\\sqrt{2}} = \\boxed{7}$" ]

[ "We could count the area contributed by each square on the $3 \\times 3$ grid:\nTop-left: $0$\nTop: Triangle with area $\\frac{1}{2}$\nTop-right: $0$\nLeft: Square with area $1$\nCenter: Square with area $1$\nRight: Square with area $1$\nBottom-left: Square with area $1$\nBottom: Triangle with area $\\frac{1}{2}$\nBottom-right: Square with area $1$\nAdding all of these together, we get $6$ which is the same as $\\boxed{6}$", "By Pick's Theorem, we get the formula, $A=I+\\frac{b}{2}-1$ where $I$ is the number of lattice points in the interior and $b$ being the number of lattice points on the boundary. In this problem, we can see that $I=1$ and $B=12$ . Substituting gives us $A=1+\\frac{12}{2}-1=6$ Thus, the answer is $\\boxed{6}$", "Notice that the extra triangle on the top with area $1$ can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area $1$ . This creates a $2*3$ rectangle, with a area of $6$ . The answer is $\\boxed{6}$ ~sakshamsethi" ]

[ "Dave and Doug paid $8+2=10$ dollars in total. Doug paid for three slices of plain pizza, which cost $\\frac{3}{8}\\cdot 8=3$ . Dave paid $10-3=7$ dollars. Dave paid $7-3=\\boxed{4}$ more dollars than Doug." ]

[ "Dave and Doug paid $8+2=10$ dollars in total. Doug paid for three slices of plain pizza, which cost $\\frac{3}{8}\\cdot 8=3$ . Dave paid $10-3=7$ dollars. Dave paid $7-3=\\boxed{4}$ more dollars than Doug." ]