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[ "The best way to solve this problem is by simple brute force.\nIt is convenient to drop the usual way how exponentiation is denoted, and to write the formula as $2\\uparrow 2\\uparrow 2\\uparrow 2$ , where $\\uparrow$ denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:\nWe can note that $2\\uparrow (2\\uparrow 2) = (2\\uparrow 2)\\uparrow 2 =16$ . Therefore options 1 and 2 are equal, and options 3 and 4 are equal.\nOption 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.\n$((2\\uparrow 2)\\uparrow 2)\\uparrow 2 = 16\\uparrow 2 = 256$\n$(2\\uparrow 2)\\uparrow (2\\uparrow 2) = 4 \\uparrow 4 = 256$\nThus the only other result is $256$ , and our answer is $\\boxed{1}$" ]

[ "The best way to solve this problem is by simple brute force.\nIt is convenient to drop the usual way how exponentiation is denoted, and to write the formula as $2\\uparrow 2\\uparrow 2\\uparrow 2$ , where $\\uparrow$ denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:\nWe can note that $2\\uparrow (2\\uparrow 2) = (2\\uparrow 2)\\uparrow 2 =16$ . Therefore options 1 and 2 are equal, and options 3 and 4 are equal.\nOption 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.\n$((2\\uparrow 2)\\uparrow 2)\\uparrow 2 = 16\\uparrow 2 = 256$\n$(2\\uparrow 2)\\uparrow (2\\uparrow 2) = 4 \\uparrow 4 = 256$\nThus the only other result is $256$ , and our answer is $\\boxed{1}$" ]

[ "Note that if $n$ is the number of friends each person has, then $n$ can be any integer from $1$ to $4$ , inclusive.\nOne person can have at most 4 friends since they cannot be all friends (stated in the problem).\nAlso note that the cases of $n=1$ and $n=4$ are the same, since a map showing a solution for $n=1$ can correspond one-to-one with a map of a solution for $n=4$ by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with $n=2$ when compared to configurations of $n=3$ . Thus, we have two cases to examine, $n=1$ and $n=2$ , and we count each of these combinations twice.\n(Note: If you aren’t familiar with one-to-one correspondences, think of it like this: the number of ways to choose 4 friends is equal to number of ways to exclude one friend from your friend group. Hence, since the number of ways to choose 1 friend is the same thing as choosing 1 to not be friends with, $n=1$ and $n=4$ have the same number of ways. Similarly, $n=2$ and $n=3$ have the same number of ways as well. ~peelybonehead)\nFor $n=1$ , if everyone has exactly one friend, that means there must be $3$ pairs of friends, with no other interconnections. The first person has $5$ choices for a friend. There are $4$ people left. The next person has $3$ choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are $15$ configurations with $n=1$\nFor $n=2$ , there are two possibilities. The group of $6$ can be split into two groups of $3$ , with each group creating a friendship triangle. The first person has $\\binom{5}{2} = 10$ ways to pick two friends from the other five, while the other three are forced together. Thus, there are $10$ triangular configurations.\nHowever, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon without loss of generality . This person has $\\binom{5}{2} = 10$ choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated \"across\" from one of the original three people, forming a different configuration. Thus, there are $10 \\cdot 3! = 60$ hexagonal configurations, and in total $70$ configurations for $n=2$\nAs stated before, $n=3$ has $70$ configurations, and $n=4$ has $15$ configurations. This gives a total of $(70 + 15)\\cdot 2 = 170$ configurations, which is option $\\boxed{170}$" ]

[ "Note that if $n$ is the number of friends each person has, then $n$ can be any integer from $1$ to $4$ , inclusive.\nOne person can have at most 4 friends since they cannot be all friends (stated in the problem).\nAlso note that the cases of $n=1$ and $n=4$ are the same, since a map showing a solution for $n=1$ can correspond one-to-one with a map of a solution for $n=4$ by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with $n=2$ when compared to configurations of $n=3$ . Thus, we have two cases to examine, $n=1$ and $n=2$ , and we count each of these combinations twice.\n(Note: If you aren’t familiar with one-to-one correspondences, think of it like this: the number of ways to choose 4 friends is equal to number of ways to exclude one friend from your friend group. Hence, since the number of ways to choose 1 friend is the same thing as choosing 1 to not be friends with, $n=1$ and $n=4$ have the same number of ways. Similarly, $n=2$ and $n=3$ have the same number of ways as well. ~peelybonehead)\nFor $n=1$ , if everyone has exactly one friend, that means there must be $3$ pairs of friends, with no other interconnections. The first person has $5$ choices for a friend. There are $4$ people left. The next person has $3$ choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are $15$ configurations with $n=1$\nFor $n=2$ , there are two possibilities. The group of $6$ can be split into two groups of $3$ , with each group creating a friendship triangle. The first person has $\\binom{5}{2} = 10$ ways to pick two friends from the other five, while the other three are forced together. Thus, there are $10$ triangular configurations.\nHowever, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon without loss of generality . This person has $\\binom{5}{2} = 10$ choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated \"across\" from one of the original three people, forming a different configuration. Thus, there are $10 \\cdot 3! = 60$ hexagonal configurations, and in total $70$ configurations for $n=2$\nAs stated before, $n=3$ has $70$ configurations, and $n=4$ has $15$ configurations. This gives a total of $(70 + 15)\\cdot 2 = 170$ configurations, which is option $\\boxed{170}$" ]

[ "Let $x$ be the number of people at the party before the bus arrives. We know that $x\\equiv 0\\pmod {12}$ , as $\\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \\equiv 0 \\pmod{25}$ , as $\\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 50 \\equiv 0 \\pmod{25}$ can be reduced to $x \\equiv 0 \\pmod{25}$ , and since we are looking for the minimum amount of people, $x$ is $300$ . That means there are $350$ people at the party after the bus arrives, and thus there are $350 \\cdot \\frac{11}{25} = \\boxed{154}$ adults at the party.", "Since at the beginning, adults make up $\\frac{5}{12}$ of the concert, the amount of people must be a multiple of 12.\nCall the amount of people in the beginning $x$ .Then $x$ must be divisible by 12, in other words: $x$ must be a multiple of 12.\nSince after 50 more people arrived, adults make up $\\frac{11}{25}$ of the concert, $x+50$ is a multiple of 25. This means $x+50$ must be a multiple of 5.\nNotice that if a number is divisible by 5, it must end with a 0 or 5. Since 5 is impossible (obviously, since multiples of 12 end in 2, 4, 6, 8, 0,...), $x$ must end in 0.\nNotice that the multiples of 12 that end in 0 are: 60, 120, 180, etc.. By trying out, you can clearly see that $x=300$ is the minimum number of people at the concert.\nSo therefore, after 50 more people arrive, there are $300+50=350$ people at the concert, and the number of adults is $350*\\frac{11}{25}=154$ . Therefore the answer is $\\boxed{154}$", "Let $a$ be the number of adults before the bus arrived and $x$ be the total number of people at the concert. So, $\\frac{a}{x}=\\frac{5}{12}$ . Solving for $x$ in terms of $a$ $x = \\frac{12}{5}a$ . After the bus arrives, let's say there are an additional $y$ adults out of the 50 more people who enter the concert. From that, we get $\\frac{a+y}{x+50}=\\frac{11}{25}$ . Replacing $x$ with the value of $a$ , the second equation becomes $\\frac{a+y}{\\frac{12}{5}a+50}=\\frac{11}{25}$\nBy cross-multiplying and simplifying, we get that $25(y-22)=\\frac{7a}{5}$\nObserve that we must make sure $y-22$ is positive and divisible by $7$ to have an integer value of $a$ . The smallest possible value of $y$ that satisfies this conditions is $29$ . Plugging this into the equation, $a = 125$ . The question asks for the minimum number of adults that are there after the bus arrives, which is $a+y$ . Thus, the answer is simply $125+29=\\boxed{154}$" ]

[ "Given the information above, we start with the equation $\\frac{t}{4}+\\frac{2t}{7} + 15 + x = t$ ,where $t$ is the total number of points scored and $x\\le 14$ is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation $x+15 = \\frac{13}{28}t$ , or $28x+28\\cdot 15=13t$ . Since $t$ is necessarily divisible by 28, let $t=28u$ where $u \\ge 0$ and divide by 28 to obtain $x + 15 = 13u$ . Then, it is easy to see $u=2$ $t=56$ ) is the only candidate remaining, giving $x=\\boxed{11}$", "We first start by setting the total number of points as $28$ , since $\\text{LCM}(4,7) = 28$ . However, we see that this does not work since we surpass the number of points just with the information given ( $28\\cdot\\frac{1}{4}+28\\cdot\\frac{2}{7} + 15 = 30$ $(> 28)$ ). Next, we can see that the total number of points scored is $56$ as, if it is more than or equal to $84$ , at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: $56\\cdot\\frac{1}{4}+56\\cdot\\frac{2}{7} + 15 = 45$ , and thus, the other seven players would have scored a total of $56-45 = \\boxed{11}$ . (We see that this works since we could have $4$ of them score $2$ points, and the other $3$ of them score $1$ point.)", "Adding together Alexa's and Brittany's fractions, we get $\\frac{15}{28}$ as the fraction of the total number of points they scored together. However, this is just a ratio, so we can introduce a variable: $\\frac{15x}{28x}$ where $x$ is the common ratio. Let $y$ and $z$ and $w$ be the number of people who scored 1, 2, and 0 points, respectively. Writing an equation, we have $\\frac{13x}{28x} = 15 + y + 2z + 0w.$ We want all of our variables to be integers. Thus, we want $15 + y + 2z = 0 \\pmod {13}.$ Simplifying, $y+2z = 11 \\pmod {13}.$ The only possible value, as this integer sum has to be less than $7 \\cdot 2 + 1 = 15,$ must be 11. Therefore, $y+2z = 11,$ and the answer is $\\boxed{11}$", "We can rewrite the question as an algebraic equation: $\\frac{1}{4} x + \\frac{2}{7} x + 15 + y$ , where $x$ represents the total amount of points and $y$ the amount of points the $7$ other players scored. From there, we add the two fractions to get $\\frac{15}{28} x + 15 = x$ . Subtracting $\\frac{15}{28} x$ from both sides, we get $\\frac{13}{28} x = y + 15$ . We multiply each side by $28$ to get rid of the denominator, in which we get $13x = 420 + 28y$ . Now let’s think of this logically. This equation is telling us that if you add $420$ and $28$ times the amount of points scored by the extra $7$ players, you get $13$ times the amount of points total. And since we have to have a whole number of points total, this means that $420 + 28y$ must be divisible by $13$ . Plugging in all the answer choices for $y$ , we find that the only answer that makes $420 + 28y$ divisible by $13$ is $\\boxed{11}$" ]

[ "Sally made $0.55*20=11$ shots originally. Letting $x$ be the number of shots she made, we have $\\frac{11+x}{25}=0.56$ . Solving for $x$ gives us $x=\\boxed{3}$" ]

[ "Naomi travels $6$ miles in a time of $10$ minutes, which is equivalent to $\\dfrac{1}{6}$ of an hour. Since $\\text{speed} = \\frac{\\text{distance}}{\\text{time}}$ , her speed is $\\frac{6}{\\left(\\frac{1}{6}\\right)} = 36$ mph. By a similar calculation, Maya's speed is $12$ mph, so the answer is $36-12 = \\boxed{24}$", "Naomi's speed of $6$ miles in $10$ minutes is equivalent to $6 \\cdot 6 = 36$ miles per hour, while Maya's speed of $6$ miles in $30$ minutes (i.e. half an hour) is equivalent to $6 \\cdot 2 = 12$ miles per hour. The difference is consequently $36-12=\\boxed{24}$" ]

[ "The smallest two-digit integer he can subtract from $200$ is $10$ . This will give the largest result for that first operation, and doubling it will keep it as the largest number possible.\n\\[200-10=190\\] \\[190\\times2=380\\]\n$\\boxed{380}$" ]

[ "This solution refers to the Diagram section.\nLet $D$ be the orthogonal projection of $B$ onto the equator. Note that $\\angle BDA = \\angle BDC = 90^\\circ$ and $\\angle BCD = 45^\\circ.$ Recall that $115^\\circ \\text{ W}$ longitude is the same as $245^\\circ \\text{ E}$ longitude, so $\\angle ACD=135^\\circ.$\nWithout the loss of generality, let $AC=BC=1.$ As shown below, we place Earth in the $xyz$ -plane with $C=(0,0,0)$ such that the positive $x$ -axis runs through $A,$ the positive $y$ -axis runs through $0^\\circ$ latitude and $160^\\circ \\text{ W}$ longitude, and the positive $z$ -axis runs through the North Pole. It follows that $A=(1,0,0)$ and $D=(-t,t,0)$ for some positive number $t.$ Since $\\triangle BCD$ is an isosceles right triangle, we have $B=\\left(-t,t,\\sqrt{2}t\\right).$ By the Distance Formula, we get $(-t)^2+t^2+\\left(\\sqrt{2}t\\right)^2=1,$ from which $t=\\frac12.$\nAs $\\vec{A} = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $\\vec{B} = \\begin{pmatrix}-1/2 \\\\ 1/2 \\\\ \\sqrt2/2 \\end{pmatrix},$ we obtain \\[\\cos\\angle ACB=\\frac{\\vec{A}\\bullet\\vec{B}}{\\left\\lVert\\vec{A}\\right\\rVert\\left\\lVert\\vec{B}\\right\\rVert}=-\\frac12\\] by the dot product, so $\\angle ACB=\\boxed{120}$ degrees.", "This solution refers to the diagram in Solution 2.\nIn spherical coordinates $(\\rho,\\theta,\\phi),$ note that $\\rho,\\theta,$ and $\\phi$ represent the radial distance, the polar angle, and the azimuthal angle, respectively.\nWithout the loss of generality, let $AC=BC=1.$ As shown in Solution 2, we place Earth in the $xyz$ -plane with origin $C$ such that the positive $x$ -axis runs through $A,$ the positive $y$ -axis runs through $0^\\circ$ latitude and $160^\\circ \\text{ W}$ longitude, and the positive $z$ -axis runs through the North Pole.\nIn spherical coordinates, we have $A=(1,90^\\circ,0^\\circ)$ and $B=(1,45^\\circ,135^\\circ).$ Now, we express $\\vec{A}$ and $\\vec{B}$ in Cartesian coordinates: \\[\\vec{A} = \\begin{pmatrix}\\sin90^\\circ \\cos0^\\circ \\\\ \\sin90^\\circ \\sin0^\\circ \\\\ \\cos90^\\circ \\end{pmatrix} = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix} \\text{ and } \\vec{B} = \\begin{pmatrix}\\sin45^\\circ \\cos135^\\circ \\\\ \\sin45^\\circ \\sin135^\\circ \\\\ \\cos45^\\circ \\end{pmatrix} = \\begin{pmatrix}-1/2 \\\\ 1/2 \\\\ \\sqrt2/2 \\end{pmatrix}.\\] We continue with the last paragraph of Solution 2 to get the answer $\\angle ACB=\\boxed{120}$ degrees." ]

[ "Because there are $14$ days in two weeks, Al spends $546/14 = 39$ dollars per day for the cost of a green pill and a pink pill. If the green pill costs $x$ dollars and the pink pill $x-1$ dollars, the sum of the two costs $2x-1$ should equal $39$ dollars. Then the cost of the green pill $x$ is $\\boxed{20}$" ]

[ "Because there are $14$ days in two weeks, Al spends $546/14 = 39$ dollars per day for the cost of a green pill and a pink pill. If the green pill costs $x$ dollars and the pink pill $x-1$ dollars, the sum of the two costs $2x-1$ should equal $39$ dollars. Then the cost of the green pill $x$ is $\\boxed{20}$" ]

[ "Let the total number of steps be $x$ , the speed of the escalator be $e$ and the speed of Bob be $b$\nIn the time it took Bob to climb up the escalator he saw 75 steps and also climbed the entire escalator. Thus the contribution of the escalator must have been an additional $x - 75$ steps. Since Bob and the escalator were both moving at a constant speed over the time it took Bob to climb, the ratio of their distances covered is the same as the ratio of their speeds, so $\\frac{e}{b} = \\frac{x - 75}{75}$\nSimilarly, in the time it took Al to walk down the escalator he saw 150 steps, so the escalator must have moved $150 - x$ steps in that time. Thus $\\frac{e}{3b} = \\frac{150 - x}{150}$ or $\\frac{e}{b} = \\frac{150 - x}{50}$\nEquating the two values of $\\frac{e}{b}$ we have $\\frac{x - 75}{75} = \\frac{150 - x}{50}$ and so $2x - 150 = 450 - 3x$ and $5x = 600$ and $x = \\boxed{120}$ , the answer.", "Again, let the total number of steps be $x$ , the speed of the escalator be $e$ and the speed of Bob be $b$ (all \"per unit time\").\nThen this can be interpreted as a classic chasing problem: Bob is \"behind\" by $x$ steps, and since he moves at a pace of $b+e$ relative to the escalator, it will take $\\frac{x}{b+e}=\\frac{75}{e}$ time to get to the top.\nSimilarly, Al will take $\\frac{x}{3b-e}=\\frac{150}{e}$ time to get to the bottom.\nFrom these two equations, we arrive at $150=\\frac{ex}{3b-e}=2\\cdot75=\\frac{2ex}{b+e}=\\frac{6ex}{3b+3e}=\\frac{(ex)-(6ex)}{(3b-e)-(3b+3e)}=\\frac{5x}{4}$ $\\implies600=5x\\implies x=\\boxed{120}$ (the proportion manipulations are motivated by the desire to isolate $x$ , prompting the isolation of the $150$ on one side, and the fact that if we could cancel out the $b$ 's, then the $e$ 's in the numerator and denominator would cancel out, resulting in an equation with $x$ by itself).", "Let $e$ and $b$ be the speeds of the escalator and Bob, respectively.\nWhen Al was on his way down, he took $150$ steps with a speed of $3b-e$ per step. When Bob was on his way up, he took $75$ steps with a speed of $b+e$ per step. Since Al and Bob were walking the same distance, we have \\[150(3b-e)=75(b+e)\\] Solving gets the ratio $\\frac{e}{b}=\\frac{3}{5}$\nThus while Bob took $75$ steps to go up, the escalator contributed an extra $\\frac{3}{5}\\cdot75=45$ steps.\nFinally, there is a total of $75+45=\\boxed{120}$ steps in the length of the escalator.", "Please understand the machinery of an escalator before proceeding to read this solution.\nLet the number of steps that disappear at the top of the escalator equal $x.$ Assume that Al takes $3$ steps per second and that Bob takes $1$ step per second. Since Al counts $150$ steps, it takes him \\[\\frac{150}{3}=50\\] seconds to traverse the distance of the escalator moving downwards. Since Bob counts $75$ steps, it takes him \\[\\frac{75}{1}=75\\] seconds to traverse the distance of the escalator moving downwards.\nFor the sake of this solution, we activate the emergency stop button on the escalator.\nNow, the escalator is not moving, or is simply a staircase. Imagine that Al is taking $3$ steps downwards every second, but we throw hands at him immediately after each second, such that he flinches and moves himself backwards $x$ steps. This is equivalent to Al taking $3-x$ steps downwards every second. Since we discovered that it takes him $50$ seconds to get from the top to the bottom of the escalator, and we are forcing Al to imitate the movement of the escalator, it also takes him $50$ seconds to move from the top to the bottom of the staircase. Thus, Al takes a total of \\[(3-x) \\cdot 50=150-50x \\qquad (\\heartsuit)\\] steps.\nThe explanation for Bob is similar except now we pick him up and place him forward $x$ steps immediately after he takes his usual step per second, and since we discovered he does this for $75$ seconds, it takes him \\[(1+x) \\cdot 75=75+75x \\qquad (\\clubsuit)\\] steps to get from the bottom to the top.\nNote that because the escalator is broken and is now a staircase, Al and Bob must have had to take an equal amount of steps to get from the bottom to the top or from the top to the bottom. (Clearly, there are an equal amount of steps from the bottom to the top, and from the top to the bottom.) Therefore, we may equate $\\heartsuit$ and $\\clubsuit$ to get \\[150-50x=75+75x\\] \\[x=\\frac{3}{5}.\\] Therefore, substituting $x$ in the expression we discovered in $\\heartsuit,$ Al takes a total of \\[150-50x=150-50\\left(\\frac{3}{5}\\right)=\\boxed{120}\\] steps, and we are done.", "WLOG, let Al's speed be $15$ steps per second, so Bob's speed is $5$ steps per second. Then, Al was on the escalator for $\\frac{150}{15}\\ = 10$ seconds and Bob was on the escalator for $\\frac{75}{5}\\ = 15$ seconds. Let $r$ be the rate of the escalator, in steps per second. Then, the total amount of steps is $150 - 10r = 75 + 15r$ . Al is getting $10$ seconds of resistance at rate $r$ from the escalator, while Bob is getting $15$ seconds of help at rate $r$ . Solving for $r$ , we have $r = 3$ steps per second. Then, we can plug $r$ into the previous equation or subtract/add it to Al/Bob's rate (respectively) then multiply by their respective time. Either way, we get $\\boxed{120}$ and we are done." ]

[ "Solution by e_power_pi_times_i\nDenote Al's age, Bob's age, and Carl's age by $a$ $b$ , and $c$ , respectively. Then, $a = 16 + b + c$ and $a^2 = 1632 + b^2 + c^2$ . Substituting the first equation into the second, $(16 + b + c)^2 = b^2 + c^2 + 2bc + 32b + 32c + 256 = b^2 + c^2 + 1632$ . Thus, $bc + 16b + 16c = 688$ , and $(b+16)(c+16) = 944$ . Since $944 = 2^4\\cdot59$ $(b,c) = (0,43)$ or $(43,0)$ . Then $a + b + c = 2b + 2c + 16 = \\boxed{102}$" ]

[ "For this problem, we will have to write a three-variable equation, but not necessarily solve it. Let $a, b,$ and $c$ represent the original portions of Al, Betty, and Clare, respectively. At the end of one year, they each have $a-100, 2b,$ and $2c$ . From this, we can write two equations, marked by (1) and (2).\n\\[a+b+c=1000\\] \\[(1). \\text{ }2a+2b+2c=2000\\] \\[a-100+2b+2c=1500\\] \\[(2). \\text{ }a+2b+2c=1600\\]\n(Equations (1) and (2) are derived from each equation above them.)\nSince all we need to find is $a,$ subtract the second equation from the first equation to get $a=400.$\nAl's original portion was $\\boxed{400}$", "Suppose the total amount of money Betty and Clare has is $1000-x$ and Al has $x$ dollars. Then, $(x-100)+2(1000-x)=1500$ , so Al has $\\boxed{400}$ dollars.", "Suppose if Al had not lost $\\textdollar 100$ . The total amount would be $\\textdollar 1600$ . As he has not gained any amount. So, Betty and Clare have collectively gained $\\textdollar 600$ and as they have doubled their collective fortune,\nthey must have $\\textdollar 600$ with them at the beginning. This leaves $\\boxed{400}$ for Al.\n~Anshulb" ]

[ "If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first combination Combo 1, and the second combination Combo 2.\nIn other words, Alex can exchange 5 red tokens for 4 silver tokens as long as he has at least 6 red tokens, and Alex can exchange 5 blue tokens for 3 silver tokens as long as he has at least 6 blue tokens. Hence after performing 14 Combo 1's and 14 Combo 2's, we end up with 5 red, 5 blue, and 98 silver tokens.\nFinally, Alex can visit the blue booth once, then do Combo 1, then visit the blue booth once more to end up with 1 red token, 2 blue tokens, and $\\boxed{103}$ silver tokens, at which point it is clear he cannot use the booths anymore.", "We can approach this problem by assuming he goes to the red booth first. You start with $75 \\text{R}$ and $75 \\text{B}$ and at the end of the first booth, you will have $1 \\text{R}$ and $112 \\text{B}$ and $37 \\text{S}$ . We now move to the blue booth, and working through each booth until we have none left, we will end up with: $1 \\text{R}$ $2 \\text{B}$ and $103 \\text{S}$ . So, the answer is $\\boxed{103}$", "Let $x$ denote the number of visits to the first booth and $y$ denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows: \\[R(x,y)=-2x+y+75\\] \\[B(x,y)=x-3y+75\\] There are no legal exchanges when he has fewer than $2$ red coins and fewer than $3$ blue coins, namely when he has a red coin and $2$ blue coins. We can then create a system of equations: \\[1=-2x+y+75\\] \\[2=x-3y+75\\] Solving yields $x=59$ and $y=44$ . Since he gains one silver coin per visit to each booth, he has $x+y=44+59=\\boxed{103}$ silver coins in total." ]

[ "If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first combination Combo 1, and the second combination Combo 2.\nIn other words, Alex can exchange 5 red tokens for 4 silver tokens as long as he has at least 6 red tokens, and Alex can exchange 5 blue tokens for 3 silver tokens as long as he has at least 6 blue tokens. Hence after performing 14 Combo 1's and 14 Combo 2's, we end up with 5 red, 5 blue, and 98 silver tokens.\nFinally, Alex can visit the blue booth once, then do Combo 1, then visit the blue booth once more to end up with 1 red token, 2 blue tokens, and $\\boxed{103}$ silver tokens, at which point it is clear he cannot use the booths anymore.", "We can approach this problem by assuming he goes to the red booth first. You start with $75 \\text{R}$ and $75 \\text{B}$ and at the end of the first booth, you will have $1 \\text{R}$ and $112 \\text{B}$ and $37 \\text{S}$ . We now move to the blue booth, and working through each booth until we have none left, we will end up with: $1 \\text{R}$ $2 \\text{B}$ and $103 \\text{S}$ . So, the answer is $\\boxed{103}$", "Let $x$ denote the number of visits to the first booth and $y$ denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows: \\[R(x,y)=-2x+y+75\\] \\[B(x,y)=x-3y+75\\] There are no legal exchanges when he has fewer than $2$ red coins and fewer than $3$ blue coins, namely when he has a red coin and $2$ blue coins. We can then create a system of equations: \\[1=-2x+y+75\\] \\[2=x-3y+75\\] Solving yields $x=59$ and $y=44$ . Since he gains one silver coin per visit to each booth, he has $x+y=44+59=\\boxed{103}$ silver coins in total." ]

[ "In order to begin this problem, we need to calculate the probability that Alfred will win on the first round.\nBecause he goes first, Alfred has a $\\frac{1}{2}$ chance of winning (getting heads) on his first flip.\nThen, Bonnie, who goes second, has a $\\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{4}$ , chance of winning on her first coin toss.\nTherefore Alfred’s chance of winning on his second flip is $\\frac{1}{4} \\times \\frac{1}{2} = \\frac{1}{8}$\nFrom this, we can see that Alfred’s (who goes first) chance of winning the first round is: $\\frac{1}{2} + \\frac{1}{8} + \\frac{1}{32} + \\cdots = \\frac{2}{3}$\nBonnie’s (who goes second) chance of winning the first round is then $1 - \\frac{2}{3} = \\frac{1}{3}$\nThis means that the person who goes first has a $\\frac{2}{3}$ chance of winning the round, while the person who goes second has a $\\frac{1}{3}$ chance of winning.\nNow, through casework, we can calculate Alfred’s chance of winning the second round.\nCase 1: Alfred wins twice; $\\frac{2}{3} \\times \\frac{1}{3}$ (Bonnie goes first this round) $=\\frac{2}{9}$\nCase 2: Alfred loses the first round, but wins the second; $\\frac{1}{3} \\times \\frac{2}{3} = \\frac{2}{9}$\nAdding up the cases, we get $\\frac{2}{9} + \\frac{2}{9} = \\frac{4}{9}$\nAlfred, therefore, has a $\\frac{4}{9}$ of winnning the second round, and Bonnie has a $1-\\frac{4}{9} = \\frac{5}{9}$ chance of winning this round.\nFrom here, it is not difficult to see that the probabilities alternate in a pattern.\nMake $A$ the probability that Alfred wins a round.\nThe chances of Alfred and Bonnie, respectively, winning the first round are $A$ and $A - \\frac{1}{3}$ , which can be written as $2A - \\frac{1}{3} = 1$\nThe second round’s for chances are $A$ and $A + \\frac{1}{9}$ , which can also be written as $2A + \\frac{1}{9}$\nFrom this, we can conclude that for the $n$ th even round, the probability that Alfred ( $A$ ) wins can be calculated through the equation $2A + \\frac{1}{3^n} = 1$\nSolving the equation for $n = 6$ , we get $A = \\frac{364}{729}$\n$364 + 729 = 1093$\nSo our answer is $\\boxed{093}$", "Rather than categorizing games as wins or losses, we can categorize them as starters (S), where Alfred starts, and non-starters (NS), where Bonnie starts. Game 1 is a starter, and since Alfred must win Game 6, Game 7 is a non-starter.\nAs shown in Solution 1, if a player starts a certain game, the probability $P(NS)$ that they will not start the next game (i.e. win the current game) is $\\frac{2}{3}$ , and the probability $P(S)$ that they will start the next game (i.e. lose the current game) is $\\frac{1}{3}$ . Similarly, if a player does not start a certain game, $P(NS) = \\frac{1}{3}$ and $P(S) = \\frac{2}{3}$ . We conclude that the probability of switching from S to NS or vice versa is always $\\frac{2}{3}$ , and the probability of staying the same is always $\\frac{1}{3}$\nListing out all the games from Game 1 to Game 7, we get: S, ?, ?, ?, ?, ?, NS. Between the 7 games, there are 6 opportunities to either switch or stay the same. We need to eventually switch from S to NS, so there must be an odd number of switches. Furthermore, this number is less than 6, so it must be 1, 3, or 5.\n1 switch: There are ${6 \\choose 1} = 6$ ways to place the switch, so $P = 6\\left(\\frac{1}{3}\\right)^5\\left(\\frac{2}{3}\\right) = \\frac{12}{729}$\n3 switches: There are ${6 \\choose 3} = 20$ ways to place the switches, so $P = 20\\left(\\frac{1}{3}\\right)^3\\left(\\frac{2}{3}\\right)^3 = \\frac{160}{729}$\n5 switches: There are ${6 \\choose 5} = 6$ ways to place the switch, so $P = 6\\left(\\frac{1}{3}\\right)\\left(\\frac{2}{3}\\right)^5 = \\frac{192}{729}$\nAdd up all these probabilities to get $\\frac{12+160+192}{729} = \\frac{364}{729}$ $364+729=1093$ , so the answer is $\\boxed{093}$" ]

[ "There are only $2$ people who can go in the driver's seat--Bonnie and Carlo. Any of the $3$ remaining people can go in the front passenger seat. There are $2$ people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are $2\\cdot3\\cdot2$ or $12$ ways. The answer is then $\\boxed{12}$", "If there weren't any extra requirements, there would be 24 combinations. However, there are only 2, which is half of 4, ways to put the people. Therefore, half of 24 is $\\boxed{12}$" ]

[ "We have two 30-60-90 triangles $ABC$ and $DBC$ that are perpendicular and share leg $BC$ (the altitude of the plane $p$ ). $AD=10$ The shared leg is the shortest leg of one triangle and the longest leg of the other. $A$ and $B$ are Bob and Alice respectively.\nFind $AC$ and $DC$ in terms of $p$ . Use Pythagorean Theorem on triangle $ADC$ to produce $p=\\sqrt{30}\\implies\\boxed{5.5}$", "Non-trig solution by e_power_pi_times_i\nSet the distance from Alice's and Bob's position to the point directly below the airplane to be $x$ and $y$ , respectively. From the Pythagorean Theorem, $x^2 + y^2 = 100$ . As both are $30-60-90$ triangles, the altitude of the airplane can be expressed as $\\dfrac{x\\sqrt{3}}{3}$ or $y\\sqrt{3}$ . Solving the equation $\\dfrac{x\\sqrt{3}}{3} = y\\sqrt{3}$ , we get $x = 3y$ . Plugging this into the equation $x^2 + y^2 = 100$ , we get $10y^2 = 100$ , or $y = \\sqrt{10}$ $y$ cannot be negative), so the altitude is $\\sqrt{3*10} = \\sqrt{30}$ , which is closest to $\\boxed{5.5}$", "Let Alice be at point $A$ , Bob be at point $B$ . Let the plane be at point $P$ and $X$ be the projection of $P$ onto the ground (the plane with contains $A$ and $B$ ). Let the height of the plane, or $PX$ , be $h$ . So, because of the $30-60-90$ triangles, \\[AX = \\dfrac{h}{\\sqrt{3}}, BX = h\\sqrt{3}\\] By Pythagorean Theorem on $\\triangle ABX$ \\[\\dfrac{h^2}{3} + 3h^2 = 100 \\implies \\dfrac{10h^2}{3} = 100 \\implies h = \\sqrt{30},\\] which is clossest to $\\boxed{5.5}.$" ]

[ "Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$ , is a multiple of $12$ . Since this number must be a multiple of $12$ , as stated in the previous sentence, $14$ has a factor $2$ $k$ must have a factor of $6$ . The smallest number of turns that is a multiple of $6$ is $\\boxed{6}$" ]

[ "Let's first try some experimentation. Alice obviously wins if there is one coin. She will just take it and win. If there are 2 remaining, then Alice will take one and then Bob will take one, so Bob wins. If there are $3$ , Alice will take $1$ , Bob will take one, and Alice will take the final one. If there are $4$ , Alice will just remove all $4$ at once. If there are $5$ , no matter what Alice does, Bob can take the final coins in one try. Notice that Alice wins if there are $1$ $3$ , or $4$ coins left. Bob wins if there are $2$ or $5$ coins left.\nAfter some thought, you may realize that there is a strategy for Bob. If there is n is a multiple of $5$ , then Bob will win. The reason for this is the following: Let's say there are a multiple of $5$ coins remaining in the stack. If Alice takes $1$ , Bob will take $4$ , and there will still be a multiple of $5$ . If Alice takes $4$ , Bob will take $1$ , and there will still be a multiple of $5$ . This process will continue until you get $0$ coins left. For example, let's say there are $205$ coins. No matter what Alice does, Bob can simply just do the complement. After each of them make a turn, there will always be a multiple of $5$ left. This will continue until there are $5$ coins left, and Bob will end up winning.\nAfter some more experimentation, you'll realize that any number that is congruent to $2$ mod $5$ will also work. This is because Bob can do the same strategy, and when there are $2$ coins left, Alice is forced to take $1$ and Bob takes the final coin. For example, let's say there are $72$ coins. If Alice takes $1$ , Bob will take $4$ . If Alice takes $4$ , Bob will take $1$ . So after they each make a turn, the number will always be equal to $2$ mod $5$ . Eventually, there will be only $2$ coins remaining, and we've established that Alice will simply take $1$ and Bob will take the final coin.\nSo we have to find the number of numbers less than or equal to $2024$ that are either congruent to $0$ mod $5$ or $2$ mod $5$ . There are $404$ numbers in the first category: $5, 10, 15, \\dots, 2020$ . For the second category, there are $405$ numbers. $2, 7, 12, 17, \\dots, 2022$ . So the answer is $404 + 405 = \\boxed{809}$", "We will use winning and losing positions, where a $W$ marks when Alice wins and an $L$ marks when Bob wins.\n$1$ coin: $W$\n$2$ coins: $L$\n$3$ coins: $W$\n$4$ coins: $W$\n$5$ coins: $L$\n$6$ coin: $W$\n$7$ coins: $L$\n$8$ coins: $W$\n$9$ coins: $W$\n$10$ coins: $L$\n$11$ coin: $W$\n$12$ coins: $L$\n$13$ coins: $W$\n$14$ coins: $W$\n$15$ coins: $L$\nWe can see that losing positions occur when $n$ is congruent to $0, 2 \\mod{5}$ and winning positions occur otherwise. In other words, there will be $2$ losing positions out of every $5$ consecutive values of n. As $n$ ranges from $1$ to $2020$ $\\frac{2}{5}$ of these values are losing positions where Bob will win. As $n$ ranges from $2021$ to $2024$ $2022$ is the only value where Bob will win. Thus, the answer is $2020\\times\\frac{2}{5}+1=\\boxed{809}$", "Denote by $A_i$ and $B_i$ Alice's or Bob's $i$ th moves, respectively.\nCase 1: $n \\equiv 0 \\pmod{5}$\nBob can always take the strategy that $B_i = 5 - A_i$ .\nThis guarantees him to win.\nIn this case, the number of $n$ is $\\left\\lfloor \\frac{2024}{5} \\right\\rfloor = 404$\nCase 2: $n \\equiv 1 \\pmod{5}$\nIn this case, consider Alice's following strategy: $A_1 = 1$ and $A_i = 5 - B_{i-1}$ for $i \\geq 2$ .\nThus, under Alice's this strategy, Bob has no way to win.\nCase 3: $n \\equiv 4 \\pmod{5}$\nIn this case, consider Alice's following strategy: $A_1 = 4$ and $A_i = 5 - B_{i-1}$ for $i \\geq 2$ .\nThus, under Alice's this strategy, Bob has no way to win.\nCase 4: $n \\equiv 2 \\pmod{5}$\nBob can always take the strategy that $B_i = 5 - A_i$ .\nTherefore, after the $\\left\\lfloor \\frac{n}{5} \\right\\rfloor$ th turn, there are two tokens leftover.\nTherefore, Alice must take 1 in the next turn that leaves the last token on the table.\nTherefore, Bob can take the last token to win the game.\nThis guarantees him to win.\nIn this case, the number of $n$ is $\\left\\lfloor \\frac{2024 - 2}{5} \\right\\rfloor +1 = 405$\nCase 5: $n \\equiv 3 \\pmod{5}$\nConsider Alice's following strategy: $A_1 = 1$ and $A_i = 5 - B_{i-1}$ for $i \\geq 2$ .\nBy doing so, there will finally be 2 tokens on the table and Bob moves first. Because Bob has the only choice of taking 1 token, Alice can take the last token and win the game.\nTherefore, in this case, under Alice's this strategy, Bob has no way to win.\nPutting all cases together, the answer is $404 + 405 = \\boxed{809}$", "Since the game Alice and Bob play is impartial (the only difference between player 1 and player 2 is that player 1 goes first (note that games like chess are not impartial because each player can only move their own pieces)), we can use the Sprague-Grundy Theorem to solve this problem. We will use induction to calculate the Grundy Values for this game.\nWe claim that heaps of size congruent to $0,2 \\bmod{5}$ will be in outcome class $\\mathcal{P}$ (win for player 2 = Bob), and heaps of size equivalent to $1,3,4 \\bmod{5}$ will be in outcome class $\\mathcal{N}$ (win for player 1 = Alice). Note that the mex (minimal excludant) of a set of nonnegative integers is the least nonnegative integer not in the set. e.g. mex $(1, 2, 3) = 0$ and mex $(0, 1, 2, 4) = 3$\n$\\text{heap}(0) = \\{\\} = *\\text{mex}(\\emptyset) = 0$\n$\\text{heap}(1) = \\{0\\} = *\\text{mex}(0) = *$\n$\\text{heap}(2) = \\{*\\} = *\\text{mex}(1) = 0$\n$\\text{heap}(3) = \\{0\\} = *\\text{mex}(0) = *$\n$\\text{heap}(4) = \\{0, *\\} = *\\text{mex}(0, 1) = *2$\n$\\text{heap}(5) = \\{*, *2\\} = *\\text{mex}(1, 2) = 0$\n$\\text{heap}(6) = \\{0, 0\\} = *\\text{mex}(0, 0) = *$\n$\\text{heap}(7) = \\{*, *\\} = *\\text{mex}(1, 1) = 0$\n$\\text{heap}(8) = \\{*2, 0\\} = *\\text{mex}(0, 2) = *$\n$\\text{heap}(9) = \\{0, *\\} = *\\text{mex}(0, 1) = *2$\n$\\text{heap}(10) = \\{*, *2\\} = *\\text{mex}(1, 2) = 0$\nWe have proven the base case. We will now prove the inductive hypothesis: If $n \\equiv 0 \\bmod{5}$ $\\text{heap}(n) = 0$ $\\text{heap}(n+1) = *$ $\\text{heap}(n+2) = 0$ $\\text{heap}(n+3) = *$ , and $\\text{heap}(n+4) = *2$ , then $\\text{heap}(n+5) = 0$ $\\text{heap}(n+6) = *$ $\\text{heap}(n+7) = 0$ $\\text{heap}(n+8) = *$ , and $\\text{heap}(n+9) = *2$\n$\\text{heap}(n+5) = \\{\\text{heap}(n+1), \\text{heap}(n+4)\\} = \\{*, *2\\} = *\\text{mex}(1, 2) = 0$\n$\\text{heap}(n+6) = \\{\\text{heap}(n+2), \\text{heap}(n+5)\\} = \\{0, 0\\} = *\\text{mex}(0, 0) = *$\n$\\text{heap}(n+7) = \\{\\text{heap}(n+3), \\text{heap}(n+6)\\} = \\{*, *\\} = *\\text{mex}(1, 1) = 0$\n$\\text{heap}(n+8) = \\{\\text{heap}(n+4), \\text{heap}(n+7)\\} = \\{*2, 0\\} = *\\text{mex}(2, 1) = *$\n$\\text{heap}(n+9) = \\{\\text{heap}(n+5), \\text{heap}(n+8)\\} = \\{0, *\\} = *\\text{mex}(0, 1) = *2$\nWe have proven the inductive hypothesis. QED.\nThere are $2020*\\frac{2}{5}=808$ positive integers congruent to $0,2 \\bmod{5}$ between 1 and 2020, and 1 such integer between 2021 and 2024. $808 + 1 = \\boxed{809}$" ]

[ "Let $k$ be one of the elements in Alices set $A$ of positive integers. The number of sets that Bob lists with the property that their maximum element is k is $2^{k-1}$ , since every positive integer less than k can be in the set or out. Thus, for the number of sets bob have listed to be 2024, we want to find a sum of unique powers of two that can achieve this. 2024 is equal to $2^{10}+2^9+2^8+2^7+2^6+2^5+2^3$ . We must increase each power by 1 to find the elements in set $A$ , which are $(11,10,9,8,7,6,4)$ . Add these up to get $\\boxed{055}$ . -westwoodmonster", "Let $A = \\left\\{ a_1, a_2, \\cdots, a_n \\right\\}$ with $a_1 < a_2 < \\cdots < a_n$\nIf the maximum element of $B$ is $a_i$ for some $i \\in \\left\\{ 1, 2, \\cdots , n \\right\\}$ , then each element in $\\left\\{ 1, 2, \\cdots, a_i- 1 \\right\\}$ can be either in $B$ or not in $B$ .\nTherefore, the number of such sets $B$ is $2^{a_i - 1}$\nTherefore, the total number of sets $B$ is\n\\begin{align*}\n\\sum_{i=1}^n 2^{a_i - 1} & = 2024 .\n\\end{align*}\nThus\n\\begin{align*}\n\\sum_{i=1}^n 2^{a_i} & = 4048 .\n\\end{align*}\nNow, the problem becomes writing 4048 in base 2, say, $4048 = \\left( \\cdots b_2b_1b_0 \\right)_2$ .\nWe have $A = \\left\\{ j \\geq 1: b_j = 1 \\right\\}$\nWe have $4048 = \\left( 111,111,010,000 \\right)_2$ .\nTherefore, $A = \\left\\{ 4, 6, 7, 8, 9, 10, 11 \\right\\}$ .\nTherefore, the sum of all elements in $A$ is $\\boxed{55}$" ]

[ "Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).\nThis method uses the counting method of stars and bars (non-negative version). Since each person must have at least $2$ apples, we can remove $2*3$ apples from the total that need to be sorted. With the remaining $18$ apples, we can use stars and bars to determine the number of possibilities. Assume there are $18$ stars in a row, and $2$ bars, which will be placed to separate the stars into groups of $3$ . In total, there are $18$ spaces for stars $+ 2$ spaces for bars, for a total of $20$ spaces. We can now do $20 \\choose 2$ . This is because if we choose distinct $2$ spots for the bars to be placed, each combo of $3$ groups will be different, and all apples will add up to $18$ . We can also do this because the apples are indistinguishable. $20 \\choose 2$ is $190$ , therefore the answer is $\\boxed{190}$", "Consider an unordered triple $(a,b,c)$ where $a+b+c=24$ and $a,b,c$ are not necessarily distinct. Then, we will either have $1$ $3$ , or $6$ distinguishable ways to assign $a$ $b$ , and $c$ to Alice, Becky, and Chris. Thus, our answer will be $x+3y+6z$ for some nonnegative integers $x,y,z$ . Notice that we only have $1$ way to assign the numbers $a,b,c$ to Alice, Becky, and Chris when $a=b=c$ . As this only happens $1$ way ( $a=b=c=8$ ), our answer is $1+3y+6z$ for some $y,z$ . Finally, notice that this implies the answer is $1$ mod $3$ . The only answer choice that satisfies this is $\\boxed{190}$", "Since each person needs to have at least two apples, we can simply give each person two, leaving $24 - 2\\times3=18$ apples. For the remaining apples, if Alice is going to have $a$ apples, Becky is going to have $b$ apples, and Chris is going to have $c$ apples, we have indeterminate equation $a+b+c=18$ . Currently, we can see that $0 \\leq a\\leq 18$ where $a$ is an integer, and when $a$ equals any number in the range, there will be $18-a+1=19-a$ sets of values for $b$ and $c$ . Thus, there are $19 + 18 + 17 + \\cdots + 1 = \\boxed{190}$ possible sets of values in total." ]

[ "To get how many cups we need, we realize that we simply need to divide the number of cups needed by the number of cups collected in her measuring cup each time. Thus, we need to evaluate the fraction $\\frac{2\\frac{1}{2}}{\\frac{1}{4}}$ . Simplifying, this is equal to $\\frac{5}{2}(4) = \\boxed{10}$" ]

[ "At any point in the game, Alice should guess whichever color has come up less frequently thus far (although if both colors have come up equally often, she may guess whichever she likes); using this strategy, her probability of guessing correctly is at least $\\frac{1}{2}$ on any given card, as desired.\nThere are ${6 \\choose 3} = 20$ possible orderings of cards, all equally likely (since any of the $6! = 720$ permutations of the cards is equally likely, and each ordering covers $3!^2 = 6^2 = 36$ permutations).\nEach of the $10$ orderings that start with red cards corresponds with one that starts with a black card; the problem is symmetrical with respect to red and black cards, so we can, without loss of generality, consider only the orderings that start with red cards.\nWe then generate a tally table showing whether Alice's guesses are correct for each ordering; for a given card, she guesses correctly if fewer than half the previously shown cards were the same color, guesses incorrectly if more than half were the same color, and guesses correctly with probability $\\frac{1}{2}$ if exactly half were the same color.\nIn this table, $\\mid$ denotes a correct guess, $\\--$ denotes an incorrect guess, and $/$ denotes a guess with $\\frac{1}{2}$ probability of being correct.\nNow we sum the tallies across orderings, obtaining $41$ , and finally divide by the number of orderings ( $10$ ) to obtain the expected number of correct guesses, $\\frac{41}{10}$ , which yields an answer of $41 + 10 = \\boxed{051}.$", "Denote by $N \\left( a, b \\right)$ the optimal expected number of cards that Alice guesses correctly, where the number of red and black cards are $a$ and $b$ , respectively.\nThus, for $a, b \\geq 1$ , we have \\begin{align*} N \\left( a, b \\right) & = \\max \\left\\{ \\frac{a}{a+b} \\left( 1 + N \\left( a - 1 , b \\right) \\right) + \\frac{b}{a+b} N \\left( a , b - 1 \\right) , \\right. \\\\ & \\hspace{1cm} \\left. \\frac{a}{a+b} N \\left( a - 1 , b \\right) + \\frac{b}{a+b} \\left( 1 + N \\left( a , b - 1 \\right) \\right) \\right\\} . \\end{align*}\nFor $a = 0$ , Alice always guesses black. So $N \\left( 0 , b \\right) = b$\nFor $b = 0$ , Alice always guesses red. So $N \\left( a , 0 \\right) = a$\nTo solve this dynamic program, we can also exploit its symmetry that $N \\left( a , b \\right) = N \\left( b , a \\right)$\nBy solving this dynamic program, we get $N \\left( 1, 1 \\right) = \\frac{3}{2}$ $N \\left( 1, 2 \\right) = \\frac{7}{3}$ $N \\left( 1 , 3 \\right) = \\frac{13}{4}$ $N \\left( 2 , 2 \\right) = \\frac{17}{6}$ $N \\left( 2 , 3 \\right) = \\frac{18}{5}$ $N \\left( 3, 3 \\right) = \\frac{41}{10}$\nTherefore, the answer is $41 + 10 = \\boxed{051}$", "Denote by $N_{i,j}$ the optimal expected number of cards that Alice guesses correctly, where the number of cards are $i$ and $j \\ge i.$\nIf $i = 0$ then Alice guesses correctly all cards, so $N_{0,j} = j.$\nIf $j = i$ then Alice guesses next card with probability $\\frac {1}{2} \\implies N_{i,i} = \\frac {1}{2} + N_{i-1,i}.$\nIf $j = i+1$ then Alice guesses next card with probability $\\frac {i+1}{2i+1} \\implies N_{i,i+1} = \\frac {i+1}{2i+1} (1+ N_{i,i}) + \\frac{i}{2i+1} N_{i-1,i+1}.$\nIf $j = i+2$ then Alice guesses next card with probability $\\frac {i+2}{2i+2} \\implies N_{i,i+2} = \\frac {i+2}{2i+2} (1+ N_{i,i+1}) + \\frac{i}{2i+2} N_{i-1,i+2}.$\nOne can find consistently: $N_{1,1} = \\frac {1}{2} + N_{0,1} = \\frac {3}{2},$ \\[N_{1,2} = \\frac {2}{3} (1 + N_{1,1}) + \\frac {1}{3} N_{0,2} = \\frac {7}{3}.\\] \\[N_{2,2} = \\frac {1}{2} + N_{1,2} = \\frac {17}{6}.\\] \\[N_{1,3} = \\frac {3}{4} (1 + N_{1,2}) + \\frac {1}{4} N_{0,3} = \\frac {13}{4}.\\] \\[N_{2,3} = \\frac {3}{5} (1 + N_{2,2}) + \\frac {2}{5} N_{1,3} = \\frac {18}{5}.\\] \\[N_{3,3} = \\frac {1}{2} + N_{2,3} = \\frac {41}{10}.\\] Therefore, the answer is $41 + 10 = \\boxed{051}$", "Denote $E_n$ the expected number of cards Alice guesses correctly given $n$ red cards and $n$ black cards. We want to find $E_3$\nAlice has a $\\frac{1}{2}$ chance of guessing the first card. WLOG assume the first card color is red. For the next card, Alice has a $\\frac{3}{5}$ chance of guessing the card if she chooses black; if they guess right, there's one less red and black card, so the expected number of cards Alice guesses from here is $E_2$ . If Alice does not guess correctly (which occurs with probability $\\frac{2}{5}$ ), this means that there's 3 black cards and 1 red card left, so Alice should guess black next with a $\\frac{3}{4}$ chance of being right. If Alice is wrong (with probability $\\frac{1}{4}$ ), there are only 3 black cards left, so Alice can guess these with certainty; if Alice is right, there are 2 blacks and 1 red left, so Alice should again guess black. If Alice is right (with probability $\\frac{2}{3}$ ), there is now 1 black and red card each, so the expected number of cards guessed is $E_1$ ; if she is wrong (with probability $\\frac{1}{3}$ ), there are 2 black cards left, so Alice can guess these with certainty.\nSumming this up into a formula: \\[E_3 = \\frac{1}{2} + \\frac{3}{5} \\left(1 + E_2 \\right) + \\frac{2}{5} \\left( \\frac{1}{4}(3) + \\frac{3}{4}\\left(1 + \\frac{2}{3}\\left(1 + E_1 \\right) + \\frac{1}{3}(2)\\right) \\right)\\]\nWe can apply similar logic to compute $E_2$ and get \\[E_2 = \\frac{1}{2} + \\frac{2}{3}(1 + E_1) + \\frac{1}{3}(2)\\]\nTo compute $E_1$ , we know that Alice can guess the last card with certainty, and there's a $\\frac{1}{2}$ chance they get the first card as well, so $E_1 = \\frac{3}{2}$\nThus, $E_2 = \\frac{17}{6}$ , and after long computation, we get $E_3 = \\frac{41}{10}$ . The requested answer is $41 + 10 = \\boxed{51}$" ]

[ "Convert everything to the same unit. Since the answer is in centimeters, change meters to centimeters by moving the decimal place two places to the right.\nThe ceiling is $240$ centimeters above the floor. The combined height of Alice and the light bulb when she reaches for it is $10+150+46=206$ centimeters. That means the stool's height needs to be $240-206=\\boxed{34}$" ]

[ "Alice may sit in the center chair, in an end chair, or in a next-to-end chair. Suppose she sits in the center chair. The 2nd and 4th chairs (next to her) must be occupied by Derek and Eric, in either order, leaving the end chairs for Bob and Carla in either order; this yields $2! * 2! = 4$ ways to seat the group.\nNext, suppose Alice sits in one of the end chairs. Then the chair beside her will be occupied by either Derek or Eric. The center chair must be occupied by Bob or Carla, leaving the last two people to fill the last two chairs in either order. $2$ ways to seat Alice times $2$ ways to fill the next chair times $2$ ways to fill the center chair times $2$ ways to fill the last two chairs yields $2 * 2 * 2 * 2 = 16$ ways to fill the chairs.\nFinally, suppose Alice sits in the second or fourth chair. Then the chairs next to her must be occupied by Derek and Eric in either order, and the other two chairs must be occupied by Bob and Carla in either order. This yields $2 * 2 * 2 = 8$ ways to fill the chairs.\nIn total, there are $4 + 8 + 16$ ways to fill the chairs, so the answer is $\\boxed{28}$" ]

[ "Let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.\nWe can split this problem up into two cases:\n$\\textbf{Case 1: }$ A sits on an edge seat.\nSince B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of $2 \\cdot 2 \\cdot 2 \\cdot 2 = 16$\n$\\textbf{Case 2: }$ A does not sit in an edge seat.\nStill, the only two people that can sit next to A are either D or E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are $3 \\cdot 2 \\cdot 2 = 12$ seatings in this case.\nAdding up all the seatings, we have $16+12 = \\boxed{28}$", "Label the seats (from left to right) $1$ through $5$ . The number of ways to seat Derek and Eric in the five seats with no restrictions is $5 \\cdot 4=20$ . The number of ways to seat Derek and Eric such that they sit next to each other is $8$ (we can treat Derek and Eric as a \"block\". There are four ways to seat this \"block\", and two ways to permute Derek and Eric, for a total of $4\\cdot 2=8$ ), so the number of ways such that Derek and Eric don't sit next to each other is $20-8=12$ . Note that once Derek and Eric are seated, we can divide into three cases.\nThe first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us $0$ ways.\nAnother possible case is if Derek and Eric sit in seats $2$ and $4$ in some order. There are $2$ possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are $3!=6$ ways to do this. So the second case gives us $2 \\cdot 6=12$ total ways.\nThe last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are $12-2-2=8$ ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us $8 \\cdot 2=16$ ways.\nSo in total there are $12+16=28$ ways. Our answer is $\\boxed{28}$ Minor $\\LaTeX$ edits by fasterthanlight", "We start with complementary counting. After all, it's much easier to count the cases where some of these restraints are true, than when they aren't.\nPIE:\nLet's count the total number of cases where one of these is true:\nWhen Alice is with Bob: $2\\cdot4!=48$\nWhen Alice is with Carla: $2\\cdot4!=48$\nWhen Derek is with Eric: $2\\cdot4!=48$\nThen, we count the cases where two of these are true.\nAlice is next to Carla, and Alice is also next to Bob.\nThere are two ways to rearrange Alice, Bob, and Carla so that this is true:\nBAC and CAB. $2\\cdot3!=12$\nAlice is next to Carla, and Derek is also next to Eric. $2\\cdot2\\cdot3!=24$\nAlice is next to Bob, and Derek is also next to Eric. $2\\cdot2\\cdot3!=24$\nFinally, we count the cases where all three of these are true: $2\\cdot2\\cdot2=8$\nWe add up the cases where one of these are true: $48\\cdot3=144$\nSubtract the cases where two of these are true: $144-60=84$\nAnd finally add back the cases where three of these are true: $84+8=92$\nThus, our answer is $5!-92=28$ , or $\\boxed{28}$", "To find the number of ways, we do casework.\nCase 1: Alice sits in the first seat (leftmost)\nSince Alice refuses to sit with Bob and Carla, then the seat on her immediate right must be Derek or Eric. The middle seat must be Bob or Carla (because Derek and Eric refuse to sit together). The seat to the right of the middle seat could be whoever is left over from Derek and Eric, or whoever is left together from Bob and Carla. The last seat only has one person left. There are $4$ ways to permute Bob, Carla, Derek, and Eric, and $2$ ways to pick who goes in the seat to the right of the middle seat, for $4\\cdot2=8$ seating arrangements here.\nCase 2: Alice sits in the second seat\nDerek and Eric must be on both sides of Alice because otherwise, we would have to put Bob or Carla next to Alice which is forbidden. Then Bob and Carla take the remaining two seats. There are $4$ ways to permute and $4$ arrangements here.\nCase 3: Alice sits in the middle seat\nOnce again, Derek and Eric must be on both sides of Alice, and Bob and Carla need to take the two remaining seats. There are also $4$ ways to permute and $4$ arrangements here.\nCase 4: Alice sits in the fourth seat\nBy symmetry, this is the same as case 2. There are $4$ arrangements.\nCase 5: Alice sits in the last seat (rightmost)\nBy symmetry, this is the same as case 1. There are $8$ arrangements.\nAdding up the cases, there are $8+4+4+4+8=28$ total seating arrangements $\\Longrightarrow \\boxed{28}$" ]

[ "If $x$ is the list price, then $10\\%(x-10)=20\\%(x-20)$ . Solving this equation gives $x=30$ , so the answer is $\\boxed{30}$" ]

[ "For each of the false statements, we identify its corresponding true statement. Note that:\nWe construct the following table: \\[\\begin{array}{c||c|c} & & \\\\ [-2.5ex] \\textbf{Hiker} & \\textbf{False Statement} & \\textbf{True Statement} \\\\ [0.5ex] \\hline & & \\\\ [-2ex] \\textbf{Alice} & [6,\\infty) & [0,6) \\\\ & & \\\\ [-2.25ex] \\textbf{Bob} & [0,5] & (5,\\infty) \\\\ & & \\\\ [-2.25ex] \\textbf{Charlie} & [0,4] & (4,\\infty) \\end{array}\\] Taking the intersection of the true statements, we have \\[[0,6)\\cap(5,\\infty)\\cap(4,\\infty)=(5,6)\\cap(4,\\infty)=\\boxed{5,6}.\\] ~MRENTHUSIASM", "Think of the distances as if they are on a number line. Alice claims that $d > 6$ , Bob says $d < 5$ , while Charlie thinks $d < 4$ . This means that all possible numbers less than $5$ and greater than $6$ are included. However, since the three statements are actually false, the distance to the nearest town is one of the numbers not covered. Therefore, the answer is $\\boxed{5,6}$" ]

[ "For each of the false statements, we identify its corresponding true statement. Note that:\nWe construct the following table: \\[\\begin{array}{c||c|c} & & \\\\ [-2.5ex] \\textbf{Hiker} & \\textbf{False Statement} & \\textbf{True Statement} \\\\ [0.5ex] \\hline & & \\\\ [-2ex] \\textbf{Alice} & [6,\\infty) & [0,6) \\\\ & & \\\\ [-2.25ex] \\textbf{Bob} & [0,5] & (5,\\infty) \\\\ & & \\\\ [-2.25ex] \\textbf{Charlie} & [0,4] & (4,\\infty) \\end{array}\\] Taking the intersection of the true statements, we have \\[[0,6)\\cap(5,\\infty)\\cap(4,\\infty)=(5,6)\\cap(4,\\infty)=\\boxed{5,6}.\\] ~MRENTHUSIASM", "Think of the distances as if they are on a number line. Alice claims that $d > 6$ , Bob says $d < 5$ , while Charlie thinks $d < 4$ . This means that all possible numbers less than $5$ and greater than $6$ are included. However, since the three statements are actually false, the distance to the nearest town is one of the numbers not covered. Therefore, the answer is $\\boxed{5,6}$" ]

[ "$20$ dollars is the same as $2000$ cents, and $1.45\\%$ of $2000$ is $0.0145\\times2000=29$ cents. $\\Rightarrow\\boxed{29}$", "Since there can't be decimal values of cents, the answer must be $\\Rightarrow\\boxed{29}$" ]

[ "$20$ dollars is the same as $2000$ cents, and $1.45\\%$ of $2000$ is $0.0145\\times2000=29$ cents. $\\Rightarrow\\boxed{29}$", "Since there can't be decimal values of cents, the answer must be $\\Rightarrow\\boxed{29}$" ]

[ "Let $x$ be the total amount of votes casted. From the chart, Brenda received $30\\%$ of the votes and had $36$ votes. We can express this relationship as $\\frac{30}{100}x=36$ . Solving for $x$ , we get $x=\\boxed{120}.$", "We're being asked for the total number of votes cast -- that represents $100\\%$ of the total number of votes. Brenda received $36$ votes, which is $\\frac{30}{100} = \\frac{3}{10}$ of the total number of votes. Multiplying $36$ by $\\frac{10}{3},$ we get the total number of votes, which is $\\boxed{120}.$", "If $36$ votes is $\\frac{3}{10}$ of all the votes, we can divide that by $3$ to get $12$ as 10%, and then we can multiply the $12$ by $10$ to get to $120$ . So, the answer is $\\boxed{120}.$" ]

[ "First, we need to find the sum of each group when split. This is the total sum of all the elements divided by the # of groups. $1 + 2 \\cdots + 9 = \\frac{9(10)}{2} = 45$ . Then, dividing by $3$ , we have $\\frac{45}{3} = 15$ , so each group of $3$ must have a sum of 15. To make the counting easier, we will just see the possible groups 9 can be with. The possible groups 9 can be with 2 distinct numbers are $(9, 2, 4)$ and $(9, 1, 5)$ . Going down each of these avenues, we will repeat the same process for $8$ using the remaining elements in the list. Where there is only 1 set of elements getting the sum of $7$ $8$ needs in both cases. After $8$ is decided, the remaining 3 elements are forced in a group, yielding us an answer of $\\boxed{2}$ as our sets are $(9, 1, 5) (8, 3, 4) (7, 2, 6)$ and $(9, 2, 4) (8, 1, 6) (7, 3 ,5)$", "The group with $5$ must have the two other numbers adding up to $10$ , since the sum of all the numbers is $(1 + 2 \\cdots + 9) = \\frac{9(10)}{2} = 45$ . The sum of the numbers in each group must therefore be $\\frac{45}{3}=15$ . We can have $(1, 5, 9)$ $(2, 5, 8)$ $(3, 5, 7)$ , or $(4, 5, 6)$ . With the first group, we have $(2, 3, 4, 6, 7, 8)$ left over. The only way to form a group of $3$ numbers that add up to $15$ is with $(3, 4, 8)$ or $(2, 6, 7)$ . One of the possible arrangements is therefore $(1, 5, 9) (3, 4, 8) (2, 6, 7)$ . Then, with the second group, we have $(1, 3, 4, 6, 7, 9)$ left over. With these numbers, there is no way to form a group of $3$ numbers adding to $15$ . Similarly, with the third group there is $(1, 2, 4, 6, 8, 9)$ left over and we can make a group of $3$ numbers adding to $15$ with $(1, 6, 8)$ or $(2, 4, 9)$ . Another arrangement is $(3, 5, 7) (1, 6, 8) (2, 4, 9)$ . Finally, the last group has $(1, 2, 3, 7, 8, 9)$ left over. There is no way to make a group of $3$ numbers adding to $15$ with this, so the arrangements are $(1, 5, 9) (3, 4, 8) (2, 6, 7)$ and $(3, 5, 7) (1, 6, 8) (2, 4, 9)$ . So，there are $\\boxed{2}$ sets that can be formed.", "The sum of the numbers across all equally valued sets is $(1 + 2 \\cdots + 9) = \\frac{9(10)}{2} = 45$ . The value of the numbers in each set would be $\\frac{45}{3} = \\textbf{15}$ . We know that the numbers $9$ $8$ , and $7$ must belong in different sets, as putting any $2$ numbers in $1$ set will either pass or match the limit of $15$ per set, and we would then still need to add $1$ more number after that. Note that these numbers must be distinct, as Alina only has $1$ of each number, and order does not matter in the sets. Starting with the set that includes the number $9$ , the next two numbers must add up to $6$ , and there are $\\textbf{2}$ ways of doing this $(2,4) (1,5)$ . Note we cannot use any number past $6$ , as those numbers must be used in the other sets. The next set, which includes the number $8$ , must have two numbers that add up to $7$ , and there are $\\textbf{3}$ ways to do this $(2,5) (1,6) (3,4)$ . The final set, which includes the number $7$ , must have $2$ numbers that sum up to $8$ , and there are $\\textbf{2}$ ways to do this $(2,6) (3,5)$ . Now we have found the number of ways in which each set sums up to $15$ . To find the number of ways in which all three sets sum up to $15$ concurrently, we must take the minimum of $2$ $3$ , and $2$ , which gives us an answer of $\\boxed{2}$ triplets of sets with 3 values, in which each set sum to the same amount.", "Note that each group of numbers should sum to $\\frac{1+2+3+4+5+6+7+8+9}{3} = 15.$ Thus, this is equivalent to asking, “How many ways can you fill in a three by three magic square with the integers $1$ through $9$ ?” since we can take the three rows of the magic square as our three groups. If you have closely studied magic squares, you might know that in a three by three magic square that is to be filled in with the integers $1$ through $9$ , the center of the square would be $5$ (the average of the numbers), and the numbers in the corners should be even(*). The such pairs (disregarding order) are $(2,8)$ and $(4,6).$ Let’s fix the position of $2$ to be the top left corner. This would make $8$ in the bottom right corner. We can have either $4$ or $6$ to be in the top right corner, for a total of $\\boxed{2}$ such groups of three. (The groups are $(8,3,4) (1,5,7) (6,9,2)$ and $(8,1,6) (3,5,7) (4,9,2).$ )\nNote that if we had instead fixed the position of $4$ $6$ , or $8$ , they would correspond to one of the two cases, just in a different configuration." ]

[ "If you draw a clear diagram like the one below, it is easy to see that there are $\\boxed{49}$ points.", "Let the number of intersections be $x$ . We know that $x\\le \\dbinom{8}{4} = 70$ , as every $4$ vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. However, four diagonals intersect in the center, so we need to subtract $\\dbinom{4}{2} -1 = 5$ from this count, $70-5 = 65$ . Note that diagonals like $\\overline{AD}$ $\\overline{CG}$ , and $\\overline{BE}$ all intersect at the same point. There are $8$ of this type with three diagonals intersecting at the same point, so we need to subtract $2$ of the $\\dbinom{3}{2}$ (one is kept as the actual intersection). In the end, we obtain $65 - 16 = \\boxed{49}$", "We know that the amount of intersection points is at most $\\dbinom{8}{4} = 70$ , as in solution $2$ . There's probably going to be more than $5$ intersections counted multiple times (to get $\\textbf{(B) }65$ ), leading us to the only reasonable answer, $\\boxed{49}$ .\n-Lcz", "Like solution one, we may draw. Except note that the octagon has eight regions, and each region has an equal number of points, so drawing only one of the eight regions and the intersection points suffices. One of the eight regions contains $8$ points (not including the octagon center). However each adjacent region share one side in common and that side contains $2$ intersection points, so in actuality there are $8 - 2 = 6$ points per region. We multiply this by $8$ to get $6\\cdot 8 = 48$ and add the one center point to get $48 + 1 = \\boxed{49}$", "\nThis problem is a counting problem of combinatoric geometry. There are 2 cases for the above diagram:\nCase 1: Red Dots\nThe red dots are the intersection of 3 or more lines. It consists of 8 dots that make up an octagon and 1 dot in the center. Hence, there are $9$ red dots.\nCase 2: Blue Dots\nThe blue dots are the intersection of 2 lines. Each vertex of the octagon has 2 purple lines, 2 green lines, and 1 orange line coming out of it. There are 5 dots of intersection on a purple line, 6 dots on a green line, and 5 dots on an orange line. There are $2 \\cdot 5+2 \\cdot 6+5=27$ dots that come out of 1 vertex, which includes 7 red dots already counted. So there are $27-7 = 20$ blue dots coming out of 1 vertex. There are 8 vertices, but each blue dot is the intersection of 2 lines, corresponding to $2 \\cdot 2 = 4$ vertices. So there are $\\frac{20 \\cdot 8}{4} = 40$ blue dots.\nThe number of intersection dots are the sum of the number of red and blue dots. Hence, the answer is $40 + 9 = \\boxed{49}$" ]

[ "If all lines satisfy the condition, then we can just plug in values for $a$ $b$ , and $c$ that form an arithmetic progression. Let's use $a=1$ $b=2$ $c=3$ , and $a=1$ $b=3$ $c=5$ . Then the two lines we get are: \\[x+2y=3\\] \\[x+3y=5\\] Use elimination to deduce \\[y = 2\\] and plug this into one of the previous line equations. We get \\[x+4 = 3 \\Rightarrow x=-1\\] Thus the common point is $\\boxed{1,2}$", "We know that $a$ $b$ , and $c$ form an arithmetic progression, so if the common difference is $d$ , we can say $a,b,c = a, a+d, a+2d.$ Now we have $ax+ (a+d)y = a+2d$ , and expanding gives $ax + ay + dy = a + 2d.$ Factoring gives $a(x+y-1)+d(y-2) = 0$ . Since this must always be true (regardless of the values of $a$ and $d$ ), we must have $x+y-1 = 0$ and $y-2 = 0$ , so $x,y = -1, 2,$ and the common point is $\\boxed{1,2}$" ]

[ "The only digits available to use in the phone number are $2$ $3$ $4$ $5$ $6$ $7$ $8$ , and $9$ . There are only $7$ spots left among the $8$ numbers, so we need to find the number of ways to choose $7$ numbers from $8$ . The answer is then $\\dbinom{8}{7}=\\dfrac{8!}{7!\\,(8-7)!}=\\boxed{8}$" ]

[ "The $6$ green marbles and yellow marbles form $1 - \\frac{1}{3} - \\frac{1}{4} = \\frac{5}{12}$ of the total marbles. Now, suppose the total number of marbles is $x$ . We know the number of yellow marbles is $\\frac{5}{12}x - 6$ and a positive integer. Therefore, $12$ must divide $x$ . Trying the smallest multiples of $12$ for $x$ , we see that when $x = 12$ , we get there are $-1$ yellow marbles, which is impossible. However when $x = 24$ , there are $\\frac{5}{12} \\cdot 24 - 6 = \\boxed{4}$ yellow marbles, which must be the smallest possible.", "Since $\\frac{1}{3}$ of the marbles are blue and $\\frac{1}{4}$ are red, it is clear that the total number of marbles must be divisible by $12$ . If there are $12$ marbles, then $4$ are blue, $3$ are red, and $6$ are green, meaning that there are $-1$ yellow marbles. This is impossible. Trying the next multiple of $12$ $24$ , we find that $8$ are green, $6$ are red, and $6$ are green, meaning that the minimum number of yellow marbles is $\\boxed{4}$" ]

[ "All the tens digits of the product will be irrelevant to finding the units digit. Thus, we are searching for the units digit of $(2\\cdot 4\\cdot 6 \\cdot 8) \\cdot (2 \\cdot 4 \\cdot 6 \\cdot 8) \\cdot (2\\cdot 4\\cdot 6 \\cdot 8) \\cdot ...$\nThere will be $10$ groups of $4$ numbers. The number now can be rewritten as $(2\\cdot 4 \\cdot 6 \\cdot 8)^{10}$\nMultiplying, we get $384^{10}$\nAgain, we can disregard the tens and hundreds digit of $384$ , since we only want the units digit of the number, leaving $4^{10}$\nNow, we try to find a pattern to the units digit of $4^n$ . To compute this quickly, we once again discard all tens digits and higher.\n$4^1 = 4$\n$4^2 = 4\\cdot 4 = 1\\underline{6}$ , discard the $1$\n$4^3 = 1 \\cdot 4 = \\underline{4}$\n$4^4 = 4 \\cdot 4 = 1\\underline{6}$ , discard the $1$\n$4^5 = 1 \\cdot 4 = \\underline{4}$\nThose equalities are, in reality, congruences $\\mod {10}$\nThus, the pattern of the units digits is $\\{4, 6, 4, 6, 4, 6, 4, 6, 4, 6\\}$ . The cycle repeats so that term $n$ is the same as term $n+2$ . The tenth number in the cycle is $6$ , giving an answer of $\\boxed{6}$", "Again, the value we seek is equal to $(2\\cdot 4\\cdot 6\\cdot 8)^{10}\\mod 10$ .\nWe can use equivalence to simplify.\n$(2\\cdot 4\\cdot 6\\cdot 8)^{10}\\mod 10$\n$\\equiv (2\\cdot 4\\cdot (-4)\\cdot (-2))^{10} \\mod 10$\n$\\equiv 64^{10} \\mod 10$\n$\\equiv 4^{10} \\mod 10$\n$\\equiv 2^{20} \\mod 10$\n$\\equiv 8^6\\cdot 2^2 \\mod 10$\n$\\equiv (-2)^6\\cdot 2^2 \\mod 10$\n$\\equiv 2^8 \\mod 10$\n$\\equiv 256 \\mod 10$\n$\\equiv 6 \\mod 10$\nThus our answer is $\\boxed{6}$" ]

[ "Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is $1$ , and thus its base is $2.$\nLet $h$ be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and $\\triangle{ADE}$ . The top triangle has a base of $3\\cdot{2}=6$ , and $DE=4\\cdot{2}=8.$ The height of $\\triangle{ADE}$ is $h+1$ , therefore our ratio is $\\frac{h}{6}=\\frac{h+1}{8}$ , which yields $h=3$ as our answer.\nTo find the area of the trapezoid, we can take the area of $\\triangle{ABC}$ and subtract the area of $\\triangle{ADE},$ whose base is $8$ and height $3+1=4$ . It follows that the area of $\\triangle{ADE}=16$ , and subtracting this from $40$ gives us $40-16=\\boxed{24}$" ]

[ "Let $x$ be the area of $ADE$ . Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$ ). Thus, we have \\[x=7+\\dfrac{9}{16}x.\\] This gives $x=16$ , so the area of $DBCE=40-x=\\boxed{24}$" ]

[ "Let the base length of the small triangle be $x$ . Then, there is a triangle $ADE$ encompassing the 7 small triangles and sharing the top angle with a base length of $4x$ . Because the area is proportional to the square of the side, let the base $BC$ be $\\sqrt{40}x$ . The ratio of the area of triangle $ADE$ to triangle $ABC$ is $\\left(\\frac{4x}{\\sqrt{40}x}\\right)^2 = \\frac{16}{40}$ . The problem says the area of triangle $ABC$ is $40$ , so the area of triangle $ADE$ is $16$ . So the area of trapezoid $DBCE$ is $40 - 16 = \\boxed{24}$", "Notice $\\big[DBCE\\big]=\\big[ABC\\big]-\\big[ADE\\big]$ .\nLet the base of the small triangles of area 1 be $x$ , then the base length of $\\Delta ADE=4x$ . Notice, $\\left(\\frac{DE}{BC}\\right)^2=\\frac{1}{40}\\implies \\frac{x}{BC}=\\frac{1}{\\sqrt{40}}$ , then $4x=\\frac{4BC}{\\sqrt{40}}\\implies \\big[ADE\\big]=\\left(\\frac{4}{\\sqrt{40}}\\right)^2\\cdot \\big[ABC\\big]=\\frac{2}{5}\\big[ABC\\big]$ Thus, $\\big[DBCE\\big]=\\big[ABC\\big]-\\big[ADE\\big]=\\big[ABC\\big]\\left(1-\\frac{2}{5}\\right)=\\frac{3}{5}\\cdot 40=\\boxed{24}.$", "The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$ . Therefore the area of the trapezoid is $40-16=\\boxed{24}$", "You can see that we can create a \"stack\" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$ , so to find the area of such trapezoid $BCED$ , we just take $40-16=\\boxed{24}$ , like so.", "The combined area of the small triangles is $7$ , and from the fact that each small triangle has an area of $1$ , we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\\frac{1}{3}$ , so will their areas have a proportion that is the square of the proportion of their sides, or $\\frac {1}{9}$ ). Thus, the combined area of the top triangle and the trapezoid immediately below is $7 + 9 = 16$ . The area of trapezoid $BCED$ is thus the area of triangle $ABC-16 =\\boxed{24}$", "You can assume for the base of one of the smaller triangles to be $\\frac{1}{a}$ and the height to be $2a$ , giving an area of 1. The larger triangle above the 7 smaller ones then has base $\\frac{3}{a}$ and height $6a$ , giving it an area of $9$ . Then the area of triangle $ADE$ is $16$ and $40-16=\\boxed{24}$", "You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence $7\\cdot\\left(1+\\left(\\frac{9}{16}\\right)+\\left(\\frac{9}{16}\\right)^2+\\left(\\frac{9}{16}\\right)^3\\dots\\right)$ . Using the infinite geometric series sum formula gives us $7\\cdot\\left(\\frac{1}{1-\\frac{9}{16}}\\right)=7\\cdot\\frac{16}{7}=16$ . The triangle's area would thus be $40-16=\\boxed{24}$" ]

[ "Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is $1$ , and thus its base is $2.$\nLet $h$ be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and $\\triangle{ADE}$ . The top triangle has a base of $3\\cdot{2}=6$ , and $DE=4\\cdot{2}=8.$ The height of $\\triangle{ADE}$ is $h+1$ , therefore our ratio is $\\frac{h}{6}=\\frac{h+1}{8}$ , which yields $h=3$ as our answer.\nTo find the area of the trapezoid, we can take the area of $\\triangle{ABC}$ and subtract the area of $\\triangle{ADE},$ whose base is $8$ and height $3+1=4$ . It follows that the area of $\\triangle{ADE}=16$ , and subtracting this from $40$ gives us $40-16=\\boxed{24}$" ]

[ "Let $x$ be the area of $ADE$ . Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$ ). Thus, we have \\[x=7+\\dfrac{9}{16}x.\\] This gives $x=16$ , so the area of $DBCE=40-x=\\boxed{24}$" ]

[ "Let the base length of the small triangle be $x$ . Then, there is a triangle $ADE$ encompassing the 7 small triangles and sharing the top angle with a base length of $4x$ . Because the area is proportional to the square of the side, let the base $BC$ be $\\sqrt{40}x$ . The ratio of the area of triangle $ADE$ to triangle $ABC$ is $\\left(\\frac{4x}{\\sqrt{40}x}\\right)^2 = \\frac{16}{40}$ . The problem says the area of triangle $ABC$ is $40$ , so the area of triangle $ADE$ is $16$ . So the area of trapezoid $DBCE$ is $40 - 16 = \\boxed{24}$", "Notice $\\big[DBCE\\big]=\\big[ABC\\big]-\\big[ADE\\big]$ .\nLet the base of the small triangles of area 1 be $x$ , then the base length of $\\Delta ADE=4x$ . Notice, $\\left(\\frac{DE}{BC}\\right)^2=\\frac{1}{40}\\implies \\frac{x}{BC}=\\frac{1}{\\sqrt{40}}$ , then $4x=\\frac{4BC}{\\sqrt{40}}\\implies \\big[ADE\\big]=\\left(\\frac{4}{\\sqrt{40}}\\right)^2\\cdot \\big[ABC\\big]=\\frac{2}{5}\\big[ABC\\big]$ Thus, $\\big[DBCE\\big]=\\big[ABC\\big]-\\big[ADE\\big]=\\big[ABC\\big]\\left(1-\\frac{2}{5}\\right)=\\frac{3}{5}\\cdot 40=\\boxed{24}.$", "The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$ . Therefore the area of the trapezoid is $40-16=\\boxed{24}$", "You can see that we can create a \"stack\" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$ , so to find the area of such trapezoid $BCED$ , we just take $40-16=\\boxed{24}$ , like so.", "The combined area of the small triangles is $7$ , and from the fact that each small triangle has an area of $1$ , we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\\frac{1}{3}$ , so will their areas have a proportion that is the square of the proportion of their sides, or $\\frac {1}{9}$ ). Thus, the combined area of the top triangle and the trapezoid immediately below is $7 + 9 = 16$ . The area of trapezoid $BCED$ is thus the area of triangle $ABC-16 =\\boxed{24}$", "You can assume for the base of one of the smaller triangles to be $\\frac{1}{a}$ and the height to be $2a$ , giving an area of 1. The larger triangle above the 7 smaller ones then has base $\\frac{3}{a}$ and height $6a$ , giving it an area of $9$ . Then the area of triangle $ADE$ is $16$ and $40-16=\\boxed{24}$", "You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence $7\\cdot\\left(1+\\left(\\frac{9}{16}\\right)+\\left(\\frac{9}{16}\\right)^2+\\left(\\frac{9}{16}\\right)^3\\dots\\right)$ . Using the infinite geometric series sum formula gives us $7\\cdot\\left(\\frac{1}{1-\\frac{9}{16}}\\right)=7\\cdot\\frac{16}{7}=16$ . The triangle's area would thus be $40-16=\\boxed{24}$" ]

[ "$AB = EC$ because they are opposite sides of a square. Also, $ED = DC = AB$ because all sides of the convex pentagon are of equal length. Since $ABCE$ is a square and $\\triangle CED$ is an equilateral triangle, $\\angle AEC = 90$ and $\\angle CED = 60.$ Use angle addition: \\[\\angle E = \\angle AEC + \\angle CED = 90 + 60 = \\boxed{150}\\]" ]

[ "Since she has half as many red marbles as green, we can call the number of red marbles $x$ , and the number of green marbles $2x$ .\nSince she has half as many green marbles as blue, we can call the number of blue marbles $4x$ . \nAdding them up, we have $7x$ marbles. The number of marbles therefore must be a multiple of $7$ . The only possible answer is $\\boxed{28}.$", "Suppose Maria has $g$ green marbles and let $t$ be the total number of marbles. She then has $\\frac{g}{2}$ red marbles and $2g$ blue marbles. Altogether, Maria has \\[g + \\frac{g}{2} + 2g = \\frac{7g}{2} = t\\] marbles, implying that $g = \\dfrac{2t}{7},$ so $t$ must be a multiple of $7$ . The only multiple of $7$ is $\\boxed{28}.$" ]

[ "Consecutive numbers share an edge. That means that it is possible to walk from $1$ to $9$ by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity: But since there are only four even numbers in the set, the five darker squares must contain the odd numbers, which sum to $1+3+5+7+9=25.$ Therefore if the sum of the numbers in the corners is $18$ , the number in the center must be $\\boxed{7}$", "Quick testing shows that \\[3~2~1\\] \\[4~7~8\\] \\[5~6~9\\] is a valid solution. $3+1+5+9 = 18$ , and the numbers follow the given condition. The center number is found to be $\\boxed{7}$ .. — @adihaya talk ) 12:27, 21 February 2016 (EST) ~edited", "First let the numbers be \\[1 ~8~ 7\\] \\[2 ~ 9 ~6\\] \\[3 ~ 4~ 5\\] with the numbers $1-8$ around the outsides and $9$ in the middle. We see that the sum of the four corner numbers is $16$ . If we switch $7$ and $9$ , then the corner numbers will add up to $18$ and the consecutive numbers will still be touching each other. The answer is $\\boxed{7}$ . ~edited", "Testing out the box with the center square taking on the value of 5 and 6, we find that they either do not satisfy the first or the second condition. Testing 7, we find that a valid configuration is \\[1 ~8~ 9\\] \\[2 ~ 7 ~6\\] \\[3 ~ 4~ 5\\]\n$\\boxed{7}$" ]

[ "Let us call the six faces of our cube $a,b,c,d,e,$ and $f$ (where $a$ is opposite $d$ $c$ is opposite $e$ , and $b$ is opposite $f$ .\nThus, for the eight vertices, we have the following products: $abc,abe,bcd,bde,acf,cdf,aef,$ and $def$ .\nLet us find the sum of these products: \\[abc+abe+bcd+bde+acf+cdf+aef+def\\] We notice $b$ is a factor of the first four terms, and $f$ is a factor of the last four terms. \\[b(ac+ae+cd+de)+f(ac+ae+cd+de)\\] Now, we can factor even more:\n\\begin{align*} & (b+f)(ac+ae+cd+de) \\\\ = &(b+f)(a(c+e)+d(c+e)) \\\\ = &(b+f)(a+d)(c+e) \\end{align*} We have the product. Notice how the factors are sums of opposite faces. The greatest sum possible is $(7+2)$ $(6+3)$ , and $(5+4)$ all factors. \\begin{align*} & (7+2)(6+3)(5+4) \\\\ = & 9 \\cdot 9 \\cdot 9 \\\\ = & 729. \\end{align*} Thus our answer is $\\boxed{729}$", "We will use parity. If we attempt to maximize this cube in any given way, for example making sure that the sides with 5,6 and 7 all meet at one single corner, the first two answers clearly are out of bounds. Now notice the fact that any three given sides will always meet at one of the eight points. Also note the fact that there are 3 odd numbers. This means that there must be one side that has an odd area! Any odd number added with even numbers is always odd. Given that both c) and e) are both even, d) is our only choice.\nThus our answer is $\\boxed{729}$", "We first find the factorization $(b+f)(a+d)(c+e)$ using the method in Solution 1. By using AM-GM, we get, $(b+f)(a+d)(c+e) \\le \\left( \\frac{a+b+c+d+e+f}{3} \\right)^3$ . To maximize the factorization, we get the answer is $\\left( \\frac{27}{3} \\right)^3 = \\boxed{729}$", "First, we intuitivly notice that multiplying large numbers together and smaller numbers together tends to produce larger sums, while multiplying large numbers with smaller numbers tends to produce smaller sums. From this, we guess that it is optimal to have $7, 6,$ and $5$ to be around one vertex to produce at least one large product of $210$ . We can immediately eliminate a) and b) as answer choices since they are very close to this product, as well as e) since it is exactly $8 \\cdot 210$ and we know that every following product will be smaller than this. We also guess that that is is most likely optimal to create a pairing where a number and its complement (the number that's the difference of 9 and this number) are on opposite sides. Using these two guesses, we can construct a net of sides and brute force the solution (or estimate) leaving us with $\\boxed{729}$" ]

[ "First assign each face the letters $a,b,c,d,e,f$ . The sum of the product of the faces is $abc+acd+ade+aeb+fbc+fcd+fde+feb$ . We can factor this into $(a+f)(b+c)(d+e)$ which is the product of the sum of each pair of opposite faces. In order to maximize $(a+f)(b+c)(d+e)$ we use the numbers $(7+2)(6+3)(5+4)$ or $\\boxed{729}$", "We'll proceed from the factoring process above.\nBy the AM-GM inequality,\n\\[\\frac{a_1+a_2+a_3}{3}\\geq\\sqrt[3]{a_1a_2a_3}\\]\nCubing both sides,\n\\[\\left(\\frac{a_1+a_2+a_3}{3}\\right)^3\\geq{a_1a_2a_3}\\]\nLet $a_1=(a+f)$ $a_2=(b+c)$ , and $a_3=(d+e)$ . Let's substitute in these values.\n\\[\\left(\\frac{a+b+c+d+e+f}{3}\\right)^3\\geq{(a+f)(b+c)(d+e)}\\]\n$a+b+c+d+e+f$ is fixed at 27.\n\\[\\left(\\frac{27}{3}\\right)^3\\geq{(a+f)(b+c)(d+e)}\\]\n\\[\\boxed{729}\\]", "First, we see that we want to try and maximize each vertex. Since the multiplication of each vertex is the product of three values, we want to maximize those three values. Doing so, we see that we want them to be as close as possible, giving $4.5^3$ (the average of all the values). This gives us the maximum for each vertex, multiplied by the 8 vertices, yields our answer $\\boxed{729}$ Also note that if you cannot evaluate $4.5^3$ quickly, a rough approximation of $5*4.5*4*8$ will yield 720, very close to our answer. -rayprati", "It is obvious to put $5$ $6$ , and $7$ on the faces that share the same vertex. As $4$ is the next biggest number, the face with $4$ has to be next to the faces with $6$ and $7$ . As $4$ is the next biggest number, the face with $3$ has to be next to the faces with $5$ and $7$ . making the face with $2$ next to the faces with $6$ and $5$\nTherefore the answer is $7 \\cdot 5 \\cdot 6 + 7 \\cdot 4 \\cdot 6 + 7 \\cdot 3 \\cdot 5 + 7 \\cdot 3 \\cdot 4 + 2 \\cdot 5 \\cdot 6 + 2 \\cdot 4 \\cdot 6 + 2 \\cdot 3 \\cdot 5 + 2 \\cdot 3 \\cdot 4 = \\boxed{729}$" ]

[ "By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$ . By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$ , so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$ . Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \\boxed{88}$", "Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$ . Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\\binom {6}{3} = 20$ products $r_a \\cdot r_b \\cdot r_c,$ we obtain \\[B= - \\left(\\binom {4}{3} \\binom {2}{0} \\cdot 2^{3} + \\binom {4}{2} \\binom{2}{1} \\cdot 2^{2} \\cdot 1 + \\binom {4}{1} \\binom {2}{2} \\cdot 2 \\right) = -\\left(32+48+8 \\right) = \\boxed{88}.\\] ~ike.chen" ]

[ "By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$ . By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$ , so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$ . Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \\boxed{88}$", "Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$ . Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\\binom {6}{3} = 20$ products $r_a \\cdot r_b \\cdot r_c,$ we obtain \\[B= - \\left(\\binom {4}{3} \\binom {2}{0} \\cdot 2^{3} + \\binom {4}{2} \\binom{2}{1} \\cdot 2^{2} \\cdot 1 + \\binom {4}{1} \\binom {2}{2} \\cdot 2 \\right) = -\\left(32+48+8 \\right) = \\boxed{88}.\\] ~ike.chen" ]

[ "Let the number of students be $n\\geq 5$ . Then the sum of their scores is at least $5\\cdot 100 + (n-5)\\cdot 60$ . At the same time, we need to achieve the mean $76$ , which is equivalent to achieving the sum $76n$\nHence we get a necessary condition on $n$ : we must have $5\\cdot 100 + (n-5)\\cdot 60 \\leq 76n$ .\nThis can be simplified to $200 \\leq 16n$ . The smallest integer $n$ for which this is true is $n=13$\nTo finish our solution, we now need to find one way how $13$ students could have scored on the test. We have $13\\cdot 76 = 988$ points to divide among them. The five $100$ s make $500$ , hence we must divide the remaining $488$ points among the other $8$ students. This can be done e.g. by giving $61$ points to each of them.\nHence the smallest possible number of students is $\\boxed{13}$" ]

[ "Let the point where the height of the triangle intersects with the base be $D$ . Now we can guess what $x$ is and find $y$ . If $x$ is $3$ , then $y$ is $9$ . The cords of $B$ and $C$ would be $(-3,9)$ and $(3,9)$ , respectively. The distance between $B$ and $C$ is $6$ , meaning the area would be $\\frac{6 \\cdot 9}{2}=27$ , not $64$ . Now we let $x=4$ $y$ would be $16$ . The cords of $B$ and $C$ would be $(-4,16)$ and $(4,16)$ , respectively. $BC$ would be $8$ , and the height would be $16$ . The area would then be $\\frac{8 \\cdot 16}{2}$ which is $64$ , so $BC$ is $\\boxed{8}$" ]

[ "By: Albert471\nPlotting points $B$ and $C$ on the graph shows that they are at $\\left( -x,x^2\\right)$ and $\\left( x,x^2\\right)$ , which is isosceles. By setting up the triangle area formula you get: $64=\\frac{1}{2}*2x*x^2 = 64=x^3$ Making x=4, and the length of $BC$ is $2x$ , so the answer is $\\boxed{8}$" ]

[ "Let the three points be at $A = (x_1, x_1^2)$ $B = (x_2, x_2^2)$ , and $C = (x_3, x_3^2)$ , such that the slope between the first two is $2$ , and $A$ is the point with the least $y$ -coordinate.\nThen, we have $\\textrm{Slope of }AC = \\frac{x_1^2 - x_3^2}{x_1 - x_3} = x_1 + x_3$ . Similarly, the slope of $BC$ is $x_2 + x_3$ , and the slope of $AB$ is $x_1 + x_2 = 2$ . The desired sum is $x_1 + x_2 + x_3$ , which is equal to the sum of the slopes divided by $2$\nTo find the slope of $AC$ , we note that it comes at a $60^{\\circ}$ angle with $AB$ . Thus, we can find the slope of $AC$ by multiplying the two complex numbers $1 + 2i$ and $1 + \\sqrt{3}i$ . What this does is generate the complex number that is at a $60^{\\circ}$ angle with the complex number $1 + 2i$ . Then, we can find the slope of the line between this new complex number and the origin: \\[(1+2i)(1+\\sqrt{3}i)\\] \\[= 1 - 2\\sqrt{3} + 2i + \\sqrt{3}i\\] \\[\\textrm{Slope } = \\frac{2 + \\sqrt{3}}{1 - 2\\sqrt{3}}\\] \\[= \\frac{2 + \\sqrt{3}}{1 - 2\\sqrt{3}} \\cdot \\frac{1 + 2\\sqrt{3}}{1 + 2\\sqrt{3}}\\] \\[= \\frac{8 + 5\\sqrt{3}}{-11}\\] \\[= \\frac{-8 - 5\\sqrt{3}}{11}.\\] The slope $BC$ can also be solved similarly, noting that it makes a $120^{\\circ}$ angle with $AB$ \\[(1+2i)(-1+\\sqrt{3}i)\\] \\[= -1 - 2\\sqrt{3} - 2i + \\sqrt{3}i\\] \\[\\textrm{Slope } = \\frac{\\sqrt{3} - 2}{-2\\sqrt{3} - 1}\\] \\[= \\frac{2 - \\sqrt{3}}{1 + 2\\sqrt{3}} \\cdot \\frac{1 - 2\\sqrt{3}}{1 - 2\\sqrt{3}}\\]\nAt this point, we start to notice a pattern: This expression is equal to $\\frac{2 + \\sqrt{3}}{1 - 2\\sqrt{3}} \\cdot \\frac{1 + 2\\sqrt{3}}{1 + 2\\sqrt{3}}$ , except the numerators of the first fractions are conjugates! Notice that this means that when we multiply out, the rational term will stay the same, but the coefficient of $\\sqrt{3}$ will have its sign switched. This means that the two complex numbers are conjugates, so their irrational terms cancel out.\nOur sum is simply $2 - 2\\cdot\\frac{8}{11} = \\frac{6}{11}$ , and thus we can divide by $2$ to obtain $\\frac{3}{11}$ , which gives the answer $\\boxed{14}$", "Using the slope formula and differences of squares, we find:\n$a+b$ = the slope of $AB$\n$b+c$ = the slope of $BC$\n$a+c$ = the slope of $AC$\nSo the value that we need to find is the sum of the slopes of the three sides of the triangle divided by $2$ . Without loss of generality, let $AB$ be the side that has the smallest angle with the positive $x$ -axis. Let $J$ be an arbitrary point with the coordinates $(1, 0)$ . Translate the triangle so $A$ is at the origin. Then $\\tan(BOJ) = 2$ . Since the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the answer is $\\dfrac{\\tan(BOJ) + \\tan(BOJ+60) + \\tan(BOJ-60)}{2}$\nUsing $\\tan(BOJ) = 2$ , and the tangent addition formula, this simplifies to $\\dfrac{3}{11}$ , so the answer is $3 + 11 = \\boxed{14}$" ]

[ "Suppose that the race is $d$ miles long. The water stations are located at \\[\\frac{d}{8}, \\frac{2d}{8}, \\ldots, \\frac{7d}{8}\\] miles from the start, and the repair stations are located at \\[\\frac{d}{3}, \\frac{2d}{3}\\] miles from the start.\nWe are given that $\\frac{3d}{8}=\\frac{d}{3}+2,$ from which \\begin{align*} \\frac{9d}{24}&=\\frac{8d}{24}+2 \\\\ \\frac{d}{24}&=2 \\\\ d&=\\boxed{48} ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM" ]

[ "Let $q$ be the number of questions Beta takes on day 1 and $a$ be the number he gets right. Let $b$ be the number he gets right on day 2.\nThese inequalities follow: \\[\\frac{a}{q} < \\frac{160}{300} = \\frac{8}{15}\\] \\[\\frac{b}{500-q} < \\frac{140}{200} = \\frac{7}{10}\\] Solving for a and b and adding the two inequalities: \\[a + b < \\frac{8}{15}q + (350 - \\frac{7}{10}q)\\] \\[a + b < 350 - \\frac{1}{6}q\\] From here, we see the largest possible value of $a+b$ is $349$\nChecking our conditions, we know that $a$ must be positive so therefore $q$ must be positive. A quick check shows that for $2\\le q \\le 5$ $q$ follows all the conditions and results in $a+b=349$\nThis makes Beta's success ratio $\\frac{349}{500}$ . Thus, the answer is $m+n = 349 + 500 = \\boxed{849}$" ]

[ "Let $P$ be the probability Amelia wins. Note that $P = \\text{chance she wins on her first turn} + \\text{chance she gets to her turn again}\\cdot P$ , since if she gets to her turn again, she is back where she started with probability of winning $P$ . The chance she wins on her first turn is $\\frac{1}{3}$ . The chance she makes it to her turn again is a combination of her failing to win the first turn - $\\frac{2}{3}$ and Blaine failing to win - $\\frac{3}{5}$ . Multiplying gives us $\\frac{2}{5}$ . Thus, \\[P = \\frac{1}{3} + \\frac{2}{5}P\\] Therefore, $P = \\frac{5}{9}$ , so the answer is $9-5=\\boxed{4}$", "Let $P$ be the probability Amelia wins. Note that $P = \\text{chance she wins on her first turn} + \\text{chance she gets to her second turn}\\cdot \\frac{1}{3} + \\text{chance she gets to her third turn}\\cdot \\frac{1}{3} \\cdots$ This can be represented as an infinite geometric series: \\[P=\\frac{\\frac{1}{3}}{1-\\frac{2}{3}\\cdot \\frac{3}{5}} = \\frac{\\frac{1}{3}}{1-\\frac{2}{5}} = \\frac{\\frac{1}{3}}{\\frac{3}{5}} = \\frac{1}{3}\\cdot \\frac{5}{3} = \\frac{5}{9}.\\] Therefore, $P = \\frac{5}{9}$ , so the answer is $9-5 = \\boxed{4}.$" ]

[ "We can solve this by using 'casework,' the cases being:\nCase 1: Amelia wins on her first turn.\nCase 2 Amelia wins on her second turn.\nand so on.\nThe probability of her winning on her first turn is $\\dfrac13$ . The probability of all the other cases is determined by the probability that Amelia and Blaine all lose until Amelia's turn on which she is supposed to win. So, the total probability of Amelia winning is: \\[\\dfrac{1}{3}+\\left(\\dfrac{2}{3}\\cdot\\dfrac{3}{5}\\right)\\cdot\\dfrac{1}{3}+\\left(\\dfrac{2}{3}\\cdot\\dfrac{3}{5}\\right)^2\\cdot\\dfrac{1}{3}+\\cdots.\\] Factoring out $\\dfrac13$ we get a geometric series: \\[\\dfrac{1}{3}\\left(1+\\dfrac{2}{5}+\\left(\\dfrac{2}{5}\\right)^2+\\cdots\\right) = \\dfrac{1}{3}\\cdot\\dfrac{1}{3/5} = \\boxed{59}.\\]" ]

[ "Let the ordered triple $(a,b,c)$ denote that $a$ songs are liked by Amy and Beth, $b$ songs by Beth and Jo, and $c$ songs by Jo and Amy. The only possible triples are $(1,1,1), (2,1,1), (1,2,1)(1,1,2)$\nTo show this, observe these are all valid conditions. Second, note that none of $a,b,c$ can be bigger than 3. Suppose otherwise, that $a = 3$ . Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either $b$ or $c$ to be at least 1. In fact, we require either $b$ or $c$ to equal 1, otherwise there will be a song liked by all three. Suppose $b = 1$ . Then we must have $c=0$ since no song is liked by all three girls, a contradiction.\nCase 1 : How many ways are there for $(a,b,c)$ to equal $(1,1,1)$ ? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So $(a,b,c)=(1,1,1)$ in $4\\cdot3\\cdot2\\cdot4 = 96$ ways.\nCase 2 : To find the number of ways for $(a,b,c) = (2,1,1)$ , observe there are $\\binom{4}{2} = 6$ choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are $6\\cdot2\\cdot3=36$ ways for the girls to like the songs.\nThat gives a total of $96 + 36 = 132$ ways for the girls to like the songs, so the answer is $\\boxed{132}$", "Let $AB, BJ$ , and $AJ$ denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let $A, B, J,$ and $N$ denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least $1\\: AB, BJ$ , and $AJ$ , they must be $3$ songs out of the $4$ that Amy, Beth, and Jo listened to. The fourth song can be of any type $N, A, B, J, AB, BJ$ , and $AJ$ (there is no $ABJ$ because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange $AB, BJ, AJ$ , and a song from the set $\\{N, A, B, J, AB, BJ, AJ\\}$\nCase 1: Fourth song = $N, A, B, J$\nNote that in Case 1, all four of the choices for the fourth song are different from the first three songs.\nNumber of ways to rearrange = $(4!)$ rearrangements for each choice $*\\: 4$ choices = $96$\nCase 2: Fourth song = $AB, BJ, AJ$\nNote that in Case $2$ , all three of the choices for the fourth song repeat somewhere in the first three songs.\nNumber of ways to rearrange = $(4!/2!)$ rearrangements for each choice $*\\: 3$ choices = $36$\n$96 + 36 = \\boxed{132}$", "There are $\\binom{4}{3}$ ways to choose the three songs that are liked by the three pairs of girls.\nThere are $3!$ ways to determine how the three songs are liked, or which song is liked by which pair of girls.\nIn total, there are $\\binom{4}{3}\\cdot3!$ possibilities for the first $3$ songs.\nThere are $3$ cases for the 4th song, call it song D.\nCase $1$ : D is disliked by all $3$ girls $\\implies$ there is only $1$ possibility.\nCase $2$ : D is liked by exactly $1$ girl $\\implies$ there are $3$ possibility.\nCase $3$ : D is liked by exactly $2$ girls $\\implies$ there are $3$ pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first $3$ songs liked by the same girls.\nCounting the overlaps, there are $3$ ways to choose the pair with overlaps and $4\\cdot3=12$ ways to choose what the other $2$ pairs like independently. In total, there are $3\\cdot12=36$ overlapped possibilities.\nFinally, there are $\\binom{4}{3}\\cdot3!\\cdot(3+1+3)-36=132$ ways for the songs to be likely by the girls. $\\boxed{132}$" ]

[ "Let the ordered triple $(a,b,c)$ denote that $a$ songs are liked by Amy and Beth, $b$ songs by Beth and Jo, and $c$ songs by Jo and Amy. The only possible triples are $(1,1,1), (2,1,1), (1,2,1)(1,1,2)$\nTo show this, observe these are all valid conditions. Second, note that none of $a,b,c$ can be bigger than 3. Suppose otherwise, that $a = 3$ . Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either $b$ or $c$ to be at least 1. In fact, we require either $b$ or $c$ to equal 1, otherwise there will be a song liked by all three. Suppose $b = 1$ . Then we must have $c=0$ since no song is liked by all three girls, a contradiction.\nCase 1 : How many ways are there for $(a,b,c)$ to equal $(1,1,1)$ ? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So $(a,b,c)=(1,1,1)$ in $4\\cdot3\\cdot2\\cdot4 = 96$ ways.\nCase 2 : To find the number of ways for $(a,b,c) = (2,1,1)$ , observe there are $\\binom{4}{2} = 6$ choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are $6\\cdot2\\cdot3=36$ ways for the girls to like the songs.\nThat gives a total of $96 + 36 = 132$ ways for the girls to like the songs, so the answer is $\\boxed{132}$", "Let $AB, BJ$ , and $AJ$ denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let $A, B, J,$ and $N$ denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least $1\\: AB, BJ$ , and $AJ$ , they must be $3$ songs out of the $4$ that Amy, Beth, and Jo listened to. The fourth song can be of any type $N, A, B, J, AB, BJ$ , and $AJ$ (there is no $ABJ$ because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange $AB, BJ, AJ$ , and a song from the set $\\{N, A, B, J, AB, BJ, AJ\\}$\nCase 1: Fourth song = $N, A, B, J$\nNote that in Case 1, all four of the choices for the fourth song are different from the first three songs.\nNumber of ways to rearrange = $(4!)$ rearrangements for each choice $*\\: 4$ choices = $96$\nCase 2: Fourth song = $AB, BJ, AJ$\nNote that in Case $2$ , all three of the choices for the fourth song repeat somewhere in the first three songs.\nNumber of ways to rearrange = $(4!/2!)$ rearrangements for each choice $*\\: 3$ choices = $36$\n$96 + 36 = \\boxed{132}$", "There are $\\binom{4}{3}$ ways to choose the three songs that are liked by the three pairs of girls.\nThere are $3!$ ways to determine how the three songs are liked, or which song is liked by which pair of girls.\nIn total, there are $\\binom{4}{3}\\cdot3!$ possibilities for the first $3$ songs.\nThere are $3$ cases for the 4th song, call it song D.\nCase $1$ : D is disliked by all $3$ girls $\\implies$ there is only $1$ possibility.\nCase $2$ : D is liked by exactly $1$ girl $\\implies$ there are $3$ possibility.\nCase $3$ : D is liked by exactly $2$ girls $\\implies$ there are $3$ pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first $3$ songs liked by the same girls.\nCounting the overlaps, there are $3$ ways to choose the pair with overlaps and $4\\cdot3=12$ ways to choose what the other $2$ pairs like independently. In total, there are $3\\cdot12=36$ overlapped possibilities.\nFinally, there are $\\binom{4}{3}\\cdot3!\\cdot(3+1+3)-36=132$ ways for the songs to be likely by the girls. $\\boxed{132}$" ]

[ "By counting the number of green cubes $2$ different ways, we have $12a=20b$ , or $a=\\dfrac{5}{3} b$ . Notice that there are only $3$ possible colors for unit cubes, so for each of the $1 \\times b \\times c$ layers, there are $bc-21$ yellow cubes, and similarly there are $ac-45$ red cubes in each of the $1 \\times a \\times c$ layers. Therefore, we have $a(bc-21)=25b$ and $b(ac-45)=9a$ . We check a few small values of $a,b$ and solve for $c$ , checking $(a,b)=(5,3)$ gives $c=12$ with a volume of $180$ $(a,b)=(10,6)$ gives $c=6$ with a volume of $360$ , and $(a,b)=(15,9)$ gives $c=4$ , with a volume of $540$ . Any higher $(a,b)$ will $ab>180$ , so therefore, the minimum volume is $\\boxed{180}$", "The total number of green cubes is given by $12a=20b\\Longrightarrow a=\\frac{5}{3}b$\nLet $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$ . Substitute $a=\\frac{5}{3}b$ gives $r=15$\nRepeating the procedure on the number of yellow cubes $y$ on each of the $a$ layers gives $y=15$\nTherefore $bc=9+12+15=36$ and $ac=15+20+25=60$ . Multiplying yields $abc^2=2160$\nSince $abc^2$ is fixed, $abc$ is minimized when $c$ is maximized, which occurs when $a$ $b$ are minimized (since each of $ac$ $bc$ is fixed). Thus $(a,b,c)=(3,5,12)\\Longrightarrow abc=\\boxed{180}$" ]

[ "\\[2mnp = (m+2)(n+2)(p+2)\\]\nLet’s solve for $p$\n\\[(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)\\] \\[[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)\\] \\[p = \\frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \\frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}\\]\nClearly, we want to minimize the denominator, so we test $(m-2)(n-2) - 8 = 1 \\Longrightarrow (m-2)(n-2) = 9$ . The possible pairs of factors of $9$ are $(1,9)(3,3)$ . These give $m = 3, n = 11$ and $m = 5, n = 5$ respectively. Substituting into the numerator, we see that the first pair gives $130$ , while the second pair gives $98$ . We now check that $130$ is optimal, setting $a=m-2$ $b=n-2$ in order to simplify calculations. Since \\[0 \\le (a-1)(b-1) \\implies a+b \\le ab+1\\] We have \\[p = \\frac{2(a+4)(b+4)}{ab-8} = \\frac{2ab+8(a+b)+32}{ab-8} \\le \\frac{2ab+8(ab+1)+32}{ab-8} = 10 + \\frac{120}{ab-8} \\le 130\\] Where we see $(m,n)=(3,11)$ gives us our maximum value of $\\boxed{130}$", "Similarly as above, we solve for $p,$ but we express the denominator differently:\n\\[p=\\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \\implies \\dfrac{1}{p}=\\dfrac{1}{2}-\\dfrac{2(m+n+2)}{(m+2)(n+2)}.\\] Hence, it suffices to maximize $\\dfrac{m+n+2}{(m+2)(n+2)},$ under the conditions that $p$ is a positive integer.\nThen since $\\dfrac{m+n+2}{(m+2)(n+2)}>\\dfrac{1}{2}$ for $m=1,2,$ we fix $m=3.$ \\[\\implies \\dfrac{1}{p}=\\dfrac{1}{2}-\\dfrac{2(n+5)}{5(n+2)}=\\dfrac{n-10}{10(n+2)},\\] where we simply let $n=11$ to achieve $p=\\boxed{130}.$" ]

[ "For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the final digit, any of the $10$ numbers are allowed. $999 \\cdot 10 = 9990 \\rightarrow \\boxed{9990}$", "Counting the prohibited cases, we find that there are 10 of them. This is because, when we start with 9,1, and 1, we can have any of the 10 digits for the last digit. So, our answer is $10^4-10=\\boxed{9990}.$" ]

[ "$\\frac{3000 \\text{ lire}}{1.6 \\text{ dollars}} = \\frac{x \\text{ lire}}{1 \\text{ dollar}}$\n$x = 1875$ , and the answer is $\\boxed{1875}$" ]

[ "If $80\\%$ of the people in the science club of $15$ people are in the Math Club, $\\frac{4}{5}\\times15=12$ people are in the both the Math Club and the Science Club.\nThese $12$ people make up $30\\%$ of the Math Club.\nSetting up a proportion:\n$12\\cdot 1.00 =0.30\\cdot x$\n$\\frac{12}{0.3} = 40=x$\nThere are $40=\\boxed{40}$ people in the Math Club." ]

[ "You can see that since the ratio of real building's heights to the model building's height is $1:20$ . We also know that the U.S Capitol is $289$ feet in real life, so to find the height of the model, we divide by 20. That gives us $14.45$ which rounds to 14. Therefore, to the nearest whole number, the duplicate is $\\boxed{14}$", "We can compute $\\frac{289}{20}$ and round our answer to get $\\boxed{14}$ .\nIt is basically Solution 1 without the ratio calculation. However, Solution 1 is referring further to the problem.", "We know that $20 \\cdot 14 = 280 ,$ and that $20 \\cdot 15 = 300 .$ These are the multiples of $20$ around $289 ,$ and the closest one of those is $280.$ Therefore, the answer is $\\dfrac {280} {20} = \\boxed{14} .$" ]

[ "Note that the three expressions are symmetric with respect to interchanging $\\sin$ and $\\cos$ , and so the probability is symmetric around $45^\\circ$ . Thus, take $0 < x < 45$ so that $\\sin x < \\cos x$ . Then $\\cos^2 x$ is the largest of the three given expressions and those three lengths not forming a triangle is equivalent to a violation of the triangle inequality\n\\[\\cos^2 x > \\sin^2 x + \\sin x \\cos x\\]\nThis is equivalent to\n\\[\\cos^2 x - \\sin^2 x > \\sin x \\cos x\\]\nand, using some of our trigonometric identities , we can re-write this as $\\cos 2x > \\frac 12 \\sin 2x$ . Since we've chosen $x \\in (0, 45)$ $\\cos 2x > 0$ so\n\\[2 > \\tan 2x \\Longrightarrow x < \\frac 12 \\arctan 2.\\]\nThe probability that $x$ lies in this range is $\\frac 1{45} \\cdot \\left(\\frac 12 \\arctan 2\\right) = \\frac{\\arctan 2}{90}$ so that $m = 2$ $n = 90$ and our answer is $\\boxed{092}$" ]

[ "Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof\nLet the set of parallel lines be perpendicular to the x-axis , such that they cross it at $0, 1, 2 \\ldots$ . The base of region $\\mathcal{A}$ is on the line $x = 1$ . The bigger base of region $\\mathcal{D}$ is on the line $x = 7$ . \nLet the top side of the angle be $y = x - s$ and the bottom side be x-axis, as dividing the angle doesn't change the problem.\nSince the area of the triangle is equal to $\\frac{1}{2}bh$\n\\[\\frac{\\textrm{Region\\ }\\mathcal{C}}{\\textrm{Region\\ }\\mathcal{B}} = \\frac{11}{5} = \\frac{\\frac 12(5-s)^2 - \\frac 12(4-s)^2}{\\frac 12(3-s)^2 - \\frac12(2-s)^2}\\]\nSolve this to find that $s = \\frac{5}{6}$\nUsing the same reasoning as above, we get $\\frac{\\textrm{Region\\ }\\mathcal{D}}{\\textrm{Region\\ }\\mathcal{A}} = \\frac{\\frac 12(7-s)^2 - \\frac 12(6-s)^2}{\\frac 12(1-s)^2}$ , which is $\\boxed{408}$", "Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be $x$ and the area of it be $x^2$ . Also, let all sections of the line on the same side as the side with length $x$ on a trapezoid be equal to $1$\nMove on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is ${\\left(\\frac{x+1}{x}\\right)}^2$ . Multiplying, we get $(x+1)^2$ as the area of the triangle, so the area of the trapezoid is $2x+1$ . Repeating this process, we get that the area of B is $2x+3$ , the area of C is $2x+7$ , and the area of D is $2x+11$\nWe can now use the given condition that the ratio of C and B is $\\frac{11}{5}$\n$\\frac{11}{5} = \\frac{2x+7}{2x+3}$ gives us $x = \\frac{1}{6}$\nSo now we compute the ratio of D and A, which is $\\frac{(2)(\\frac{1}{6}) + 11}{(\\frac{1}{6})^2} = \\boxed{408.}$", "Let the distances from the apex to the parallel lines be $x$ and $y$ and the distance between the intersections be $a,b.$ We know the area ratio means $\\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\\frac{5}{11}$ which simplifying yields $ab = 3ay+3bx.$ The ratio we seek is $\\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} =\\frac{ay+yx+11ab}{xy}.$ We know that $ab = 3ay+3bx$ so the ratio we seed is $\\frac{33(ay+yx)}{11xy}.$ Finally note that by similar triangles $\\frac{x}{x+a} =\\frac{y}{y+b} \\implies bx = ya.$ Therefore the ratio we seek is $\\frac{66(ay)}{11xy} =\\frac{66a}{11x}.$ Finally note that $ab=3ay+3bx \\implies ab = 6bx \\implies a = 6x$ so the final ratio is $6 \\cdot 68 = \\boxed{408}.$" ]

[ "The area of the large circle is $\\pi b^2$ , the area of the small one is $\\pi c^2$ , hence the shaded area is $\\pi(b^2-c^2)$\nFrom the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$ , hence $b^2-c^2=a^2$ and thus the shaded area is $\\boxed{2}$", "Set $c=0,$ then the shaded area is just the area of a circle with radius $a,$ which is $\\boxed{2}$" ]

[ "The area of the large circle is $\\pi b^2$ , the area of the small one is $\\pi c^2$ , hence the shaded area is $\\pi(b^2-c^2)$\nFrom the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$ , hence $b^2-c^2=a^2$ and thus the shaded area is $\\boxed{2}$", "Set $c=0,$ then the shaded area is just the area of a circle with radius $a,$ which is $\\boxed{2}$" ]

[ "For all positive integers $k,$ let\nThe base case occurs at $k=1,$ from which $\\left(N(1,\\mathrm{BB}),N(1,\\mathrm{BT}),N(1,\\mathrm{TB}),N(1,\\mathrm{TT})\\right)=(2,1,0,0).$\nSuppose the ant makes exactly $k$ moves for some $k\\geq2.$ We perform casework on its last move:\nAlternatively, this recursion argument is illustrated below, where each dashed arrow indicates $1$ way, and each solid arrow indicates $2$ ways: Therefore, we have the following relationships: \\begin{align*} N(1,\\mathrm{BB})&=2, \\\\ N(1,\\mathrm{BT})&=1, \\\\ N(1,\\mathrm{TB})&=0, \\\\ N(1,\\mathrm{TT})&=0, \\\\ N(k+1,\\mathrm{BB})&=N(k,\\mathrm{BB})+2\\cdot N(k,\\mathrm{TB}), \\\\ N(k+1,\\mathrm{BT})&=N(k,\\mathrm{BB}), \\\\ N(k+1,\\mathrm{TB})&=N(k,\\mathrm{TT}), \\\\ N(k+1,\\mathrm{TT})&=N(k,\\mathrm{TT})+2\\cdot N(k,\\mathrm{BT}). \\end{align*} Using these equations, we recursively fill out the table below: \\[\\begin{array}{c||c|c|c|c|c|c|c|c} \\hspace{7mm}&\\hspace{6.5mm}&\\hspace{6.5mm}&\\hspace{6.75mm}&\\hspace{6.75mm}&\\hspace{6.75mm}&\\hspace{6.75mm}&& \\\\ [-2.5ex] \\boldsymbol{k} & \\boldsymbol{1} & \\boldsymbol{2} & \\boldsymbol{3} & \\boldsymbol{4} & \\boldsymbol{5} & \\boldsymbol{6} & \\boldsymbol{7} & \\boldsymbol{8} \\\\ \\hline \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{BB})} &2&2&2&6&18&38&66&118 \\\\ \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{BT})} &1&2&2&2&6&18&38&66 \\\\ \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{TB})} &0&0&2&6&10&14&26&62 \\\\ \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{TT})} &0&2&6&10&14&26&62&138 \\\\ \\hline \\hline &&&&&&&& \\\\ [-2.25ex] \\textbf{Total}&\\boldsymbol{3}&\\boldsymbol{6}&\\boldsymbol{12}&\\boldsymbol{24}&\\boldsymbol{48}&\\boldsymbol{96}&\\boldsymbol{192}&\\boldsymbol{384} \\end{array}\\] By the Multiplication Principle, there are $3\\cdot2^{k-1}$ ways to make exactly $k$ moves. So, we must get \\[N(k,\\mathrm{BB})+N(k,\\mathrm{BT})+N(k,\\mathrm{TB})+N(k,\\mathrm{TT})=3\\cdot2^{k-1}\\] for all values of $k.$\nFinally, the requested probability is \\[\\frac{N(8,\\mathrm{BT})+N(8,\\mathrm{TT})}{N(8,\\mathrm{BB})+N(8,\\mathrm{BT})+N(8,\\mathrm{TB})+N(8,\\mathrm{TT})}=\\frac{66+138}{118+66+62+138}=\\frac{204}{384}=\\frac{17}{32},\\] from which the answer is $17+32=\\boxed{049}.$", "Let the state from bottom to top be $B2T,$ from top to top be $T2T,$ from top to bottom be $T2B,$ and from bottom to bottom be $B2B.$ We can draw the following State Transition Diagram with Markov Chain . The numbers on the transition arc are the transition probabilities.\nThe probabilities of being in a state after $n$ steps and after $n-1$ steps has the following relationships: \\begin{align*} B2T(n) &= B2B(n-1) \\cdot \\frac12\\\\ T2T(n) &= B2T(n-1) + T2T(n-1) \\cdot \\frac12\\\\ T2B(n) &= T2T(n-1) \\cdot \\frac12\\\\ B2B(n) &= T2B(n-1) + B2B(n-1) \\cdot \\frac12 \\end{align*} Those probabilities are calculated by Dynamic Programming in the following table: \\[\\begin{array}{c|cccc} & & & & \\\\ [-2ex] n & B2T(n) & T2T(n) & T2B(n) & B2B(n) \\\\ [1ex] \\hline & & & & \\\\ [-1ex] 1 & \\frac13 & 0 & 0 & \\frac23\\\\ & & & & \\\\ 2 & \\frac23 \\cdot \\frac12 = \\frac13 & \\frac13 & 0 & \\frac23 \\cdot \\frac12 = \\frac13 \\\\ & & & & \\\\ 3 & \\frac13 \\cdot \\frac12 = \\frac16 & \\frac13 + \\frac13 \\cdot \\frac12 = \\frac12 & \\frac13 \\cdot \\frac12 = \\frac16 & \\frac13 \\cdot \\frac12 = \\frac16 \\\\ & & & & \\\\ 4 & \\frac16 \\cdot \\frac12 = \\frac{1}{12} & \\frac16 + \\frac12 \\cdot \\frac12 = \\frac{5}{12} & \\frac12 \\cdot \\frac12 = \\frac14 & \\frac16 + \\frac16 \\cdot \\frac12 = \\frac14 \\\\ & & & & \\\\ 5 & \\frac14 \\cdot \\frac12 = \\frac18 & \\frac{1}{12} + \\frac{5}{12} \\cdot \\frac{1}{2} = \\frac{7}{24} & \\frac{5}{12} \\cdot \\frac12 = \\frac{5}{24} & \\frac14 + \\frac14 \\cdot \\frac12 = \\frac38 \\\\ & & & & \\\\ 6 & \\frac38 \\cdot \\frac12 = \\frac{3}{16} & \\frac18 + \\frac{7}{24} \\cdot \\frac12 = \\frac{13}{48} & \\frac{7}{24} \\cdot \\frac12 = \\frac{7}{48} & \\frac{5}{24} + \\frac38 \\cdot \\frac12 = \\frac{19}{48} \\\\ & & & & \\\\ 7 & \\frac{19}{48} \\cdot \\frac12 = \\frac{19}{96}& \\frac{3}{16} + \\frac{13}{48} \\cdot \\frac{1}{2} = \\frac{31}{96}& \\frac{13}{48} \\cdot \\frac{1}{2} = \\frac{13}{96} & \\frac{7}{48} + \\frac{19}{48} \\cdot \\frac12 = \\frac{11}{32}\\\\ & & & & \\\\ 8 & \\frac{11}{32} \\cdot \\frac{1}{2} = \\frac{11}{64} & \\frac{19}{96} + \\frac{31}{96} \\cdot \\frac12 = \\frac{23}{64} & \\frac{31}{96} \\cdot \\frac12 = \\frac{31}{192} & \\frac{13}{96} + \\frac{11}{32} \\cdot \\frac12 = \\frac{59}{192} \\\\ [1ex] \\end{array}\\] Finally, the requested probability is $\\frac{11}{64} + \\frac{23}{64} = \\frac{17}{32},$ from which the answer is $17 + 32 = \\boxed{049}.$", "Note that we don't care which exact vertex the ant is located at, just which level (either top face or bottom face).\nConsider the ant to be on any of the two levels and having moved at least one move. Define $p_i$ to be the probability that after $i$ moves, the ant ends up on the level it started on.\nOn the first move, the ant can stay on the bottom level with $\\frac{2}{3}$ chance and $7$ moves left. Or, it can move to the top level with $\\frac{1}{3}$ chance and $6$ moves left (it has to spend another on the top as it can not return immediately). So the requested probability is $P = \\frac{2}{3}(1 - p_7) + \\frac{1}{3}p_6$\nConsider when the ant has $i$ moves left (and it's not the ant's first move). It can either stay on its current level with $\\frac{1}{2}$ chance and $i - 1$ moves left, or travel to the opposite level with $\\frac{1}{2}$ chance, then move to another vertex on the opposite level, to have $i - 2$ moves left. Thus we obtain the recurrence \\[p_i = \\frac{1}{2}p_{i - 1} + \\frac{1}{2}(1 - p_{i - 2})\\]\nComputing $p_i$ with the starting conditions $p_0 = 1$ and $p_1 = \\frac{1}{2}$ , we obtain $p_6 = \\frac{33}{64}$ and $p_7 = \\frac{59}{128}$ . Hence $P = \\frac{2}{3}(1 - p_7) + \\frac{1}{3}p_6= \\frac{17}{32}$ as desired; $\\boxed{049}$", "On each move, we can either stay on the level we previously were (stay on the bottom/top) or switch levels (go from top to bottom and vise versa). Since we start on the bottom, ending on the top means that we will have to switch an odd number of times; since we cannot switch twice in a row, over an eight-move period we can either make one or three switches. Furthermore, once we switch to a level we can choose one of two directions of traveling on that level: clockwise or counterclockwise (since we can't go back to our previous move, our first move on the level after switching determines our direction).\nOur probability is then $\\frac{176 + 28}{3 \\cdot 2^7} = \\frac{17}{32}$ , so the answer is $17+32=\\boxed{049}$" ]

[ "The water level will rise $1$ cm for every $100 \\cdot 40 = 4000\\text{cm}^2$ . Since $1000$ is $\\frac{1}{4}$ of $4000$ , the water will rise $\\frac{1}{4}\\cdot1 = \\boxed{0.25}$" ]

[ "Consider a circle whose center lies on the positive $y$ -axis and which passes through the origin. As the radius of this circle becomes arbitrarily large, its curvature near the $x$ -axis becomes almost flat, and so it can intersect the curve $y = \\sin x$ arbitrarily many times (since the $x$ -axis itself intersects the curve infinitely many times). Hence, in particular, we can choose a radius sufficiently large that the circle intersects the curve at $\\boxed{16}$" ]

[ "The pillar at $B$ has height $9$ and the pillar at $A$ has height $12.$ Since the solar panel is flat, the inclination from pillar $B$ to pillar $A$ is $3.$ Call the center of the hexagon $G.$ Since $\\overrightarrow{CG}\\parallel\\overrightarrow{BA},$ it follows that the solar panel has height $13$ at $G.$ Since the solar panel is flat, the heights of the solar panel at $B,G,$ and $E$ are collinear. Therefore, the pillar at $E$ has height $9+4+4=\\boxed{17}.$", "Let the height of the pillar at $D$ be $x.$ Notice that the difference between the heights of pillar $C$ and pillar $D$ is equal to the difference between the heights of pillar $A$ and pillar $F.$ So, the height at $F$ is $x+2.$ Now, doing the same thing for pillar $E$ we get the height is $x+3.$ Therefore, we can see the difference between the heights at pillar $C$ and pillar $D$ is half the difference between the heights at $B$ and $E,$ so \\begin{align*} x+3-9&=2 \\cdot (x-10) \\\\ x-6&=2 \\cdot (x-10) \\\\ x&=14. \\end{align*} The answer is $x+3=\\boxed{17}.$", "We can extend $BA$ and $BC$ to $G$ and $H$ , respectively, such that $AG = CH$ and $E$ lies on $\\overline{GH}$ Because of hexagon proportions, $\\frac{BA}{AG} = \\frac{1}{3}$ and $\\frac{BC}{CH} = \\frac{1}{3}$ . Let $g$ be the height of $G$ . Because $A$ $B$ and $G$ lie on the same line, $\\frac{12-9}{g-12} = \\frac{1}{3}$ , so $g-12 = 9$ and $g = 21$ . Similarly, the height of $H$ is $13$ $E$ is the midpoint of $GH$ , so we can take the average of these heights to get our answer, $\\boxed{17}$", "Denote by $h_X$ the height of any point $X$\nDenote by $M$ the midpoint of $A$ and $C$ .\nHence, \\[h_M = \\frac{h_A + h_C}{2} = 11.\\] Denote by $O$ the center of $ABCDEF$ . Because $ABCDEF$ is a regular hexagon, $O$ is the midpoint of $B$ and $E$ .\nHence, \\[h_O = \\frac{h_E + h_B}{2} = \\frac{h_E + 9}{2}.\\] Because $ABCDEF$ is a regular hexagon, $M$ is the midpoint of $B$ and $O$ .\nHence, \\[h_M = \\frac{h_B + h_O}{2} = \\frac{9 + h_O}{2}.\\] Solving these equations, we get $h_E = \\boxed{17}$", "In this solution, we define rise as the change of height (in meters) from the solar panel to the ground. It follows that the rise from $B$ to $A$ is $12-9=3,$ and the rise from $B$ to $C$ is $10-9=1.$ Note that $\\vec{BE}=2\\vec{BA}+2\\vec{BC},$ so the rise from $B$ to $E$ is $2\\cdot3+2\\cdot1=8.$\nTogether, the height of the pillar at $E$ is $9+8=\\boxed{17}$ meters.", "WLOG, let the side length of the hexagon be $6$\nEstablish a 3D coordinate system, in which $A=(0,0,0)$ . Let the coordinates of $B$ and $C$ be $(6,0,0)$ $\\left(9,-3\\sqrt{3},0\\right)$ , respectively. Then, the solar panel passes through $P=(0,0,12), Q=(6,0,9), R=\\left(9,-3\\sqrt{3},10\\right)$\nThe vector $\\vec{PQ}=\\langle 6,0,-3\\rangle$ and $\\vec{PR}=\\left\\langle 9,-3\\sqrt{3}, -2\\right\\rangle$ . Computing $\\vec{PQ} \\times \\vec{PR}$ by the matrix \\[\\begin{bmatrix} i & j & k \\\\ 6 & 0 & -3 \\\\ 9 & -3\\sqrt{3} & -2 \\end{bmatrix}\\] gives the result $-9\\sqrt{3}i -15j -18\\sqrt{3} k$ . Therefore, a normal vector of the plane of the solar panel is $\\left\\langle -9\\sqrt{3},-15,-18\\sqrt{3}\\right\\rangle$ , and the equation of the plane is $-9\\sqrt{3}x-15y-18\\sqrt{3}z=k$ . Substituting $(x,y,z)=(0,0,12)$ , we find that $k=-216\\sqrt{3}$\nSince $E=\\left(0,-6\\sqrt{3}\\right)$ , we substitute $(x,y)=\\left(0,-6\\sqrt{3}\\right)$ into $-9\\sqrt{3}x-15y-18\\sqrt{3}z=-216\\sqrt{3}$ , which gives $z=\\boxed{17}$", "Let the pillars be $AA', BB', \\ldots, FF'$ . Since solar panel $A'B'C'D'E'F'$ is a hexagon, the line $B'E'$ hits the midpoint $M$ of $A'C'$ . So, the 3D slope (change in $x$ : change in $y$ : change in $z$ ) of $BE$ is same as $BD$ . If $a$ is side of the hexagonal solar panel, \\[B'M' = \\frac{1}{2}a, B'E' = a+2\\cdot \\frac{1}{2}a = 2a\\] . So, $B'M:B'E'$ $1:4$ . Since the height of $M$ to ground $ABCDEF$ is $(10+12)/2 = 11$ , the rise (in z) from $B'$ to $M$ is 2 meaning the rise from $B'$ to $E'$ is $8$ . Thus, $EE' = 8+BB' = \\boxed{17}$", "Set the midpoint of $\\overline{AC}$ as $M$ We know that the height of $M$ is $11$ as it is the midpoint of $\\overline{AC}$ , so the height is the average of $A$ and $C$ , which is $\\frac{10 + 12}{2}= 11$ . Since $ABCDEF$ is a regular hexagon, $BE = 4\\cdot BM$ . Because the increase in height is proportional to the length of the line segments, and the increase in height from $B$ to $M$ is $2$ , the increase in height from $B$ to $E$ is $2\\cdot4=8.$ Adding to the height of $B$ , we get $8+9=\\boxed{17}$", "Because the three points given are integers, it is likely that the answer is also an integer. This leaves us with $9$ or $17$ . Because both $A$ and $C$ are greater than $9$ and closer to $E$ than $B$ , we can assume that the height increases as the point gets closer to $E$ . Thus, we know the answer is greater than $9$ . The only choice that satisfies both these criteria is $\\boxed{17}$" ]

[ "We begin filling in the table. The top row has a first term $1$ and a fifth term $25$ , so we have the common difference is $\\frac{25-1}4=6$ . This means we can fill in the first row of the table: \nThe fifth row has a first term of $17$ and a fifth term of $81$ , so the common difference is $\\frac{81-17}4=16$ . We can fill in the fifth row of the table as shown: \nWe must find the third term of the arithmetic sequence with a first term of $13$ and a fifth term of $49$ . The common difference of this sequence is $\\frac{49-13}4=9$ , so the third term is $13+2\\cdot 9=\\boxed{31}$", "The middle term of the first row is $\\frac{25+1}{2}=13$ , since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is $\\frac{17+81}{2}=49$ . Applying this again for the middle column, the answer is $\\frac{49+13}{2}=\\boxed{31}$", "The value of $X$ is simply the average of the average values of both diagonals that contain $X$ . This is $\\frac{\\frac{1+81}{2}+\\frac{17+25}{2}}{2} =\\frac{\\frac{82}{2}+\\frac{42}{2}}{2} = \\frac{41+21}{2} = \\boxed{31}$" ]

[ "Since one of the terms was either $1$ more or $1$ less than it should have been, the sum should have been $222-1=221$ or $222+1=223.$\nThe formula for an arithmetic series is $an+d\\left(\\dfrac{(n-1)n}2\\right)=\\dfrac n2\\left(2a+d(n-1)\\right).$ This can quickly be rederived by noticing that the sequence goes $a,a+d,a+2d,a+3d,\\dots,a+(n-1)d$ , and grouping terms.\nWe know that $\\dfrac n2(2a+d(n-1))=221$ or $223$ . Let us now show that $223$ is not possible.\nIf $\\dfrac n2(2a+d(n-1))=223$ , we can simplify this to be $n(a+d(n-1))=223\\cdot2.$ Since every expression in here should be an integer, we know that either $n=2$ and $a+d(n-1)=223$ or $n=223$ and $a+d(n-1)=2.$ The latter is not possible, since $n\\ge3,d>1,$ and $a>0.$ The former is also impossible, as $n\\ge3.$ Thus, $\\dfrac n2(2a+d(n-1))\\neq223\\implies\\dfrac n2(2a+d(n-1))=221$\nWe can factor $221$ as $13\\cdot17$ . Using similar reasoning, we see that $221\\cdot2$ can not be paired as $2$ and $221$ , but rather must be paired as $13$ and $17$ with a factor of $2$ somewhere.\nLet us first try $n=13.$ Our equation simplifies to $2a+12d=34\\implies a+6d=17.$ We know that $d>1,$ so we try the smallest possible value: $d=2.$ This would give us $a=17-2\\cdot6=17-12=5.$ (Indeed, this is the only possible $d$ .)\nThere is nothing wrong with the values we have achieved, so it is reasonable to assume that this is the only valid solution (or all solutions sum to the same thing), so we answer $a+d+n=5+2+13=\\boxed{20.}$", "There are $n$ terms, the $x$ th term is $a+(x-1)d$ , summation is $an+dn(n-1)/2=n(a+\\frac{d(n-1)}{2})$\nThe summation of the set is $222 \\pm 1 = 221,223$ . First, $221$ : its only possible factors are $1,13,17,221$ , and as said by the problem, $n\\ge3$ , so $n$ must be $13,17,$ or $221$ . Let's start with $n=13$ . Then, $a+6d=17$ , and this means $a=5$ $d=2$ . Summing gives $13+5+2=\\boxed{20}$ . We don't need to test any more cases, since the problem writes that all $a+d+n$ are the same.", "We must have the sum of terms of the arithmetic sequence is $222\\pm 1$ , which is $221$ or $223$\nSince we have $223$ is prime, it cannot be the sum of the arithmetic sequence.\nWe see that $221$ is just $13\\times 17$\nWe can write any arithmetic sequence with an odd amount of terms like this: $b-md,\\cdots ,b-2d,b-d,b,b+d,b+2d,\\cdots b+md$ where b is the middle term and d is the common difference.\nBy the sum of an arithmetic sequence, we have $13b=221$ and $17b=221$ and therefore $(b,n)$ $=$ $(13,17)$ or $(17,13)$\nThen $a+d+n=b-md+d+n=30-(m-1)d$\nWe must have that m is either $\\frac{17-1}{2}$ or $\\frac{13-1}{2}$ , so m is either $6$ or $8$\nSo $a+d+n=30-5d$ or $30-7d$\nTaking $\\mod 7$ , we have no answer choices that give $2$ , and then taking $\\mod 5$ gives the only answer that works is $20$\nTherefore we have $a+d+n=\\boxed{20.}$", "The formula for the sum of an arithmetic sequence is $n(\\frac{a_1+a_n}{2})$ , where $a$ is the first term, $a_n$ is the last term, and $n$ is the number of terms. Let $a$ be the first term, $d$ be the common difference, and $n$ be the number of terms of Carl's sequence. Since the sum the sequence is $1$ less or $1$ more than $222$ , we have \\[n(\\frac{a+d(n-1)}{2})\\pm{1} = 222\\] \\[n(\\frac{a+d(n-1)}{2}) = 222\\pm{1}.\\] The right-hand side is either $221$ or $223$ . We know that it has to be divisible by $n$ so we can find the factors of $221$ and $223$ . Checking all the primes less than $15$ , we find that $223$ is prime and $221=13\\cdot17$\nBecause $n\\ge3$ , the sum must be $221$ and the only possible values of $n$ are $13$ and $17$ . We can test both cases.\nCase 1: $n=17$ Substituting for $n$ gives us $a+8d=17$ . Since the sequence consists of only positive integers, $d$ is an integer. We know that $d>1$ but if $d\\ge2$ , then $8d>13$ . Hence, this case is not possible.\nCase 2: $n=13$ Substituting for $n$ gives us $a+6d=17$ . Using the same logic from case 1, we get $3>d>1$ , so $d=2$ . Solving for $a$ , we get $a=5$ . Therefore, $a+d+n=5+2+13=20$ , so the answer is $\\boxed{20.}$ .\n~azc1027" ]

[ "We can consider the contributions of the sides of the three layers and the tops of the layers separately.\nLayer $n$ (counting from the top starting at $1$ ) has $4$ side faces each with $n$ unit squares, so the sides of the pyramid contribute $4+8+12=24$ for the surface area.\nThe tops of the layers when combined form the same arrangement of unit cubes as the bottom of the pyramid, which is a $3\\times 3$ square, hence this contributes $9$ for the surface area.\nYou do not have to count the side underneath, since it is not exposed.\nThus, the artist paints $24+9=33 \\rightarrow \\boxed{33}$ square meters." ]

[ "The maximum heart rate is $220-26=194$ beats per minute. The target heart rate is then $0.8*194 \\approx \\boxed{155}$ beats per minute." ]

[ "We first note that if a row has $n$ seats, then the maximum number of students that can be seated in that row is $\\left\\lceil \\frac{n}{2} \\right\\rceil$ , where $\\lceil x \\rceil$ is the smallest integer greater than or equal to $x$ . If a row has $2k$ seats, clearly we can only fit $k$ students in that row. If a row has $2k+1$ seats, we can fit $k+1$ students by putting students at the ends and then alternating between skipping a seat and putting a student in.\nFor each row with $10+k$ seats, there is a corresponding row with $29-k$ seats. The sum of the maximum number of students for these rows is \\[\\left\\lceil \\frac{10+k}{2}\\right\\rceil +\\left\\lceil \\frac{29-k}{2} \\right\\rceil = 20.\\] There are $20/2=10$ pairs of rows, so the maximum number of students for the exam is $20\\times 10=200\\rightarrow \\boxed{200}$" ]

[ "Use dimensional analysis. \\[\\frac{a/6 \\text{ feet}}{r \\text{ seconds}} \\cdot \\frac{1 \\text{ yard}}{3 \\text{ feet}} \\cdot \\frac{60 \\text{ seconds}}{1 \\text{ minute}} \\cdot 3 \\text{ minutes}\\] \\[\\frac{10a}{r} \\text{ yards}\\] The answer is $\\boxed{10}$" ]

[ "An ellipse is defined to be the locus of points $P$ such that the sum of the distances between $P$ and the two foci is constant. Let $F_1 = (9, 20)$ $F_2 = (49, 55)$ and $X = (x, 0)$ be the point of tangency of the ellipse with the $x$ -axis. Then $X$ must be the point on the axis such that the sum $F_1X + F_2X$ is minimal. (The last claim begs justification: Let $F'_2$ be the reflection of $F_2$ across the $x$ -axis. Let $Y$ be where the line through $F_1$ and $F’_2$ intersects the ellipse. We will show that $X=Y$ . Note that $X F_2 = X F’_2$ since $X$ is on the $x$ -axis. Also, since the entire ellipse is on or above the $x$ -axis and the line through $F_2$ and $F’_2$ is perpendicular to the $x$ -axis, we must have $F_2 Y \\leq F’_2 Y$ with equality if and only if $Y$ is on the $x$ -axis. Now, we have \\[F_1 X + F'_2 X = F_1 X + F_2 X = F_1 Y + F_2 Y \\leq F_1 Y + F’_2 Y\\] But the right most sum is the straight-line distance from $F_1$ to $F’_2$ and the left is the distance of some path from $F_1$ to $F_2$ ., so this is only possible if we have equality and thus $X = Y$ ). Finding the optimal location for $X$ is a classic problem: for any path from $F_1$ to $X$ and then back to $F_2$ , we can reflect (as above) the second leg of this path (from $X$ to $F_2$ ) across the $x$ -axis. Then our path connects $F_1$ to the reflection $F_2'$ of $F_2$ via some point on the $x$ -axis, and this path will have shortest length exactly when our original path has shortest length. This occurs exactly when we have a straight-line path, and by the above argument, this path passes through the point of tangency of the ellipse with the $x-$ axis.\nThe sum of the two distances $F_1 X$ and $F_2X$ is therefore equal to the length of the segment $F_1F_2'$ , which by the distance formula is just $d = \\sqrt{(9 - 49)^2 + (20 + 55)^2} = \\sqrt{40^2 + 75^2} = 5\\sqrt{8^2 + 15^2} = 5\\cdot 17 = 85$\nFinally, let $A$ and $B$ be the two endpoints of the major axis of the ellipse. Then by symmetry $AF_1 = F_2B$ so $AB = AF_1 + F_1B = F_2B + F_1B = d$ (because $B$ is on the ellipse), so the answer is $\\boxed{085}$", "An ellipse is defined as the set of points where the sum of the distances from the foci to the point is fixed. The length of major axis is equal to the sum of these distances $(2a)$ . Thus if we find the sum of the distances, we get the answer. Let k be this fixed sum; then we get, by the distance formula:\n$k = \\sqrt{(x - 9)^2 + 20^2} + \\sqrt{(x - 49)^2 + 55^2}$\nThis is the equation of the ellipse expressed in terms of $x$ . The line tangent to the ellipse at the given point $P(x, 0)$ will thus have slope $0$ . Taking the derivative gives us the slope of this line. To simplify, let $f(x) = (x - 9)^2 + 20^2$ and $g(x) = (x - 49)^2 + 55^2$ . Then we get:\n$0 = \\frac{f^\\prime(x)}{2\\sqrt{f(x)}} + \\frac{g^\\prime(x)}{2\\sqrt{g(x)}}$\nNext, we multiply by the conjugate to remove square roots. We next move the resulting $a^2 - b^2$ form expression into form $a^2 = b^2$\n$\\frac{(f^\\prime(x))^2}{4\\cdot f(x)} = \\frac{(g^\\prime(x))^2}{4\\cdot g(x)}$\nWe know $f^\\prime(x) = 2x - 18$ and $g^\\prime(x) = 2x - 98$ . Simplifying yields:\n$\\frac{(x - 9)^2}{(x - 9)^2 + 20^2} = \\frac{(x - 49)^2}{(x - 49)^2 + 55^2}$\nTo further simplify, let $a = (x - 9)^2$ and $b = (x - 49)^2$ . This means $\\frac{a}{a + 400} = \\frac{b}{b + 3025}$ . Solving yields that $16b = 121a$ .\nSubstituting back $a$ and $b$ yields:\n$16 \\cdot (x - 49)^2 = 121 \\cdot (x - 9)^2$\nSolving for $x$ yields $x = \\frac{59}{3}$ . Substituting back into our original distance formula, solving for $k$ yields $\\boxed{085}$" ]

[ "Place the cube in the xyz-coordinate, with the positive x-axis pointing forward, the positive y-axis pointing right, and the positive z-axis pointing up. Let the position of a unit cube be $(x, y, z)$ if it is centered at $(x, y, z)$ . Place the $2020 \\times 2020 \\times 2020$ cube so that the edges are parallel to the axes, and two of its corners are at $(1, 1, 1)$ and $(2020, 2020, 2020)$ . Now call a beam z-oriented if its endpoints differ only in z-coordinates, and similarly call it x-oriented or y-oriented if the endpoints differ in x- or y-coordinates, respectively.\nWe claim that the answer is $\\boxed{3030}$ . First we will prove 3030 suffices. Place the first beam so its endpoints lie at $(1, 1, 1)$ , and $(2020, 1, 1)$ . Place the 2nd beam so its endpoints lie at $(1, 1, 2)$ and $(1, 2020, 2)$ . Place the 3rd beam so its endpoints lie at $(2, 2, 1)$ and $(2, 2, 2020)$ . Now it is clear that the only faces among these three beams that are not touching the faces of the $2020 \\times 2020 \\times 2020$ cube of the face of another beam are the top face of beam 2, and the front and right faces of beam 3." ]

[ "The area of the octagon can be divided up into $5$ squares with side $\\frac{\\sqrt2}2$ and $4$ right triangles, which are half the area of each of the squares.\nTherefore, the area of the octagon is equal to the area of $5+4\\left(\\frac12\\right)=7$ squares.\nThe area of each square is $\\left(\\frac{\\sqrt2}2\\right)^2=\\frac12$ , so the area of $7$ squares is $\\boxed{72}$", "Refer to the following diagram:\nAMC10 2005A P20.png\n(Picture made on Geogebra)\nNote that each square has area $\\frac14$ , and each triangle has area $\\frac18$ . The total area is $12\\cdot\\frac14+4\\cdot\\frac18=\\frac72 \\Longrightarrow \\boxed{72}$" ]

[ "Let the perimeter of the equilateral triangle be $3s$ . The side length of the equilateral triangle would then be $s$ and the sidelength of the hexagon would be $\\frac{s}{2}$\nA hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio $1 : 4$ , since the sidelength of the small equilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is $1$ . The area of the hexagon is then $1 \\times 6 = \\boxed{6}$", "Let the side length of the equilateral triangle be $s$ and the side length of the hexagon be $y$ . Since the perimeters are equal, we must have $3s=6y$ which reduces to $s=2y$ . Substitute this value in to the area of an equilateral triangle to yield $\\dfrac{(2y)^2\\sqrt{3}}{4}=\\dfrac{4y^2\\sqrt{3}}{4}$\nSetting this equal to $4$ gives us $\\dfrac{4y^2\\sqrt{3}}{4}=4\\implies 4y^2\\sqrt{3}=16\\implies y^2\\sqrt{3}=4$\nSubstitute $y^2\\sqrt{3}$ into the area of a regular hexagon to yield $\\dfrac{3(4)}{2}=6$\nTherefore, our answer is $\\boxed{6}$", "Let the side length of the triangle be $s$ and the side length of the hexagon be $t$ . As explained in Solution 1, $s=2t$ , or $t=\\frac{s}{2}$ . The area of the triangle is $\\frac{s^2\\sqrt3}{4}=4$ and the area of the hexagon is $\\frac{t^2\\sqrt3}{4} \\cdot 6=\\frac{3t^2\\sqrt3}{2}$ . Substituting $\\frac{s}{2}$ in for $t$ , we get \\[\\frac{\\frac{3s^2\\sqrt3}{4}}{2}=\\frac{3s^2\\sqrt3}{8}.\\] $\\frac{s^2\\sqrt3}{4}=4 \\implies \\frac{s^2\\sqrt3}{8}=2 \\implies \\frac{3s^2\\sqrt3}{8}=\\boxed{6}$" ]

[ "The perimeter of the first triangle is $3a$ . The perimeter of the 2nd triangle is half of that, after drawing a picture. The 3rd triangle's perimeter is half the second's, and so on. Therefore, we are computing $3a+\\frac{3a}{2}+\\frac{3a}{4}+\\cdots$\nThe starting term is $3a$ , and the common ratio is $1/2$ . Therefore, the sum is $\\frac{3a}{1-\\frac{1}{2}}=\\boxed{6}$" ]

[ "Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$\nNote that the slope of $\\overline{AC}$ is $\\tan 60^\\circ = \\sqrt {3}.$ Hence, the equation of the line containing $\\overline{AC}$ is \\[y = x\\sqrt {3} + 1.\\] This will intersect the ellipse when \\begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\\sqrt {3} + 1)^{2} \\\\ & = & x^{2} + 4(3x^{2} + 2x\\sqrt {3} + 1) \\implies x(13x+8\\sqrt 3)=0\\implies x = \\frac { - 8\\sqrt {3}}{13}. \\end{eqnarray*} We ignore the $x=0$ solution because it is not in quadrant 3.\nSince the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $\\left(\\frac {8\\sqrt {3}}{13},y_{0}\\right)$ and $\\left(\\frac { - 8\\sqrt {3}}{13},y_{0}\\right),$ respectively, for some value of $y_{0}.$\nIt is clear that the value of $y_{0}$ is irrelevant to the length of $BC$ . Our answer is \\[BC = 2*\\frac {8\\sqrt {3}}{13}=\\sqrt {4\\left(\\frac {8\\sqrt {3}}{13}\\right)^{2}} = \\sqrt {\\frac {768}{169}}\\implies m + n = \\boxed{937}.\\]", "Solving for $y$ in terms of $x$ gives $y=\\sqrt{4-x^2}/2$ , so the two other points of the triangle are $(x,\\sqrt{4-x^2}/2)$ and $(-x,\\sqrt{4-x^2}/2)$ , which are a distance of $2x$ apart. Thus $2x$ equals the distance between $(x,\\sqrt{4-x^2}/2)$ and $(0,1)$ , so by the distance formula we have\n\\[2x=\\sqrt{x^2+(1-\\sqrt{4-x^2}/2)^2}.\\]\nSquaring both sides and simplifying through algebra yields $x^2=192/169$ , so $2x=\\sqrt{768/169}$ and the answer is $\\boxed{937}$", "Since the altitude goes along the $y$ axis, this means that the base is a horizontal line, which means that the endpoints of the base are $(x,y)$ and $(-x,y)$ , and WLOG, we can say that $x$ is positive.\nNow, since all sides of an equilateral triangle are the same, we can do this (distance from one of the endpoints of the base to the vertex and the length of the base):\n$\\sqrt{x^2 + (y-1)^2} = 2x$\nSquare both sides,\n$x^2 + (y-1)^2 = 4x^2\\implies (y-1)^2 = 3x^2$\nNow, with the equation of the ellipse: $x^2 + 4y^2 = 4$\n$x^2 = 4-4y^2$\n$3x^2 = 12-12y^2$\nSubstituting,\n$12-12y^2 = y^2 - 2y +1$\nMoving stuff around and solving:\n$y = \\frac{-11}{13}, 1$\nThe second is found to be extraneous, so, when we go back and figure out $x$ and then $2x$ (which is the side length), we find it to be:\n$\\sqrt{\\frac{768}{169}}$\nand so we get the desired answer of $\\boxed{937}$", "Denote $(0,1)$ as vertex $A,$ $B$ as the vertex to the left of the $y$ -axis and $C$ as the vertex to the right of the $y$ -axis. Let $D$ be the intersection of $BC$ and the $y$ -axis.\nLet $x_0$ be the $x$ -coordinate of $C.$ This implies \\[C=\\left(x_0 , \\sqrt{\\frac{4-x_0^2}{4}}\\right)\\] and \\[B=\\left(-x_0 , \\sqrt{\\frac{4-x_0^2}{4}}\\right).\\] Note that $BC=2x_0$ and \\[\\frac{BC}{\\sqrt3}=AD=1-\\sqrt{\\frac{4-x_0^2}{4}}.\\] This yields \\[\\frac{2x_0}{\\sqrt3}=1-\\sqrt{\\frac{4-x_0^2}{4}}.\\] Re-arranging and squaring, we have \\[\\frac{4-x_0^2}{4}=\\frac{4x_0^2}{3}-\\frac{4x_0}{\\sqrt3} +1.\\] Simplifying and solving for $x_0$ , we have \\[x_0=\\frac{48}{13\\sqrt 3}.\\] As the length of each side is $2x_0,$ our desired length is \\[4x_0^2=\\frac{768}{169}\\] which means our desired answer is \\[768+169=\\boxed{937}\\]", "Notice that $x^2+4y^2=4$ can be rewritten as $(x)^2+(2y)^2=2^2$ . The points of the triangle are $(0, 1)$ $(-x, 1-x\\sqrt{3})$ , and $(x, 1-x\\sqrt{3})$ . When plugging the second coordinate into the equation, we get $x^2+4-8x\\sqrt{3}+12x^2=4$ , which equals $13x^2-8x\\sqrt{3}=0$ . \nThis yields $x(13x-8\\sqrt{3})=0$ . Obviously x can't be 0, so $x=\\frac{8\\sqrt{3}}{13}$ . The side length of the equilateral triangle is twice of this, so $\\frac{16\\sqrt{3}}{13}$ .\nThis can be rewritten as $\\sqrt{\\frac{256\\cdot3}{169}}=\\sqrt{\\frac{768}{169}}$ $768+169=\\boxed{937}$ .\n~ MC413551", "Consider the transformation $(x,y)$ to $(x/2, y).$ This sends the ellipse to the unit circle. If we let $n$ be one-fourth of the side length of the triangle, the equilateral triangle is sent to an isosceles triangle with side lengths $2n, n\\sqrt{13}, n\\sqrt{13}.$ Let the triangle be $ABC$ such that $AB=AC.$ Let the foot of the altitude from A be $X.$ Then $BX=n,$ and $AX=2n\\sqrt{3}.$ Let $C$ be a point such that $AC$ is a diameter of the unit circle. Then $XC=2-2n\\sqrt{3}.$ Using power of a point on X, \\[n^2=2n\\sqrt{3}(2-2n\\sqrt{3})\\] Simplifying gets us to $13n^2=4n\\sqrt{3}.$ Then, $n=\\dfrac{4\\sqrt{3}}{13}$ which means the side length is $\\dfrac{16\\sqrt{3}}{13}=\\sqrt{\\dfrac{768}{169}}.$ Thus, the answer is $768+169=\\boxed{937}.$" ]

[ "The area of the large triangle is $\\frac{10^2\\sqrt3}{4}$ , while the area of each small triangle is $\\frac{1^2\\sqrt3}{4}$ . Dividing these two quantities results in $100$ , therefore $\\boxed{100}$ small triangles can fill the large one without overlap." ]

[ "\nThe number of triangles is $1+3+\\dots+19 = \\boxed{100}$" ]

[ "Notice that the area of the shaded region is $(2^2\\pi-1^2\\pi)+(4^2\\pi-3^2\\pi)+(6^2\\pi-5^2\\pi)+ \\cdots + (n^2\\pi-(n-1)^2 \\pi)$ for any even number $n$\nUsing the difference of squares, this simplifies to $(1+2+3+4+\\cdots+n) \\pi$ . So, we are basically finding the smallest $n$ such that $\\frac{n(n+1)}{2}>2023 \\Rightarrow n(n+1) > 4046$ . Since $60(61) > 60^2=3600$ , the only option higher than $60$ is $\\boxed{64}$", "After first observing the problem, we can work out a few of the areas.\n1st area = $4\\pi-\\pi = 3\\pi$\n2nd area = $16\\pi-9\\pi = 7\\pi$\n3rd area = $36\\pi-25\\pi = 11\\pi$\n4th area = $64\\pi-49\\pi = 15\\pi$\nWe can see that the pattern is an arithmetic sequence with first term $3$ and common difference $4$ . From here, we can start from the bottom of the answer choices and work our way up. As the question asks for the least number of circles needed total, we have to divide the number of total circles by 2.\nWe can find the sum of the first $32$ terms of the arithmetic sequence by using the formula.\nThe last term is: $3 + 4\\cdot(32-1) = 127$\nThen, we can find the sum: $(3+127)/2 \\cdot 32 = 65\\cdot32 = 2080$ . It is clear that $64$ works.\nThe next answer choice is $60$ , which we have to divide by 2 to get $30$\nThe last term is: $3 + 4\\cdot(30-1) = 119$\nThe sum is: $(3+119)/2 \\cdot 30 = 61\\cdot30 = 1830$ . This does not work.\nAs answer choice $D$ does not work and $E$ does, we can conclude that the answer is $\\boxed{64}$", "We can easily see that all of the answer choices are even, which helps us solve this problem a little.\nLets just not consider the $\\pi$ , since it is not that important, so let's just cancel that out.\nWhen we plug in 64, we get $64^2-63^2+62^2-61^2+\\cdots +4^2-3^2+2^2-1^2$ . By difference of squares, we get $1+2+3+\\cdots+62+63+64$ , which by the sum of an arithmetic sequence, is $\\frac{64(64+1)}{2}$ , which is $2080$\nSimilarly, we can use this for answer choice $D$ , and we have $\\frac{60(60+1)}{2}$ which is $1830$\nSo, we see that answer choice $D$ is too small to satisfy the requirements, so we conclude the answer is $\\boxed{64}$", "We can consider making a table.\nIf there is 1 circle, the area of the shaded region is 0π. (We can write this as 0π.)\nIf there are 2 circles, the area of the shaded region is 3π. (We can write this as (1+2)π).\nIf there are 3 circles, the area of the shaded region is 3π. (We can write this as (1+2)π).\nIf there are 4 circles, the area of the shaded region is 10π. (We can write this as (1+2+3+4)π).\nIf there are 5 circles, the area of the shaded region is 10π. (We can write this as (1+2+3+4)π).\nIf there are 6 circles, the area of the shaded region is 21π. (We can write this as (1+2+3+4+5+6)π).\nNow the pattern emerges. When $n$ is even, the area of the shaded region is $(1+2+3+\\cdots+n)\\pi$ , or $\\left( \\frac{n(n+1)}{2} \\right) \\pi$ . But remember that the problem stated that there are an even number of circles. So now we are solving the equation \\[\\left( \\frac{n(n+1)}{2} \\right) \\pi\\ge2023\\pi.\\] Dividing by $\\pi$ and multiplying by 2 on both sides, we get $n(n+1)\\ge4046$ . Now we can plug in the answer choices, and we start off with $60$ because it is the only answer choice that is a multiple of $10$ . Plugging in we get $60\\cdot61=3660$ , and this is not quite yet more than $4046$ . But only option $(\\text{E})$ is bigger, so we know that the solution can only be $\\boxed{64}$", "Denote $S$ the area of shaded region and $W$ the area of white region.\n$S > W$ and $S \\approx W$ if $R$ is big.\nTherefore \\[\\pi R^2 = S + W \\approx 2S \\ge 4046 \\pi \\implies R \\approx 64.\\] So we conclude the answer is $\\boxed{64}$" ]

[ "Let's call the first term of the original geometric series $a$ and the common ratio $r$ , so $2005 = a + ar + ar^2 + \\ldots$ . Using the sum formula for infinite geometric series, we have $\\;\\;\\frac a{1 -r} = 2005$ . Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + \\ldots$ . We know this series has sum $20050 = \\frac{a^2}{1 - r^2}$ . Dividing this equation by $\\frac{a}{1-r}$ , we get $10 = \\frac a{1 + r}$ . Then $a = 2005 - 2005r$ and $a = 10 + 10r$ so $2005 - 2005r = 10 + 10r$ $1995 = 2015r$ and finally $r = \\frac{1995}{2015} = \\frac{399}{403}$ , so the answer is $399 + 403 = \\boxed{802}$", "We can write the sum of the original series as $a + a\\left(\\dfrac{m}{n}\\right) + a\\left(\\dfrac{m}{n}\\right)^2 + \\ldots = 2005$ , where the common ratio is equal to $\\dfrac{m}{n}$ . We can also write the sum of the second series as $a^2 + a^2\\left(\\dfrac{m}{n}\\right)^2 + a^2\\left(\\left(\\dfrac{m}{n}\\right)^2\\right)^2 + \\ldots = 20050$ . Using the formula for the sum of an infinite geometric series $S=\\dfrac{a}{1-r}$ , where $S$ is the sum of the sequence, $a$ is the first term of the sequence, and $r$ is the ratio of the sequence, the sum of the original series can be written as $\\dfrac{a}{1-\\frac{m}{n}}=\\dfrac{a}{\\frac{n-m}{n}}=\\dfrac{a \\cdot n}{n-m}=2005\\;\\text{(1)}$ , and the second sequence can be written as $\\dfrac{a^2}{1-\\frac{m^2}{n^2}}=\\dfrac{a^2}{\\frac{n^2-m^2}{n^2}}=\\dfrac{a^2\\cdot n^2}{(n+m)(n-m)}=20050\\;\\text{(2)}$ . Dividing $\\text{(2)}$ by $\\text{(1)}$ , we obtain $\\dfrac{a\\cdot n}{m+n}=10$ , which can also be written as $a\\cdot n=10(m+n)$ . Substitute this value for $a\\cdot n$ back into $\\text{(1)}$ , we obtain $10\\cdot \\dfrac{n+m}{n-m}=2005$ . Dividing both sides by 10 yields $\\dfrac{n+m}{n-m}=\\dfrac{401}{2}$ we can now write a system of equations with $n+m=401$ and $n-m=2$ , but this does not output integer solutions. However, we can also write $\\dfrac{n+m}{n-m}=\\dfrac{401}{2}$ as $\\dfrac{n+m}{n-m}=\\dfrac{802}{4}$ . This gives the system of equations $m+n=802$ and $n-m=4$ , which does have integer solutions. Our answer is therefore $m+n=\\boxed{802}$ (Solving for $m$ and $n$ gives us $399$ and $403$ , respectively, which are co-prime)." ]

[ "Given any path on the surface, we can unfold the surface into a plane to get a path of the same length in the plane. Consider the net of a tetrahedron in the picture below. A pair of opposite points is marked by dots. It is obvious that in the plane the shortest path is just a segment that connects these two points. Its length is the same as the length of the tetrahedron's edge, i.e. $\\boxed{1}$" ]

[ "Notice that we can rewrite $N^{16}$ as $(N^{4})^4$ . By Fermat's Little Theorem , we know that $N^{(5-1)} \\equiv 1 \\pmod {5}$ if $N \\not \\equiv 0 \\pmod {5}$ . Therefore for all $N \\not \\equiv 0 \\pmod {5}$ we have $N^{16} \\equiv (N^{4})^4 \\equiv 1^4 \\equiv 1 \\pmod 5$ . Since $1\\leq N \\leq 2020$ , and $2020$ is divisible by $5$ $\\frac{1}{5}$ of the possible $N$ are divisible by $5$ . Therefore, $N^{16} \\equiv 1 \\pmod {5}$ with probability $1-\\frac{1}{5},$ or $\\boxed{45}$", "Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits $0-9$ .\nThe pattern for $0$ is $0$ , no matter what power, so $0$ doesn't work. Likewise, the pattern for $5$ is always $5$ . Doing the same for the rest of the digits, we find that the units digits of $1^{16}$ $2^{16}$ $3^{16}$ $4^{16}$ $6^{16}$ $7^{16}$ $8^{16}$ and $9^{16}$ all have the remainder of $1$ when divided by $5$ , so $\\boxed{45}$" ]

[ "If the common sum of the first two and last two digits is $n$ , such that $1 \\leq n \\leq 9$ , there are $n$ choices for the first two digits and $n + 1$ choices for the second two digits (since zero may not be the first digit). This gives $\\sum_{n = 1}^9 n(n + 1) = 330$ balanced numbers. If the common sum of the first two and last two digits is $n$ , such that $10 \\leq n \\leq 18$ , there are $19 - n$ choices for both pairs. This gives $\\sum_{n = 10}^{18} (19 - n)^2 = \\sum_{n = 1}^9 n^2 = 285$ balanced numbers. Thus, there are in total $330 + 285 = \\boxed{615}$ balanced numbers.", "Call the number $\\overline{abcd}$ . Then $a+b=c+d$ . Set $a+b=x$\nClearly, $0\\le x \\le18$\nIf $x=0$ $0000$ is not acceptable.\nIf $x=1$ : The only case is $1001$ or $1010$ . 2 choices.\nIf $x=2$ : then since $a\\neq0$ $a=1=b$ or $a=2, b=0$ . There are 3 choices for $(c,d)$ $(2,0), (0, 2), (1, 1)$ $2*3=6$ here.\nIf $x=3$ : Clearly, $a\\neq b$ because if so, the sum will be even, not odd. Counting $(a,b)=(3,0)$ , we have $4$ choices. Subtracting that, we have $3$ choices. Since it doesn't matter whether $c=0$ or $d=0$ , we have 4 choices for $(c,d)$ . So $3*4=12$ here.\nIf $x=4$ : Continue as above. $4$ choices for $(a,b)$ $5$ choices for $(c,d)$ $4*5=20$ here.\nIf $x=5$ : You get the point. $5*6=30$\nIf $x=6$ $6*7=42$\nIf $x=7$ $7*8=56$\nIf $x=8$ $8*9=72$\nIf $x=9$ $9*10=90$\nNow we need to be careful because if $x=10$ $(c,d)=(0,10)$ is not valid. However, we don't have to worry about $a\\neq0$\nIf $x=10$ $(a,b)=(1,9), (2, 8), ..., (9, 1)$ . Same thing for $(c,d)$ $9*9=81$\nIf $x=11$ : We start at $(a,b)= (2,9)$ . So $8*8$\nContinue this pattern until $x=18: 1*1=1$ . Add everything up: we have $\\boxed{615}$", "We ignore the requirement that the first digit is non-zero, and do casework on the sum of the sum of the pairs of digits.\nIf two digits $a$ and $b$ sum to $0$ , we have $1$ possibility: $(a,b) = (0,0)$\nIf $a+b = 1$ , we have $2$ possibilities: $(a,b) = (0,1)$ and $(a,b) = (1,0)$\n$a+b = 2$ $(0,2)$ $(2,0)$ , and $(1,1)$ are the only $3$ possibilities.\nWe notice a pattern: for all $k \\leq 9$ , there are $k+1$ ordered pairs of digits $(a,b)$ such that $a+b = k$ . Then, testing for $10 \\leq k \\leq 18$ , we notice that there are $(18 - k) + 1$ ordered pairs with a sum of $k$\nSo the number of ways to pick $4$ digits satisfying the constraint is \\[(1)(1) + (2)(2) \\dots (10)(10) + (9)(9) + \\dots + (1)(1) = \\frac{(10)(11)(21)}{6} + \\frac{(9)(10)(19)}{6} = 670\\]\nWe have to subtract out the cases where the first digit is $0$ , which is $1+2+\\dots+10 = 55$\nSo our answer is $670-55 = \\boxed{615}$" ]

[ "Each of the vertices is counted $3$ times because each vertex is shared by three different edges. \nEach of the edges is counted $2$ times because each edge is shared by two different faces. \nSince the sum of the integers assigned to all vertices is $21$ , the final answer is $21\\times3\\times2=\\boxed{126}$", "Note that each vertex is counted $2\\times 3=6$ times. Thus, the answer is $21\\times6=\\boxed{126}$", "Just set one vertice equal to $21$ , it is trivial to see that there are $3$ faces with value $42$ , and $42 \\cdot 3=\\boxed{126}$", "Since there are 8 vertices in a cube, there are $\\dfrac{21}4$ vertices for two edges. There are $4$ edges per face, and $6$ faces in a cube, so the value of the cube is $\\dfrac{21}4 \\cdot 24 = \\boxed{126}$", "Set each vertex to value 1, so the sum of the vertices is 8. We find that the value of the cube, if all vertices are 1, is 48. We conclude that the value of the cube is 6 times the value of the sum of the vertices. Therefore, we choose $21\\times6=\\boxed{126}$", "The wording of the problem implies that the answer should hold for any valid combination of integers. Thus, we choose the numbers $21, 0, 0, 0, 0, 0, 0, 0$ , which are indeed $8$ integers that add to $21$ . Doing this, we find three edges that have a value of $21$ , and from there, we get three faces with a value of $42$ (while the other three faces have a value of $0$ ). Adding the three faces together, we get $42+42+42 = \\boxed{126}$" ]

[ "This problem can be solved via recursion since we are \"building a string\" of numbers with some condition. We want to create a new string by adding a new digit at the front so we avoid complications( $0$ can't be at the front and no digit is less than $9$ ). There are $4$ options to add no matter what(try some examples if you want) so the recursion is $S_n=4S_{n-1}$ where $S_n$ stands for the number of such numbers with $n$ digits. Since $S_1=10$ the answer is $\\boxed{640}$" ]

[ "We divide the problem into two cases: one in which zero is one of the digits and one in which it is not. In the latter case, suppose we pick digits $x_1,x_2,x_3,x_4$ such that $x_1<x_2<x_3<x_4$ . There are five arrangements of these digits that satisfy the condition of being snakelike: $x_1x_3x_2x_4$ $x_1x_4x_2x_3$ $x_2x_3x_1x_4$ $x_2x_4x_1x_3$ $x_3x_4x_1x_2$ . Thus there are $5\\cdot {9\\choose 4}=630$ snakelike numbers which do not contain the digit zero.\nIn the second case we choose zero and three other digits such that $0<x_2<x_3<x_4$ . There are three arrangements of these digits that satisfy the condition of being snakelike: $x_2x_30x_4$ $x_2x_40x_3$ $x_3x_40x_2$ . Because we know that zero is a digit, there are $3\\cdot{9\\choose 3}=252$ snakelike numbers which contain the digit zero. Thus there are $630+252=\\boxed{882}$ snakelike numbers.", "Let's create the snakelike number from digits $a < b < c < d$ , and, if we already picked the digits there are 5 ways to do so, as said in the first solution. And, let's just pick the digits from 0-9. This get's a total count of $5\\cdot{10 \\choose 4}$ But, this over-counts since it counts numbers like 0213. We can correct for this over-counting. Lock the first digit as 0 and permute 3 other chosen digits $a < b < c$ . There are 2 ways to permute to make the number snakelike, b-a-c, or c-a-b. And, we pick a,b,c from 1 to 9, since 0 has already been chosen as one of the digits. So, the amount we have overcounted by is $2\\cdot{9 \\choose 3}$ . Thus our answer is $5\\cdot{10 \\choose 4} - 2\\cdot{9 \\choose 3} = \\boxed{882}$", "We will first decide the order of the 4 digits, greatest to least. To do this, we will pretend that we have selected the digits 1,2,3,4, and we need to arrange them to create a snakelike number. By testing all permutations, there are only 5 ways to make a snakelike number: (1,3,2,4),(1,4,2,3),(2,3,1,4),(2,4,1,3),(3,4,1,2).\nNow we select 4 digits to replace the 1,2,3,4.\nIn first 2 of cases: (1,3,2,4),(1,4,2,3), the leading digit is a 1, which means it is the smallest of our 4 digits. If we select a 0, the leading digit will be a zero, which is bad because all numbers between 1000 and 9999 have nonzero leading digits. So, we need to select our 4 digits only from the pool of 1-9. There are 9 choose 4 ways and there are 2 cases:\n$2 \\cdot \\dbinom{9}{4} = 252$\nThus, there are 252 ways for those 2 cases.\nFor the next 3 cases, selecting a 0 is okay, so we can select from the pool of 0-9. There are 10 choose 4 ways to select our 4 digits and there are 3 cases:\n$3 \\cdot \\dbinom{10}{4} = 630$\nFor those 3 cases there are 630 ways.\nThus, our answer is 630+252 = $\\boxed{882}$" ]