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[ "The volume of a cone is $\\frac{1}{3} \\cdot\\pi \\cdot r^2 \\cdot h$ where $r$ is the base radius and $h$ is the height. The water completely fills up the cone so the volume of the water is $\\frac{1}{3}\\cdot18\\cdot144\\pi = 6\\cdot144\\pi$\nThe volume of a cylinder is $\\pi \\cdot r^2 \\cdot h$ so the volume of the water in the cylinder would be $24\\cdot24\\cdot\\pi\\cdot h$\nWe can equate these two expressions because the water volume stays the same like this $24\\cdot24\\cdot\\pi\\cdot h = 6\\cdot144\\pi$ . We get $4h = 6$ and $h=\\frac{6}{4}$\nSo the answer is $\\boxed{1.5}.$", "The water completely fills up the cone. For now, assume the radius of both cone and cylinder are the same. Then the cone has $\\frac{1}{3}$ of the volume of the cylinder, and so the height is divided by $3$ . Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since $2^2 = 4$ ).\nTherefore, the height is divided by $3$ and divided by $4$ , which is $18 \\div 3 \\div 4 = \\boxed{1.5}.$" ]

[ "The volume of a cone is $\\frac{1}{3} \\cdot\\pi \\cdot r^2 \\cdot h$ where $r$ is the base radius and $h$ is the height. The water completely fills up the cone so the volume of the water is $\\frac{1}{3}\\cdot18\\cdot144\\pi = 6\\cdot144\\pi$\nThe volume of a cylinder is $\\pi \\cdot r^2 \\cdot h$ so the volume of the water in the cylinder would be $24\\cdot24\\cdot\\pi\\cdot h$\nWe can equate these two expressions because the water volume stays the same like this $24\\cdot24\\cdot\\pi\\cdot h = 6\\cdot144\\pi$ . We get $4h = 6$ and $h=\\frac{6}{4}$\nSo the answer is $\\boxed{1.5}.$", "The water completely fills up the cone. For now, assume the radius of both cone and cylinder are the same. Then the cone has $\\frac{1}{3}$ of the volume of the cylinder, and so the height is divided by $3$ . Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since $2^2 = 4$ ).\nTherefore, the height is divided by $3$ and divided by $4$ , which is $18 \\div 3 \\div 4 = \\boxed{1.5}.$" ]

[ "Because the smaller triangles are congruent, the shaded area take $\\frac{10}{16}$ of the largest triangles area, which is $\\frac{8 \\times 8}{2}=32$ , so the shaded area is $\\frac{10}{16} \\times 32= \\boxed{20}$", "Each of the triangle has side length of $\\frac{1}{4} \\times 8=2$ , so the area is $\\frac{1}{2} \\times 2 \\times 2=2$ . Because there are $10$ triangles is the shaded area, its area is $2 \\times 10 =\\boxed{20}$" ]

[ "$10$ moves results in a lot of possible endpoints, so we try small cases first.\nIf the object only makes $1$ move, it is obvious that there are only 4 possible points that the object can move to.\nIf the object makes $2$ moves, it can move to $(0, 2)$ $(1, 1)$ $(2, 0)$ $(1, -1)$ $(0, -2)$ $(-1, -1)$ $(-2, 0)$ as well as $(0, 0)$ , for a total of $9$ moves.\nIf the object makes $3$ moves, it can end up at $(0, 3)$ $(2, 1)$ $(1, 2)$ $(3, 0)$ $(2, -1)$ $(1, -2)$ $(0, -3)$ $(-1, -2)$ $(-2, -1)$ $(0, -3)$ etc. for a total of $16$ moves.\nAt this point we can guess that for n moves, there are $(n + 1)^2$ different endpoints. Thus, for 10 moves, there are $11^2 = 121$ endpoints, and the answer is $\\boxed{121}$" ]

[ "Let the starting point be $(0,0)$ . After $10$ steps we can only be in locations $(x,y)$ where $|x|+|y|\\leq 10$ . Additionally, each step changes the parity of exactly one coordinate. Hence after $10$ steps we can only be in locations $(x,y)$ where $x+y$ is even. It can easily be shown that each location that satisfies these two conditions is indeed reachable.\nOnce we pick $x\\in\\{-10,\\dots,10\\}$ , we have $11-|x|$ valid choices for $y$ , giving a total of $\\boxed{121}$ possible positions." ]

[ "$WLOG$ , let the object turn clockwise.\nLet $B = (0, 0)$ $A = (0, -8)$\nNote that the possible points of $C$ create a semi-circle of radius $5$ and center $B$ . The area where $AC < 7$ is enclosed by a circle of radius $7$ and center $A$ . The probability that $AC < 7$ is $\\frac{\\angle ABO}{180 ^\\circ}$\nThe function of $\\odot B$ is $x^2 + y^2 = 25$ , the function of $\\odot A$ is $x^2 + (y+8)^2 = 49$\n$O$ is the point that satisfies the system of equations: $\\begin{cases} x^2 + y^2 = 25 \\\\ x^2 + (y+8)^2 = 49 \\end{cases}$\n$x^2 + (y+8)^2 - x^2 - y^2 = 49 - 25$ $64 + 16y =24$ $y = - \\frac52$ $x = \\frac{5 \\sqrt{3}}{2}$ $O = (\\frac{5 \\sqrt{3}}{2}, - \\frac52)$\nNote that $\\triangle BDO$ is a $30-60-90$ triangle, as $BO = 5$ $BD = \\frac{5 \\sqrt{3}}{2}$ $DO = \\frac52$ . As a result $\\angle CBO = 30 ^\\circ$ $\\angle ABO = 60 ^\\circ$\nTherefore the probability that $AC < 7$ is $\\frac{\\angle ABO}{180 ^\\circ} = \\frac{60 ^\\circ}{180 ^\\circ} = \\boxed{13}$" ]

[ "Since no carrying over is allowed, the range of possible values of any digit of $m$ is from $0$ to the respective digit in $1492$ (the values of $n$ are then fixed). Thus, the number of ordered pairs will be $(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\\cdot 5\\cdot 10\\cdot 3 = \\boxed{300}$" ]

[ "We claim that to maximize the number of cables used, we isolate one computer and connect all cables for the remaining $29$ computers, then connect one more cable for the isolated computer.\nIf a brand A computer is isolated, then the technician can use at most $19\\cdot10+1=191$ cables. If a brand B computer is isolated instead, then the technician can use at most $20\\cdot9+1=181$ cables. Therefore, the answer is $\\boxed{191}.$" ]

[ "First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to $0.58$\nThe probability both are green is $\\frac{4}{10}\\cdot\\frac{16}{16+N}$ , and the probability both are blue is $\\frac{6}{10}\\cdot\\frac{N}{16+N}$ , so \\[\\frac{4}{10}\\cdot\\frac{16}{16+N}+\\frac{6}{10}\\cdot\\frac{N}{16+N}=\\frac{29}{50}\\] Solving this equation, \\[20\\left(\\frac{16}{16+N}\\right)+30\\left(\\frac{N}{16+N}\\right)=29\\] Multiplying both sides by $16+N$ , we get\n\\begin{align*} 20\\cdot16+30\\cdot N&=29(16+N)\\\\ 320+30N&=464+29N\\\\ N&=\\boxed{144}" ]

[ "Let the bag contain $n$ marbles total, with $r, w, b, g$ representing the number of red, white, blue, and green marbles, respectively. Note that $r + w + b + g = n$\nThe number of ways to select four red marbles out of the set of marbles without replacement is:\n\\[\\binom{r}{4} = \\frac{r!}{24\\cdot (r -4)!}\\]\nThe number of ways to select one white and three red marbles is:\n\\[\\binom{w}{1}\\binom{r}{3} = \\frac{w\\cdot r!}{6\\cdot (r - 3)!}\\]\nThe number of ways to select one white, one blue, and two red marbles is:\n\\[\\binom{w}{1}\\binom{b}{1} \\binom{r}{2} = \\frac{wb\\cdot r!}{2(r-2)!}\\]\nThe number of ways to select one marble of each colors is:\n\\[\\binom{w}{1}\\binom{b}{1} \\binom{g}{1}\\binom{r}{1} = wbgr\\] Setting the first and second statements equal, we find:\n\\[\\frac{r!}{24\\cdot (r -4)!} = \\frac{w\\cdot r!}{6\\cdot (r - 3)!}\\]\n\\[r - 3 = 4w\\]\nSetting the first and third statements equal, we find:\n\\[\\frac{r!}{24\\cdot (r -4)!} = \\frac{wb\\cdot r!}{2(r-2)!}\\]\n\\[(r-3)(r-2) = 12wb\\]\nSetting the last two statements equal, we find:\n\\[\\frac{wb\\cdot r!}{2(r-2)!} = wbgr\\]\n\\[r - 1 = 2g\\]\nThese are all the \"linking equations\" that are needed; the transitive property of equality makes the other three equalities unnecessary.\nFrom the first equation, we know that $r$ must be $3$ more than a multiple of $4$ , or that $r \\equiv 3 \\mod 4$\nPutting the first equation into the second equation, we find $r-2 = 3b$ . Therefore, $r \\equiv 2 \\mod 3$ . Using the Chinese Remainder Theorem , we find that $r \\equiv 11 \\mod 12$\nThe third equation gives no new restrictions on $r$ ; it is already odd by the first equation.\nThus, the minimal positive value of $r$ is $11$ . This requires $g=\\frac{r - 1}{2} = 5$ by the third equation, and $w = \\frac{r-3}{4} = 2$ by the first equation. Finally, the second equation gives $b = \\frac{(r-3)(r-2)}{12w} = 3$\nThe minimal total number of marbles is $11 + 5 + 2 + 3 = \\boxed{21}$" ]

[ "Suppose that we have a deck, currently containing just one black card. We then insert $n$ red cards one-by-one into the deck at random positions. It is easy to see using induction, that the black card is randomly situated in the deck.\nNow, suppose that we have this deck again, with only one black card. Each time we pick a red ball, we place a card above the black card, and each time we pick a blue ball, we place a card below the black card. It is easy to see that the probability that the card is inserted into the top part of the deck is simply equal to the number of red balls divided by the total number of balls, and the probability that the card is inserted into the bottom part of the deck is equal to the number of blue balls divided by the total number of balls. Therefore, this is equivalent to inserting the card randomly into the deck.\nFinally, four more red cards will be inserted into the deck, and so the black card can be in five possible positions. Only one corresponds to having three balls of each type. Our probability is thus $\\frac{1}{5}$ , and so the answer is $\\boxed{15}$", "Let $R$ denote the action where George selects a red ball and $B$ denote the action where he selects a blue one. Now, in order to get $3$ balls of each color, he needs $2$ more of both $R$ and $B$\nThere are 6 cases: $RRBB, RBRB, RBBR, BBRR, BRBR, BRRB$ (we can confirm that there are only $6$ since $\\binom{4}{2}=6$ ). However we can clump $RRBB + BBRR$ $RBRB + BRBR$ , and $RBBR + BRRB$ together since they are equivalent by symmetry.\n$\\textbf{CASE 1: }$ $RRBB$ and $BBRR$\nLet's find the probability that he picks the balls in the order of $RRBB$\nThe probability that the first ball he picks is red is $\\frac{1}{2}$\nNow there are $2$ reds and $1$ blue in the urn. The probability that he picks red again is now $\\frac{2}{3}$\nThere are $3$ reds and $1$ blue now. The probability that he picks a blue is $\\frac{1}{4}$\nFinally, there are $3$ reds and $2$ blues. The probability that he picks a blue is $\\frac{2}{5}$\nSo the probability that the $RRBB$ case happens is $\\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{1}{4} \\cdot \\frac{2}{5} = \\frac{1}{30}$ . However, since the $BBRR$ case is the exact same by symmetry, case 1 has a probability of $\\frac{1}{30} \\cdot 2 = \\frac{1}{15}$ chance of happening.\n$\\textbf{CASE 2: }$ $RBRB$ and $BRBR$\nLet's find the probability that he picks the balls in the order of $RBRB$\nThe probability that the first ball he picks is red is $\\frac{1}{2}$\nNow there are $2$ reds and $1$ blue in the urn. The probability that he picks blue is $\\frac{1}{3}$\nThere are $2$ reds and $2$ blues now. The probability that he picks a red is $\\frac{1}{2}$\nFinally, there are $3$ reds and $2$ blues. The probability that he picks a blue is $\\frac{2}{5}$\nSo the probability that the $RBRB$ case happens is $\\frac{1}{2} \\cdot \\frac{1}{3} \\cdot \\frac{1}{2} \\cdot \\frac{2}{5} = \\frac{1}{30}$ . However, since the $BRBR$ case is the exact same by symmetry, case 2 has a probability of $\\frac{1}{30} \\cdot 2 = \\frac{1}{15}$ chance of happening.\n$\\textbf{CASE 3: }$ $RBBR$ and $BRRB$\nLet's find the probability that he picks the balls in the order of $RBBR$\nThe probability that the first ball he picks is red is $\\frac{1}{2}$\nNow there are $2$ reds and $1$ blue in the urn. The probability that he picks blue is $\\frac{1}{3}$\nThere are $2$ reds and $2$ blues now. The probability that he picks a blue is $\\frac{1}{2}$\nFinally, there are $2$ reds and $3$ blues. The probability that he picks a red is $\\frac{2}{5}$\nSo the probability that the $RBBR$ case happens is $\\frac{1}{2} \\cdot \\frac{1}{3} \\cdot \\frac{1}{2} \\cdot \\frac{2}{5} = \\frac{1}{30}$ . However, since the $BRRB$ case is the exact same by symmetry, case 3 has a probability of $\\frac{1}{30} \\cdot 2 = \\frac{1}{15}$ chance of happening.\nAdding up the cases, we have $\\frac{1}{15}+\\frac{1}{15}+\\frac{1}{15}=\\boxed{15}$ ~quacker88", "We know that we need to find the probability of adding 2 red and 2 blue balls in some order.\nThere are 6 ways to do this, since there are $\\binom{4}{2}=6$ ways to arrange $RRBB$ in some order.\nWe will show that the probability for each of these 6 ways is the same.\nWe first note that the denominators should be counted by the same number. This number is $2 \\cdot 3 \\cdot 4 \\cdot 5=120$ . This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the $k-th$ step involves $k+1$ numbers to choose from.\nThe numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally.\nThe same goes for the blue ones. The numerator must equal $(1 \\cdot 2)^2$\nTherefore, the probability for each of the orderings of $RRBB$ is $\\frac{4}{120}=\\frac{1}{30}$ . There are 6 of these, so the total probability is $\\boxed{15}$", "First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a $\\frac{1}{2}$ chance each. We can assume he chooses Red(chance $\\frac{1}{2}$ ), and then multiply the final answer by two for symmetry. Now, there are two red balls and one blue ball in the urn. Then, he can either choose another Red(chance $\\frac{2}{3}$ ), in which case he must choose two blues to get three of each, with probability $\\frac{1}{4}\\cdot\\frac{2}{5}=\\frac{1}{10}$ or a blue for two blue and two red in the urn, with chance $\\frac{1}{3}$ . If he chooses blue next, he can either choose a red then a blue, or a blue then a red. Each of these has a $\\frac{1}{2}\\cdot\\frac{2}{5}$ for total of $2\\cdot\\frac{1}{5}=\\frac{2}{5}$ . The total probability that he ends up with three red and three blue is $2\\cdot\\frac{1}{2}(\\frac{2}{3}\\cdot\\frac{1}{10}+\\frac{1}{3}\\cdot\\frac{2}{5})=\\frac{1}{15}+\\frac{2}{15}=\\boxed{15}$", "Let the probability that the urn ends up with more red balls be denoted $P(R)$ . Since this is equal to the probability there are more blue balls, the probability there are equal amounts is $1-2P(R)$ $P(R) =$ the probability no more blues are chosen plus the probability only 1 more blue is chosen. The first case, $P(\\text{no more blues}) = \\frac{1}{2}\\cdot\\frac{2}{3}\\cdot\\frac{3}{4}\\cdot\\frac{4}{5}=\\frac{1}{5}$\nThe second case, $P(\\text{1 more blue}) = 4\\cdot\\frac{1\\cdot1\\cdot2\\cdot3}{2\\cdot3\\cdot4\\cdot5} = \\frac{1}{5}$ . Thus, the answer is $1-2\\left(\\frac{1}{5}+\\frac{1}{5}\\right)=1-\\frac{4}{5}=\\boxed{15}$", "Here $X$ stands for R or B, and $Y$ for the remaining color.\nAfter 3 rounds one can either have a $4+1$ configuration ( $XXXXY$ ), or $3+2$ configuration ( $XXXYY$ ). The probability of getting to $XXXYYY$ from $XXXYY$ is $\\frac{2}{5}$ . Observe that the probability of arriving to $4+1$ configuration is \\[\\frac{2}{3} \\cdot \\frac{3}{4} = \\frac{1}{2}\\] $\\frac{2}{3}$ to get from $XXY$ to $XXXY$ $\\frac{3}{4}$ to get from $XXXY$ to $XXXXY$ ). Thus the probability of arriving to $3+2$ configuration is also $\\frac{1}{2}$ , and the answer is \\[\\frac{1}{2} \\cdot \\frac{2}{5} = \\boxed{15}.\\]", "We can use dynamic programming to solve this problem.\nWe let $dp[i][j]$ be the probability that we end up with $i$ red balls and $j$ blue balls.\nNotice that there are only two ways that we can end up with $i$ red balls and $j$ blue balls: one is by fetching a red ball from the urn when we have $i - 1$ red balls and $j$ blue balls and the other is by fetching a blue ball from the urn when we have $i$ red balls and $j - 1$ blue balls.\nThen we have $dp[i][j] = \\frac{i - 1}{i - 1 + j} dp[i - 1][j] + \\frac{j - 1}{i - 1 + j} dp[i][j - 1]$\nThen we start can with $dp[1][1] = 1$ and try to compute $dp[3][3]$\n\\[\\begin{array}{|c || c | c | c | c | c |} \\hline i \\text{\\ \\textbackslash\\ } j & 1 & 2 & 3\\\\ \\hline\\hline 1 & 1 & 1/2 & 1/3\\\\ \\hline 2 & 1/2 & 1/3 & 1/4\\\\ \\hline 3 & 1/3 & 1/4 & 1/5\\\\ \\hline \\end{array}\\] The answer is $\\boxed{15}$" ]

[ "Suppose that we have a deck, currently containing just one black card. We then insert $n$ red cards one-by-one into the deck at random positions. It is easy to see using induction, that the black card is randomly situated in the deck.\nNow, suppose that we have this deck again, with only one black card. Each time we pick a red ball, we place a card above the black card, and each time we pick a blue ball, we place a card below the black card. It is easy to see that the probability that the card is inserted into the top part of the deck is simply equal to the number of red balls divided by the total number of balls, and the probability that the card is inserted into the bottom part of the deck is equal to the number of blue balls divided by the total number of balls. Therefore, this is equivalent to inserting the card randomly into the deck.\nFinally, four more red cards will be inserted into the deck, and so the black card can be in five possible positions. Only one corresponds to having three balls of each type. Our probability is thus $\\frac{1}{5}$ , and so the answer is $\\boxed{15}$", "Let $R$ denote the action where George selects a red ball and $B$ denote the action where he selects a blue one. Now, in order to get $3$ balls of each color, he needs $2$ more of both $R$ and $B$\nThere are 6 cases: $RRBB, RBRB, RBBR, BBRR, BRBR, BRRB$ (we can confirm that there are only $6$ since $\\binom{4}{2}=6$ ). However we can clump $RRBB + BBRR$ $RBRB + BRBR$ , and $RBBR + BRRB$ together since they are equivalent by symmetry.\n$\\textbf{CASE 1: }$ $RRBB$ and $BBRR$\nLet's find the probability that he picks the balls in the order of $RRBB$\nThe probability that the first ball he picks is red is $\\frac{1}{2}$\nNow there are $2$ reds and $1$ blue in the urn. The probability that he picks red again is now $\\frac{2}{3}$\nThere are $3$ reds and $1$ blue now. The probability that he picks a blue is $\\frac{1}{4}$\nFinally, there are $3$ reds and $2$ blues. The probability that he picks a blue is $\\frac{2}{5}$\nSo the probability that the $RRBB$ case happens is $\\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{1}{4} \\cdot \\frac{2}{5} = \\frac{1}{30}$ . However, since the $BBRR$ case is the exact same by symmetry, case 1 has a probability of $\\frac{1}{30} \\cdot 2 = \\frac{1}{15}$ chance of happening.\n$\\textbf{CASE 2: }$ $RBRB$ and $BRBR$\nLet's find the probability that he picks the balls in the order of $RBRB$\nThe probability that the first ball he picks is red is $\\frac{1}{2}$\nNow there are $2$ reds and $1$ blue in the urn. The probability that he picks blue is $\\frac{1}{3}$\nThere are $2$ reds and $2$ blues now. The probability that he picks a red is $\\frac{1}{2}$\nFinally, there are $3$ reds and $2$ blues. The probability that he picks a blue is $\\frac{2}{5}$\nSo the probability that the $RBRB$ case happens is $\\frac{1}{2} \\cdot \\frac{1}{3} \\cdot \\frac{1}{2} \\cdot \\frac{2}{5} = \\frac{1}{30}$ . However, since the $BRBR$ case is the exact same by symmetry, case 2 has a probability of $\\frac{1}{30} \\cdot 2 = \\frac{1}{15}$ chance of happening.\n$\\textbf{CASE 3: }$ $RBBR$ and $BRRB$\nLet's find the probability that he picks the balls in the order of $RBBR$\nThe probability that the first ball he picks is red is $\\frac{1}{2}$\nNow there are $2$ reds and $1$ blue in the urn. The probability that he picks blue is $\\frac{1}{3}$\nThere are $2$ reds and $2$ blues now. The probability that he picks a blue is $\\frac{1}{2}$\nFinally, there are $2$ reds and $3$ blues. The probability that he picks a red is $\\frac{2}{5}$\nSo the probability that the $RBBR$ case happens is $\\frac{1}{2} \\cdot \\frac{1}{3} \\cdot \\frac{1}{2} \\cdot \\frac{2}{5} = \\frac{1}{30}$ . However, since the $BRRB$ case is the exact same by symmetry, case 3 has a probability of $\\frac{1}{30} \\cdot 2 = \\frac{1}{15}$ chance of happening.\nAdding up the cases, we have $\\frac{1}{15}+\\frac{1}{15}+\\frac{1}{15}=\\boxed{15}$ ~quacker88", "We know that we need to find the probability of adding 2 red and 2 blue balls in some order.\nThere are 6 ways to do this, since there are $\\binom{4}{2}=6$ ways to arrange $RRBB$ in some order.\nWe will show that the probability for each of these 6 ways is the same.\nWe first note that the denominators should be counted by the same number. This number is $2 \\cdot 3 \\cdot 4 \\cdot 5=120$ . This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the $k-th$ step involves $k+1$ numbers to choose from.\nThe numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally.\nThe same goes for the blue ones. The numerator must equal $(1 \\cdot 2)^2$\nTherefore, the probability for each of the orderings of $RRBB$ is $\\frac{4}{120}=\\frac{1}{30}$ . There are 6 of these, so the total probability is $\\boxed{15}$", "First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a $\\frac{1}{2}$ chance each. We can assume he chooses Red(chance $\\frac{1}{2}$ ), and then multiply the final answer by two for symmetry. Now, there are two red balls and one blue ball in the urn. Then, he can either choose another Red(chance $\\frac{2}{3}$ ), in which case he must choose two blues to get three of each, with probability $\\frac{1}{4}\\cdot\\frac{2}{5}=\\frac{1}{10}$ or a blue for two blue and two red in the urn, with chance $\\frac{1}{3}$ . If he chooses blue next, he can either choose a red then a blue, or a blue then a red. Each of these has a $\\frac{1}{2}\\cdot\\frac{2}{5}$ for total of $2\\cdot\\frac{1}{5}=\\frac{2}{5}$ . The total probability that he ends up with three red and three blue is $2\\cdot\\frac{1}{2}(\\frac{2}{3}\\cdot\\frac{1}{10}+\\frac{1}{3}\\cdot\\frac{2}{5})=\\frac{1}{15}+\\frac{2}{15}=\\boxed{15}$", "Let the probability that the urn ends up with more red balls be denoted $P(R)$ . Since this is equal to the probability there are more blue balls, the probability there are equal amounts is $1-2P(R)$ $P(R) =$ the probability no more blues are chosen plus the probability only 1 more blue is chosen. The first case, $P(\\text{no more blues}) = \\frac{1}{2}\\cdot\\frac{2}{3}\\cdot\\frac{3}{4}\\cdot\\frac{4}{5}=\\frac{1}{5}$\nThe second case, $P(\\text{1 more blue}) = 4\\cdot\\frac{1\\cdot1\\cdot2\\cdot3}{2\\cdot3\\cdot4\\cdot5} = \\frac{1}{5}$ . Thus, the answer is $1-2\\left(\\frac{1}{5}+\\frac{1}{5}\\right)=1-\\frac{4}{5}=\\boxed{15}$", "Here $X$ stands for R or B, and $Y$ for the remaining color.\nAfter 3 rounds one can either have a $4+1$ configuration ( $XXXXY$ ), or $3+2$ configuration ( $XXXYY$ ). The probability of getting to $XXXYYY$ from $XXXYY$ is $\\frac{2}{5}$ . Observe that the probability of arriving to $4+1$ configuration is \\[\\frac{2}{3} \\cdot \\frac{3}{4} = \\frac{1}{2}\\] $\\frac{2}{3}$ to get from $XXY$ to $XXXY$ $\\frac{3}{4}$ to get from $XXXY$ to $XXXXY$ ). Thus the probability of arriving to $3+2$ configuration is also $\\frac{1}{2}$ , and the answer is \\[\\frac{1}{2} \\cdot \\frac{2}{5} = \\boxed{15}.\\]", "We can use dynamic programming to solve this problem.\nWe let $dp[i][j]$ be the probability that we end up with $i$ red balls and $j$ blue balls.\nNotice that there are only two ways that we can end up with $i$ red balls and $j$ blue balls: one is by fetching a red ball from the urn when we have $i - 1$ red balls and $j$ blue balls and the other is by fetching a blue ball from the urn when we have $i$ red balls and $j - 1$ blue balls.\nThen we have $dp[i][j] = \\frac{i - 1}{i - 1 + j} dp[i - 1][j] + \\frac{j - 1}{i - 1 + j} dp[i][j - 1]$\nThen we start can with $dp[1][1] = 1$ and try to compute $dp[3][3]$\n\\[\\begin{array}{|c || c | c | c | c | c |} \\hline i \\text{\\ \\textbackslash\\ } j & 1 & 2 & 3\\\\ \\hline\\hline 1 & 1 & 1/2 & 1/3\\\\ \\hline 2 & 1/2 & 1/3 & 1/4\\\\ \\hline 3 & 1/3 & 1/4 & 1/5\\\\ \\hline \\end{array}\\] The answer is $\\boxed{15}$" ]

[ "Let $A$ be the age of Ana and $B$ be the age of Bonita. Then,\n\\[A-1 = 5(B-1)\\] and \\[A = B^2.\\]\nSubstituting the second equation into the first gives us\n\\[B^2-1 = 5(B-1).\\]\nBy using difference of squares and dividing, $B=4.$ Moreover, $A=B^2=16.$\nThe answer is $16-4 = 12 \\implies \\boxed{12}$", "Simple guess and check works. Start with all the square numbers - $1$ $4$ $9$ $16$ $25$ $36$ , etc. (probably stop at around $100$ since at that point it wouldn't make sense). If Ana is $9$ , then Bonita is $3$ , so in the previous year, Ana's age was $4$ times greater than Bonita's. If Ana is $16$ , then Bonita is $4$ , and Ana's age was $5$ times greater than Bonita's in the previous year, as required. The difference in the ages is $16 - 4 = 12. \\boxed{12}$", "The second sentence of the problem says that Ana's age was once $5$ times Bonita's age. Therefore, the difference of the ages $n$ must be divisible by $4.$ The only answer choice which is divisible by $4$ is $12 \\rightarrow \\boxed{12}.$" ]

[ "Notice that a 20% raise translates to Ana's monthly salary multiplied by $(100 + 10)\\% = 120\\%$ , and a 20% pay cut translates to her salary multiplied by $100\\% - 20\\% = 80\\%$ , so\n\\[2000 \\cdot 120\\% = 2400\\] \\[2400 \\cdot 80\\% = 1920 \\rightarrow \\boxed{1920}\\]" ]

[ "\nLet $a,b,c$ be the labeled lengths as shown in the diagram. Also, assume WLOG the time taken is $1$ second.\nObserve that $\\dfrac{2a+b+c}{8.6}=1$ or $2a+b+c=8.6$ , and $\\dfrac{b+c}{6.2}=1$ or $b+c=6.2$ . Subtracting the second equation from the first gives $2a=2.4$ , or $a=1.2$\nNow, let us solve $b$ and $c$ . Note that $\\dfrac{\\sqrt{b^2+c^2}}{5}=1$ , or $b^2+c^2=25$ . We also have $b+c=6.2$\nWe have a system of equations: \\[\\left\\{\\begin{array}{l}b+c=6.2\\\\ b^2+c^2=25\\end{array}\\right.\\]\nSquaring the first equation gives $b^2+2bc+c^2=38.44$ , and subtracting the second from this gives $2bc=13.44$ . Now subtracting this from $b^2+c^2=25$ gives $b^2-2bc+c^2=(b-c)^2=11.56$ , or $b-c=3.4$ . Now we have the following two equations:\n\\[\\left\\{\\begin{array}{l}b+c=6.2\\\\ b-c=3.4\\end{array}\\right.\\]\nAdding the equations and dividing by two gives $b=4.8$ , and it follows that $c=1.4$\nThe ratios we desire are therefore $1.4:6:4.8=7:30:24$ , and our answer is $7+30+24=\\boxed{061}$" ]

[ "Since the distance between them decreases at a rate of $1$ kilometer per minute when they are both biking, their combined speed is $1$ kilometer per minute. Andrea travels three times as fast as Lauren, so they travel at speeds of $\\frac{3}{4}$ kilometers per minute and $\\frac{1}{4}$ kilometers per minute, respectively.\nAfter $5$ minutes, the distance between them will have decreased by $5$ kilometers, so they will be $20-5 = 15$ kilometers apart when Andrea stops. Then, Lauren will take $\\frac{15}{\\frac{1}{4}} = 15*4=60$ more minutes to reach Andrea.\nThey started to bike $5$ minutes before Andrea stopped, so the total time Lauren passed from the time they started biking to the time Lauren reached Andrea is $60+5=65$ minutes. Hence, the answer is $\\boxed{65}$ . ~azc1027" ]

[ "Because the speed of Andrea is 3 times as fast as Lauren and the distance between them is decreasing at a rate of 1 kilometer per minute, Andrea's speed is $\\frac{3}{4} \\textbf{km/min}$ , and Lauren's $\\frac{1}{4} \\textbf{km/min}$ . Therefore, after 5 minutes, Andrea will have biked $\\frac{3}{4} \\cdot 5 = \\frac{15}{4}km$\nIn all, Lauren will have to bike $20 - \\frac{15}{4} = \\frac{80}{4} - \\frac{15}{4} = \\frac{65}{4}km$ . Because her speed is $\\frac{1}{4} \\textbf{km/min}$ , the time elapsed will be $\\frac{\\frac{65}{4}}{\\frac{1}{4}} = \\boxed{65}$" ]

[ "Let their speeds in kilometers per hour be $v_A$ and $v_L$ . We know that $v_A=3v_L$ and that $v_A+v_L=60$ . (The second equation follows from the fact that $1\\mathrm km/min = 60\\mathrm km/h$ .) This solves to $v_A=45$ and $v_L=15$\nAs the distance decreases at a rate of $1$ kilometer per minute, after $5$ minutes the distance between them will be $20-5=15$ kilometers.\nFrom this point on, only Lauren will be riding her bike. As there are $15$ kilometers remaining and $v_L=15$ , she will need exactly an hour to get to Andrea. Therefore the total time in minutes is $5+60 = \\boxed{65}$" ]

[ "Let our denominator be $(5!)^3$ , so we consider all possible distributions.\nWe use PIE (Principle of Inclusion and Exclusion) to count the successful ones.\nWhen we have at $1$ box with all $3$ balls the same color in that box, there are $_{5} C _{1} \\cdot _{5} P _{1} \\cdot (4!)^3$ ways for the distributions to occur ( $_{5} C _{1}$ for selecting one of the five boxes for a uniform color, $_{5} P _{1}$ for choosing the color for that box, $4!$ for each of the three people to place their remaining items).\nHowever, we overcounted those distributions where two boxes had uniform color, and there are $_{5} C _{2} \\cdot _{5} P _{2} \\cdot (3!)^3$ ways for the distributions to occur ( $_{5} C _{2}$ for selecting two of the five boxes for a uniform color, $_{5} P _{2}$ for choosing the color for those boxes, $3!$ for each of the three people to place their remaining items).\nAgain, we need to re-add back in the distributions with three boxes of uniform color... and so on so forth.\nOur success by PIE is \\[_{5} C _{1} \\cdot _{5} P _{1} \\cdot (4!)^3 - _{5} C _{2} \\cdot _{5} P _{2} \\cdot (3!)^3 + _{5} C _{3} \\cdot _{5} P _{3} \\cdot (2!)^3 - _{5} C _{4} \\cdot _{5} P _{4} \\cdot (1!)^3 + _{5} C _{5} \\cdot _{5} P _{5} \\cdot (0!)^3 = 120 \\cdot 2556.\\] \\[\\frac{120 \\cdot 2556}{120^3}=\\frac{71}{400},\\] yielding an answer of $\\boxed{471}$", "As In Solution 1, the probability is \\[\\frac{\\binom{5}{1}\\cdot 5\\cdot (4!)^3 - \\binom{5}{2}\\cdot 5\\cdot 4\\cdot (3!)^3 + \\binom{5}{3}\\cdot 5\\cdot 4\\cdot 3\\cdot (2!)^3 - \\binom{5}{4}\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot (1!)^3 + \\binom{5}{5}\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1}{(5!)^3}\\] \\[= \\frac{5\\cdot 5\\cdot (4!)^3 - 10\\cdot 5\\cdot 4\\cdot (3!)^3 + 10\\cdot 5\\cdot 4\\cdot 3\\cdot (2!)^3 - 5\\cdot 5! + 5!}{(5!)^3}.\\] Dividing by $5!$ , we get \\[\\frac{5\\cdot (4!)^2 - 10\\cdot (3!)^2 + 10\\cdot (2!)^2 - 5 + 1}{(5!)^2}.\\] Dividing by $4$ , we get \\[\\frac{5\\cdot 6\\cdot 24 - 10\\cdot 9 + 10 - 1}{30\\cdot 120}.\\] Dividing by $9$ , we get \\[\\frac{5\\cdot 2\\cdot 8 - 10 + 1}{10\\cdot 40} = \\frac{71}{400} \\implies \\boxed{471}\\]", "Use complementary counting. \nDenote $T_n$ as the total number of ways to put $n$ colors to $n$ boxes by 3 people out of which $f_n$ ways are such that no box has uniform color. Notice $T_n = (n!)^3$ . From this setup we see the question is asking for $1-\\frac{f_5}{(5!)^3}$ . To find $f_5$ we want to exclude the cases of a) one box of the same color, b) 2 boxes of the same color, c) 3 boxes of same color, d) 4 boxes of the same color, and e) 5 boxes of the same color. Cases d) and e) coincide. From this, we have\n\\[f_5=T_5 -{\\binom{5}{1}\\binom{5}{1}\\cdot f_4 - \\binom{5}{2}\\binom{5}{2}\\cdot 2!\\cdot f_3 - \\binom{5}{3}\\binom{5}{3}\\cdot 3!\\cdot f_2 - 5!}\\]\nIn case b), there are $\\binom{5}{2}$ ways to choose 2 boxes that have the same color, $\\binom{5}{2}$ ways to choose the two colors, 2! ways to permute the 2 chosen colors, and $f_3$ ways so that the remaining 3 boxes don’t have the same color. The same goes for cases a) and c). In case e), the total number of ways to permute 5 colors is $5!$ . Now, we just need to calculate $f_2$ $f_3$ and $f_4$\nWe have $f_2=T_2-2 = (2!)^3 - 2 = 6$ , since we subtract the number of cases where the boxes contain uniform colors, which is 2.\nIn the same way, $f_3=T_3-\\Big[3!+ \\binom{3}{1}\\binom{3}{1}\\cdot f_2 \\Big] = 156$ - again, we must subtract the number of ways at least 1 box has uniform color. There are $3!$ ways if 3 boxes each contains uniform color. Two boxes each contains uniform color is the same as previous. If one box has the same color, there are $\\binom{3}{1}$ ways to pick a box, and $\\binom{3}{1}$ ways to pick a color for that box, 1! ways to permute the chosen color, and $f_2$ ways for the remaining 2 boxes to have non-uniform colors. Similarly, $f_4=(4!)^3-\\Big[ 4!+ \\binom{4}{2}\\binom{4}{2}\\cdot 2! \\cdot f_2+ \\binom{4}{1}\\binom{4}{1}\\cdot f_3\\Big] = 10,872$\nThus, $f_5 = f_5=(5!)^3-\\Big[\\binom{5}{1}\\binom{5}{1}\\cdot f_4+ \\binom{5}{2}\\binom{5}{2}\\cdot 2!\\cdot f_3+\\binom{5}{3}\\binom{5}{3}\\cdot 3!\\cdot f_2 + 5!\\Big] = (5!)^3 - 306,720$\nThus, the probability is $\\frac{306,720}{(5!)^3} = 71/400$ and the answer is $\\boxed{471}$", "WLOG fix which block Ang places into each box. (This can also be thought of as labeling each box by the color of Ang's block.) There are then $(5!)^2$ total possibilities.\nAs in Solution 1 , we use PIE. With $1$ box of uniform color, there are ${}_{5} C _{1} \\cdot (4!)^2$ possibilities ( ${}_{5} C _{1}$ for selecting one of the five boxes (whose color is fixed by Ang), $4!$ for each of Ben and Jasmin to place their remaining items). We overcounted cases with $2$ boxes, of which there are ${}_{5} C _{2} \\cdot (3!)^2$ possibilities, and so on.\nThe probability is thus \\[\\frac{{}_{5} C _{1} \\cdot (4!)^2 - {}_{5} C _{2} \\cdot (3!)^2 + {}_{5} C _{3} \\cdot (2!)^2 - {}_{5} C _{4} \\cdot (1!)^2 + {}_{5} C _{5} \\cdot (0!)^2}{(5!)^2}\\] \\[= \\frac{5 \\cdot (4!)^2 - 10 \\cdot (3!)^2 + 10 \\cdot (2!)^2 - 5 + 1}{(5!)^2}\\] at which point we can proceed as in #Alternate simplification to simplify to $\\frac{71}{400} \\implies \\boxed{471}$", "$!n$ denotes the number of derangements of $n$ elements, i.e. the number of permutations where no element appears in its original position. Recall the recursive formula $!0 = 1, !1 = 0, !n = (n-1)(!(n-1)+{!}(n-2))$\nWe will consider the number of candidate boxes (ones that currently remain uniform-color) after each person places their blocks.\nAfter Ang, all $5$ boxes are candidates, and each has a fixed color its blocks must be to remain uniform-color.\nBen has $5!$ total ways to place blocks. There are $\\tbinom{5}{k}\\cdot{!}k$ ways to disqualify $k$ candidates, $0 \\le k \\le 5$ . (There are $\\tbinom{5}{k}$ ways to choose the candidates to disqualify, and $!k$ ways to arrange the disqualified candidates' same-color blocks.)\nJasmin has $5!$ total ways to place blocks. Currently, $k$ boxes are no longer candidates. Consider the number of ways for Jasmin to place blocks such that no box remains a candidate, where $n$ is the number of boxes and $k$ is the number of non-candidates, and call it $D(n,k)$ (not to be confused with partial derangements). We wish to find $5!-D(5,k)$ for each $k$ $0 \\le k \\le 5$\nNotice that $D(n,0) = {!}n$ , since all boxes are candidates and no block can be placed in the same-colored box. Also notice the recursive formula $D(n,k) = D(n-1,k-1)+D(n,k-1)$ , since the first non-candidate box can either contain its same-color block (in which case it can be ignored), or a different-color block (in which case it can be treated as a candidate).\nWe can make a table ( $n$ horizontal, $k$ vertical): \\[\\begin{array}{r|rrrrrr} D(n,k) & 0 & 1 & 2 & 3 & 4 & 5 \\\\ \\hline 0 & 1 & 0 & 1 & 2 & 9 & 44 \\\\ 1 & & 1 & 1 & 3 & 11 & 53 \\\\ 2 & & & 2 & 4 & 14 & 64 \\\\ 3 & & & & 6 & 18 & 78 \\\\ 4 & & & & & 24 & 96 \\\\ 5 & & & & & & 120 \\\\ \\end{array}\\] (Note that $D(n,n) = n!$ since there are no candidates to begin with, which checks out with the diagonal.)\nThe desired values of $D(5,k)$ are in the rightmost column.\nThe probability that at least one box is of uniform color is thus \\begin{align*} & \\frac{\\sum_{k=0}^{5}{\\left(\\binom{5}{k}\\cdot{!}k\\right)(5!-D(5,k))}}{(5!)^2} \\\\ ={}& \\frac{(1\\cdot1)(120-44)+(5\\cdot0)(120-53)+(10\\cdot1)(120-64)+(10\\cdot2)(120-78)+(5\\cdot9)(120-96)+(1\\cdot44)(120-120)}{(5!)^2} \\\\ ={}& \\frac{76+0+560+840+1080+0}{(5!)^2} \\\\ ={}& \\frac{19+140+210+270}{(5!)(5\\cdot3\\cdot2)} \\\\ ={}& \\frac{639}{(5!)(5\\cdot3\\cdot2)} \\\\ ={}& \\frac{71}{(5\\cdot4\\cdot2)(5\\cdot2)} \\\\ ={}& \\frac{71}{400} \\implies \\boxed{471}" ]

[ "If we designate a person to be on a certain side, then all placements of the other people can be considered unique. WLOG, assign Angie to be on the side. There are then $3!=6$ total seating arrangements. If Carlos is across from Angie, there are only $2!=2$ ways to fill the remaining two seats. Then the probability Angie and Carlos are seated opposite each other is $\\frac26=\\boxed{13}$", "If we seat Angie first, there would be only one out of three ways Carlos can sit across from Angie. So the final answer is $\\boxed{13}$" ]

[ "We could see that both $374$ and $319$ are divisible by $11$ in the outset, and that $34$ and $29$ , the quotients, are relatively prime. Both are the $average$ number of minutes across the $11$ days, so we need to subtract $\\left \\lfloor{\\frac{11}{2}}\\right \\rfloor=5$ from each to get $(n,t)=(29,24)$ and $29+24=\\boxed{053}$", "If we let $k$ be equal to the number of days it took to read the book, the sum of $n$ through $n+k$ is equal to $(2n+k)(k+1)=748$ Similarly, $(2t+k)(k+1)=638$ We know that both factors must be integers and we see that the only common multiple of $748$ and $638$ not equal to $1$ that will get us positive integer solutions for $n$ and $t$ is $11$ . We set $k+1=11$ so $k=10$ . We then solve for $n$ and $t$ in their respective equations, getting $2n+10=68$ $n=29$ We also get $2t+10=58$ $t=24$ . Our final answer is $29+24=\\boxed{053}$", "Notice $374=34\\cdot 11$ and $319=29\\cdot 11$ . Also, note the sum of an arithmetic series is $\\frac{2n+k}{2} \\cdot b$ , where $n$ is our first term, $n+k$ is our final term, and $b$ is the number of terms. Since we know both sequences of $n$ and $t$ have the same length, and since $11$ is prime and shared by both $319$ and $374$ , we deduce that $b=11$ . Thus from here we know $2n+k=68$ and $2t+k=58$ by using our other factors $34$ and $29$ . Finally, we add the two systems up and we get $2t+2k+2n=126$ . But, notice that $k=b-1$ , since the first term has $k=0$ , and our last term has $k=b-1$ . Plugging this back into our equation we get $2n+2t=106 \\implies n+t=\\boxed{053}$", "We list two equations:\n\\begin{align*}\nn+(n+1)+...+(n+k)&=374\\\\\nt+(t+1)+...+(t+k)&=319.\n\\end{align*}\nSubtracting the two, we get: \\[(n-t)(k+1)=374-319=55.\\] Manipulating the first and second equation, we get:\n\\begin{align*}\nn(k+1)+\\frac{k(k+1)}{2}&=374 \\\\\nt(k+1)+\\frac{k(k+1)}{2}&=319.\n\\end{align*}\nWe factor out the common factor $k+1$ :\n\\begin{align*}\n(k+1)\\left(n+\\frac{k}{2}\\right)&=374 \\\\\n(k+1)\\left(t+\\frac{k}{2}\\right)&=319.\n\\end{align*}\nNote that $374$ and $319$ have a GCD of $11,$ now combining this with our equation that $(n-t)(k+1)=55,$ we see that $k+1$ has to equal $11.$ Thus, we get: \\[(n,t)=(29,24) \\Rightarrow n+t=29+24=\\boxed{053}.\\] Note: We see that because n-t=5, it becomes impossible for n+k/2 and t+k/2 to both be multiples of 11. Thus, this satisfies our condition. Thus k+1 must be 11 to satisfy the common factor 11 constraint. mathboy282" ]

[ "This problem is very wordy. Nonetheless, let $a$ and $b$ be Ann and Barbara's current ages, respectively. We are given that $a+b=44$ . Let $y$ equal the difference between their ages, so $y=a-b$ . Know that $y$ is constant because the difference between their ages will always be the same.\nNow, let's tackle the equation: $b=$ Ann's age when Barbara was Ann's age when Barbara was $\\frac{a}{2}$ . When Barbara was $\\frac{a}{2}$ years old, Ann was $\\frac{a}{2}+y$ years old. So the equation becomes $b=$ Ann's age when Barbara was $\\frac{a}{2}+y$ . Adding on their age difference again, we get $b = \\frac{a}{2} + y + y \\Rightarrow b = \\frac{a}{2} + 2y$ . Substitute $a-b$ back in for $y$ to get $b = \\frac{a}{2} + 2(a-b)$ . Simplify: $2b = a + 4(a-b) \\Rightarrow 6b = 5a$ . Solving $b$ in terms of $a$ , we have $b = \\frac{5a}{6}$ . Substitute that back into the first equation of $a+b=44$ to get $\\frac{11a}{6}=44$ . Solve for $a$ , and the answer is $\\boxed{24}$ . ~ jiang147369" ]

[ "We can see that a $1$ -step staircase requires $4$ toothpicks and a $2$ -step staircase requires $10$ toothpicks. Thus, to go from a $1$ -step to $2$ -step staircase, $6$ additional toothpicks are needed and to go from a $2$ -step to $3$ -step staircase, $8$ additional toothpicks are needed. Applying this pattern, to go from a $3$ -step to $4$ -step staircase, $10$ additional toothpicks are needed and to go from a $4$ -step to $5$ -step staircase, $12$ additional toothpicks are needed. Our answer is $10+12=\\boxed{22}$" ]

[ "Five years ago, Bella was $6$ years old, and the kitten was $0$ years old.\nToday, Bella is $11$ years old, and the kitten is $5$ years old. It follows that Anna is $30-11-5=14$ years old.\nTherefore, Anna is $14-11=\\boxed{3}$ years older than Bella." ]

[ "Let $x$ be the cost of her dinner.\n$27.50=x+\\frac{1}{10}*x+\\frac{3}{20}*x$\n$27+\\frac{1}{2}=\\frac{5}{4}*x$\n$\\frac{55}{2}=\\frac{5}{4}x$\n$\\frac{55}{2}*\\frac{4}{5}=x$\n$x=22$\n$\\boxed{22}$" ]

[ "Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run $\\boxed{5}$ laps.", "Call $x$ the distance Annie runs. If Annie is $25\\%$ faster than Bonnie, then Bonnie will run a distance of $\\frac{4}{5}x$ . For Annie to meet Bonnie, she must run an extra $400$ meters, the length of the track. So $x-\\left(\\frac{4}{5}\\right)x=400 \\implies x=2000$ , which is $\\boxed{5}$ laps." ]

[ "$70 \\% \\cdot 10=7$\n$80 \\% \\cdot 20=16$\n$90 \\% \\cdot 30=27$\nAdding them up gets $7+16+27=50$ . The overall percentage correct would be $\\frac{50}{60}=\\frac{5}{6}=5 \\cdot 16.\\overline{6}=83.\\overline{3} \\approx \\boxed{83}$" ]

[ "The only equilateral triangles that can be formed are through the diagonals of the faces of the square. From P you have $3$ possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are $3$ possible triangles. So the answer is $\\boxed{3}$ ~Math645\n~andliu766\n~e___", "Each other compatible point must be an even number of edges away from P, so the compatible points are R, V, and T. Therefore, we must choose two of the three points, because P must be a point in the triangle. So, the answer is ${3 \\choose 2} = \\boxed{3}$", "List them out- you get $PRV$ $PRT$ , and $PVT$ . Therefore, the answer is $\\boxed{3}$" ]

[ "Taking the discount of $40$ % means you're only paying $60$ % of the bill. That results in $10,000\\cdot0.6=\\textdollar{6,000}$\nLikewise, taking two discounts of $36$ % and $4$ % means taking $64$ % of the original amount and then $96$ % of the result. That results in $10,000\\cdot0.64\\cdot0.96=\\textdollar{6,144}$\nTaking the difference results in $6,144-6,000=\\textdollar{144}$ , or answer choice $\\boxed{144}$" ]

[ "Shea has grown $20\\%$ , if x was her original height, then $1.2x = 60$ , so she was originally $\\frac{60}{1.2}=50$ inches tall which is a $60 - 50 = 10$ inch increase. Ara also started off at $50$ inches. Since Ara grew half as much as Shea, Ara grew $\\frac{10}{2} = 5$ inches. Therefore, Ara is now $50+5=55$ inches tall which is choice $\\boxed{55}.$" ]

[ "Since $\\left(a_n\\right)$ and $\\left(b_n\\right)$ have integer terms with $a_1=b_1=1$ , we can write the terms of each sequence as\n\\begin{align*}&\\left(a_n\\right) \\Rightarrow \\{1, x+1, 2x+1, 3x+1, ...\\}\\\\ &\\left(b_n\\right) \\Rightarrow \\{1, y+1, 2y+1, 3y+1, ...\\}\\end{align*}\nwhere $x$ and $y$ $x\\leq y$ ) are the common differences of each, respectively.\nSince\n\\begin{align*}a_n &= (n-1)x+1\\\\ b_n &= (n-1)y+1\\end{align*}\nit is easy to see that\n$a_n \\equiv b_n \\equiv 1 \\mod{(n-1)}$\nHence, we have to find the largest $n$ such that $\\frac{a_n-1}{n-1}$ and $\\frac{b_n-1}{n-1}$ are both integers; equivalently, we want to maximize $\\gcd(a_n-1, b_n-1)$\nThe prime factorization of $2010$ is $2\\cdot{3}\\cdot{5}\\cdot{67}$ . We list out all the possible pairs that have a product of $2010$ , noting that these are the possible values of $(a_n, b_n)$ and we need $a_n \\leq b_n$\n\\[(2,1005), (3, 670), (5,402), (6,335), (10,201),(15,134),(30,67)\\]\nand soon find that the largest $n-1$ value is $7$ for the pair $(15, 134)$ , and so the largest $n$ value is $\\boxed{8}$", "Since\n$a_n*b_n = 2010,$\nand\n$a_n \\le b_n$ ,\nblue+yellow=green\nit follows that\n$a_n \\le \\sqrt{2010} \\Rightarrow a_n \\le 44$\nBut $a_n$ and $b_n$ are also integers, so $a_n$ must be a factor of $2010$ smaller than $44$ . Notice that $2010 = 2*3*5*67$ . Therefore $a_n = 2, 3, 5, 6, 112, 15,$ or $30$ and $b_n = 1005, 670, 402, 335, 201, 134,$ or $67$ ; respectively.\nNotice that the term $a_m$ is equivalent to the first term $a_1 = 1$ plus $(m-1)$ times the common difference for that particular arithmetic sequence. Let the common difference of $(a_n)$ be $k$ and the common difference of $(b_n)$ be $i$ (not $\\sqrt{-1}$ ). Then\n$a_n$ (the $n$ th term, not the sequence itself) $=1 + k(n-1)$\nand\n$b_n = 1 + i(n-1)$\nSubtracting one from all the possible values listed above for $a_n$ and $b_n$ , we get\n$k(n-1) = 1, 2, 4, 5, 9, 14, 29$\nand\n$i(n-1) = 1004, 669, 401, 334, 200, 133, 66$\nIn order to maximize $n$ , we must maximize $n-1$ . Therefore $k$ and $i$ are coprime and $n-1$ is the GCF of any corresponding pair. Inspecting all of the pairs, we see that the GCF is always $1$ except for the pair $(14, 133),$ which has a GCF of $7$ . Therefore the maximum value of $n$ is $8 \\Rightarrow \\boxed{8}$" ]

[ "We say that a game state is an N-position if it is winning for the next player (the player to move), and a P-position if it is winning for the other player. We are trying to find which of the given states is a P-position.\nFirst we note that symmetrical positions are P-positions, as the second player can win by mirroring the first player's moves. It follows that $(6, 1, 1)$ is an N-position, since we can win by moving to $(2, 2, 1, 1)$ ; this rules out $\\textbf{(A)}$ . We next look at $(6, 2, 1)$ . The possible next states are \\[(6, 2), \\enskip (6, 1, 1), \\enskip (6, 1), \\enskip (5, 2, 1), \\enskip (4, 2, 1, 1), \\enskip (4, 2, 1), \\enskip (3, 2, 2, 1), \\enskip (3, 2, 1, 1), \\enskip (2, 2, 2, 1).\\] None of these are symmetrical, so we might reasonably suspect that they are all N-positions. Indeed, it just so happens that for all of these states except $(6, 2)$ and $(6, 1)$ , we can win by moving to $(2, 2, 1, 1)$ ; it remains to check that $(6, 2)$ and $(6, 1)$ are N-positions.\nTo save ourselves work, it would be nice if we could find a single P-position directly reachable from both $(6, 2)$ and $(6, 1)$ . We notice that $(3, 2, 1)$ is directly reachable from both states, so it would suffice to show that $(3, 2, 1)$ is a P-position. Indeed, the possible next states are \\[(3, 2), \\enskip (3, 1, 1), \\enskip (3, 1), \\enskip (2, 2, 1), \\enskip (2, 1, 1, 1), \\enskip (2, 1, 1),\\] which allow for the following refutations:\n\\begin{align*} &(3, 2) \\to (2, 2), && &&(3, 1, 1) \\to (1, 1, 1, 1), && &&(3, 1) \\to (1, 1), \\\\ &(2, 2, 1) \\to (2, 2), && &&(2, 1, 1, 1) \\to (1, 1, 1, 1), && &&(2, 1, 1) \\to (1, 1). \\end{align*}\nHence, $(3, 2, 1)$ is a P-position, so $(6, 2)$ and $(6, 1)$ are both N-positions, along with all other possible next states from $(6, 2, 1)$ as noted before. Thus, $(6, 2, 1)$ is a P-position, so our answer is $\\boxed{6,2,1}$ as in Solution 2.)", "$(6,1,1)$ can be turned into $(2,2,1,1)$ by Arjun, which is symmetric, so Beth will lose.\n$(6,3,1)$ can be turned into $(3,1,3,1)$ by Arjun, which is symmetric, so Beth will lose.\n$(6,2,2)$ can be turned into $(2,2,2,2)$ by Arjun, which is symmetric, so Beth will lose.\n$(6,3,2)$ can be turned into $(3,2,3,2)$ by Arjun, which is symmetric, so Beth will lose.\nThat leaves $(6,2,1)$ or $\\boxed{6,2,1}$", "Consider the following, much simpler game.\nArjun and Beth can each either take 1 or 2 bricks from the right-hand-side of a continuous row of initially $n$ bricks.\nIt is easy to see that for $n$ a multiple of 3, Beth can \"mirror\" whatever Arjun plays: if he takes 1, she takes 2, and if he takes 2, she takes 1. With this strategy, Beth always takes the last brick. If $n$ is not a multiple of 3, then Arjun takes whichever amount puts Beth in the losing position.\nThe total number of bricks in the initial states given by the answer choices is $8, 9, 10, 10, 11$ . Thus, answer choice $\\textbf{(B)}$ appears promising as a winning position for Beth. The difference between this game and the simplified game is that in certain positions, namely those consisting of fragments of size only 1, taking 2 bricks is not allowed. We can assume that for the starting position $(6, 2, 1)$ , Beth always has a move to ensure that she can continue to mirror Arjun throughout the game. (This could be proven rigorously with lots of casework. In particular, she must avoid providing Arjun with a position of 3 continuous bricks, because then he could take the middle block and force a win.) The assumption seems reasonable, so the answer is $\\boxed{6,2,1}$", "We can start by guessing and checking our solutions. Let's start with $A$ $(6,1,1)$ . Arjun goes first, and he has a winning strategy. His strategy is to cut the 6 block in half, so whatever Beth does, Arjun will copy. If we see this in practice, Arjun has made the block in to $(2,2,1,1)$ . If Beth takes 1, he will take one. If Beth takes 2, he will take 2 as well.\nThen, we can guess solution $B$ $(6,2,1)$ . If Arjun starts by taking 1 of the blocks, he will either have $(6,1,1)$ $(6,2)$ $(x,y,2,1)$ (Where $x+y = 5$ ). If he has $(6,1,1)$ , Beth can take 2 out of the middle of $(6,1,1)$ to get $(2,2,1,1)$ which we know Arjun will lose because we know by symmetry from answer choice $A$ . If he has $(6,2)$ , Beth will take out one from the edge, and we get $(5,2)$ . Beth can now copy whatever Arjun does, so she will win.\nNext, if he takes one out to get $(x,y,2,1)$ , we can either have $(4,1,2,1)$ or $(3,2,2,1)$ . If it is the first case, Beth can take the edge 2 out of the 4, which gives us $(2,1,2,1)$ and whatever Arjun does, Beth can do, so she has established a win. If it is the 2nd case, Beth can take out 2 from the 3 to get $(1,2,2,1)$ , in which, Beth can copy whatever Arjun does, so Beth will win. So in all 4 cases, Beth wins. So our answer is $\\boxed{6,2,1}$" ]

[ "We say that a game state is an N-position if it is winning for the next player (the player to move), and a P-position if it is winning for the other player. We are trying to find which of the given states is a P-position.\nFirst we note that symmetrical positions are P-positions, as the second player can win by mirroring the first player's moves. It follows that $(6, 1, 1)$ is an N-position, since we can win by moving to $(2, 2, 1, 1)$ ; this rules out $\\textbf{(A)}$ . We next look at $(6, 2, 1)$ . The possible next states are \\[(6, 2), \\enskip (6, 1, 1), \\enskip (6, 1), \\enskip (5, 2, 1), \\enskip (4, 2, 1, 1), \\enskip (4, 2, 1), \\enskip (3, 2, 2, 1), \\enskip (3, 2, 1, 1), \\enskip (2, 2, 2, 1).\\] None of these are symmetrical, so we might reasonably suspect that they are all N-positions. Indeed, it just so happens that for all of these states except $(6, 2)$ and $(6, 1)$ , we can win by moving to $(2, 2, 1, 1)$ ; it remains to check that $(6, 2)$ and $(6, 1)$ are N-positions.\nTo save ourselves work, it would be nice if we could find a single P-position directly reachable from both $(6, 2)$ and $(6, 1)$ . We notice that $(3, 2, 1)$ is directly reachable from both states, so it would suffice to show that $(3, 2, 1)$ is a P-position. Indeed, the possible next states are \\[(3, 2), \\enskip (3, 1, 1), \\enskip (3, 1), \\enskip (2, 2, 1), \\enskip (2, 1, 1, 1), \\enskip (2, 1, 1),\\] which allow for the following refutations:\n\\begin{align*} &(3, 2) \\to (2, 2), && &&(3, 1, 1) \\to (1, 1, 1, 1), && &&(3, 1) \\to (1, 1), \\\\ &(2, 2, 1) \\to (2, 2), && &&(2, 1, 1, 1) \\to (1, 1, 1, 1), && &&(2, 1, 1) \\to (1, 1). \\end{align*}\nHence, $(3, 2, 1)$ is a P-position, so $(6, 2)$ and $(6, 1)$ are both N-positions, along with all other possible next states from $(6, 2, 1)$ as noted before. Thus, $(6, 2, 1)$ is a P-position, so our answer is $\\boxed{6,2,1}$ as in Solution 2.)", "$(6,1,1)$ can be turned into $(2,2,1,1)$ by Arjun, which is symmetric, so Beth will lose.\n$(6,3,1)$ can be turned into $(3,1,3,1)$ by Arjun, which is symmetric, so Beth will lose.\n$(6,2,2)$ can be turned into $(2,2,2,2)$ by Arjun, which is symmetric, so Beth will lose.\n$(6,3,2)$ can be turned into $(3,2,3,2)$ by Arjun, which is symmetric, so Beth will lose.\nThat leaves $(6,2,1)$ or $\\boxed{6,2,1}$", "Consider the following, much simpler game.\nArjun and Beth can each either take 1 or 2 bricks from the right-hand-side of a continuous row of initially $n$ bricks.\nIt is easy to see that for $n$ a multiple of 3, Beth can \"mirror\" whatever Arjun plays: if he takes 1, she takes 2, and if he takes 2, she takes 1. With this strategy, Beth always takes the last brick. If $n$ is not a multiple of 3, then Arjun takes whichever amount puts Beth in the losing position.\nThe total number of bricks in the initial states given by the answer choices is $8, 9, 10, 10, 11$ . Thus, answer choice $\\textbf{(B)}$ appears promising as a winning position for Beth. The difference between this game and the simplified game is that in certain positions, namely those consisting of fragments of size only 1, taking 2 bricks is not allowed. We can assume that for the starting position $(6, 2, 1)$ , Beth always has a move to ensure that she can continue to mirror Arjun throughout the game. (This could be proven rigorously with lots of casework. In particular, she must avoid providing Arjun with a position of 3 continuous bricks, because then he could take the middle block and force a win.) The assumption seems reasonable, so the answer is $\\boxed{6,2,1}$", "We can start by guessing and checking our solutions. Let's start with $A$ $(6,1,1)$ . Arjun goes first, and he has a winning strategy. His strategy is to cut the 6 block in half, so whatever Beth does, Arjun will copy. If we see this in practice, Arjun has made the block in to $(2,2,1,1)$ . If Beth takes 1, he will take one. If Beth takes 2, he will take 2 as well.\nThen, we can guess solution $B$ $(6,2,1)$ . If Arjun starts by taking 1 of the blocks, he will either have $(6,1,1)$ $(6,2)$ $(x,y,2,1)$ (Where $x+y = 5$ ). If he has $(6,1,1)$ , Beth can take 2 out of the middle of $(6,1,1)$ to get $(2,2,1,1)$ which we know Arjun will lose because we know by symmetry from answer choice $A$ . If he has $(6,2)$ , Beth will take out one from the edge, and we get $(5,2)$ . Beth can now copy whatever Arjun does, so she will win.\nNext, if he takes one out to get $(x,y,2,1)$ , we can either have $(4,1,2,1)$ or $(3,2,2,1)$ . If it is the first case, Beth can take the edge 2 out of the 4, which gives us $(2,1,2,1)$ and whatever Arjun does, Beth can do, so she has established a win. If it is the 2nd case, Beth can take out 2 from the 3 to get $(1,2,2,1)$ , in which, Beth can copy whatever Arjun does, so Beth will win. So in all 4 cases, Beth wins. So our answer is $\\boxed{6,2,1}$" ]

[ "We first assume a population of $100$ to facilitate solving. Then we simply organize the statistics given into a Venn diagram.\n\nLet $x$ be the number of men with all three risk factors. Since \"the probability that a randomly selected man has all three risk factors, given that he has A and B is $\\frac{1}{3}$ ,\" we can tell that $x = \\frac{1}{3}(x+14)$ , since there are $x$ people with all three factors and 14 with only A and B. Thus $x=7$\nLet $y$ be the number of men with no risk factors. It now follows that \\[y= 100 - 3 \\cdot 10 - 3 \\cdot 14 - 7 = 21.\\] The number of men with risk factor A is $10+2 \\cdot 14+7 = 45$ (10 with only A, 28 with A and one of the others, and 7 with all three). Thus the number of men without risk factor $A$ is 55, so the desired conditional probability is $21/55$ . So the answer is $21+55=\\boxed{076}$" ]

[ "The radius of each semicircle is $2$ , half the sidelength of the square. The line straight down the middle of square $ABCD$ is the sum of two radii and the length of the smaller square, which is equivalent to its side length. The area of $ABCD$ is $(4+2+2)^2 = \\boxed{64}$" ]

[ "Because they are both moving in the same direction, Emily is riding relative to Emerson $12-8=4$ mph. Now we can look at it as if Emerson is not moving at all [on his skateboard] and Emily is riding at $4$ mph. It takes her\n\\[\\frac12 \\ \\text{mile} \\cdot \\frac{1\\ \\text{hour}}{4\\ \\text{miles}} = \\frac18\\ \\text{hour}\\]\nto ride the $1/2$ mile to reach him, and then the same amount of time to be $1/2$ mile ahead of him. This totals to\n\\[2 \\cdot \\frac18 \\ \\text{hour} \\cdot \\frac{60\\ \\text{minutes}}{1\\ \\text{hour}} = \\boxed{15}\\]" ]

[ "The perimeter of the polygon is $3+4+6+3+7 = 23$ . Hence as we roll the polygon to the right, every $23$ units the side $\\overline{AB}$ will be the bottom side.\nWe have $2009 = 23 \\times 87 + 8$ . Thus at some point in time we will get the situation when $A=(2001,0)$ and $\\overline{AB}$ is the bottom side. Obviously, at this moment $B=(2004,0)$\nAfter that, the polygon rotates around $B$ until point $C$ hits the $x$ axis at $(2008,0)$\nAnd finally, the polygon rotates around $C$ until point $D$ hits the $x$ axis at $(2014,0)$ .\nAt this point the side $\\boxed{2009,0}$" ]

[ "From the top, we can go down in five different ways to the five faces underneath the first face. From here we can go down or go to the adjacent faces. From the face you went down from the top face, you can either go clockwise or counterclockwise $1$ $2$ $3$ ,or $4$ times, or you can go straight down. Then from there, you go down into the lower row, which you have two choices, left or right down. From here we have $5 \\cdot 9 \\cdot 2$ ways multiplied by the ways you can move from the bottom ring to the bottom face, but we don't need to know that since from here we can see that $\\boxed{90}$ . ~Terribleteeth" ]

[ "Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.\nWe have $5$ choices for which face we visit first on the top ring. From there, we have $9$ choices for how far around the top ring we go before moving down: $1,2,3,$ or $4$ faces around clockwise, $1,2,3,$ or $4$ faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring.\nWe then have $2$ choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly $2$ lower-ring faces) and then once again $9$ choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip.\nMultiplying together all the numbers of choices we have, we get $5 \\cdot 9 \\cdot 2 \\cdot 9 = \\boxed{810}$", "Like Solution 3, we can use casework, but in a different way. We can split the caseworks into the amount of moves we spend on each row (top and bottom, where both rows have 5 faces).\nCase 1: 1 move on the top row;\nThere are 5 ways to make 1 move on the top row from the starting point (just moving on each of the five faces). Now we want to look at how many ways we can move towards the bottom. Notice how we have two faces on the bottom row that are adjacent to each of the top faces. If we spend one move on the bottom, there are 2 ways. If we spend two moves on the bottom, there are 4 ways (since order matters, we can flip the direction of these moves). If we spend three moves on the bottom, there are also 4 ways (similar reasoning). If we spend four moves on the bottom, there are 4 ways. And if we spend five moves on the bottom, there are again, 4 ways. In total there are $2 + 4 + 4 + 4 + 4 = 18$ ways to move from the bottom after the top moves are chosen. And since there are 5 ways to make 1 move on the top row, there are $5 \\cdot 18 = 90$ ways.\nCase 2: 2 moves on the top row;\nFrom just counting, we can see that there are five ways to move once on the top, and after that, two options for the second move (either left or right), so there are $5 \\cdot 2 = 10$ ways. And from using what we obtained in Case 1, there are $10 \\cdot 18 = 180$ ways for this case.\nCase 3: three moves on the top row;\nThere are, again, 10 ways to do this. And so, there are $10 \\cdot 18 = 180$ ways for this case to happen.\nNow, we can see that there will be 10 ways for the remaining cases of the first move for the top row (because 5 ways to choose the first, and then you can go either left or right, so $5 \\cdot 2 = 10$ ).\nSo for Case 4 and Case 5, there will both also be $10 \\cdot 18 = 180$ ways.\nFinally, adding all these cases up, we obtain $90 + 180 + 180 + 180 + 180 = 810$ , or $\\boxed{810}$ ~Misclicked" ]

[ "By angle subtraction, we have $\\angle ADE = 360^\\circ - \\angle ADC - \\angle CDE = 160^\\circ.$ Note that $\\triangle DEF$ is isosceles, so $\\angle DFE = \\frac{180^\\circ - \\angle ADE}{2}=10^\\circ.$ Finally, we get $\\angle AFE = 180^\\circ - \\angle DFE = \\boxed{170}$ degrees.", "We can extend $\\overline{AD}$ to $G$ , making $\\angle CDG$ a right angle. It follows that $\\angle GDE$ is $110^\\circ - 90^\\circ = 20^\\circ$ , as shown below. Since $\\angle DFE = \\angle DEF$ , we see that $\\angle DFE = \\angle DEF = \\frac{20}{2} = 10^\\circ$ . Thus, $\\angle AFE = 180^\\circ - 10^\\circ = \\boxed{170}$ degrees." ]

[ "By angle subtraction, we have $\\angle ADE = 360^\\circ - \\angle ADC - \\angle CDE = 160^\\circ.$ Note that $\\triangle DEF$ is isosceles, so $\\angle DFE = \\frac{180^\\circ - \\angle ADE}{2}=10^\\circ.$ Finally, we get $\\angle AFE = 180^\\circ - \\angle DFE = \\boxed{170}$ degrees.", "We can extend $\\overline{AD}$ to $G$ , making $\\angle CDG$ a right angle. It follows that $\\angle GDE$ is $110^\\circ - 90^\\circ = 20^\\circ$ , as shown below. Since $\\angle DFE = \\angle DEF$ , we see that $\\angle DFE = \\angle DEF = \\frac{20}{2} = 10^\\circ$ . Thus, $\\angle AFE = 180^\\circ - 10^\\circ = \\boxed{170}$ degrees." ]

[ "Divide the circle into four parts: the top semicircle by connecting E, F, and G( $A$ ); the bottom sector ( $B$ ), whose arc angle is $120^{\\circ}$ because the large circle's radius is $2$ and the short length (the radius of the smaller semicircles) is $1$ , giving a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle; the triangle formed by the radii of $A$ and the chord ( $C$ ); and the four parts which are the corners of a circle inscribed in a square ( $D$ ). Then the area is $A + B - C + D$ (in $B-C$ , we find the area of the bottom shaded region, and in $D$ we find the area of the shaded region above the semicircles but below the diameter).\nThe area of $A$ is $\\frac{1}{2} \\pi \\cdot 2^2 = 2\\pi$\nThe area of $B$ is $\\frac{120^{\\circ}}{360^{\\circ}} \\pi \\cdot 2^2 = \\frac{4\\pi}{3}$\nFor the area of $C$ , the radius of $2$ , and the distance of $1$ (the smaller semicircles' radius) to $BC$ , creates two $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangles, so $C$ 's area is $2 \\cdot \\frac{1}{2} \\cdot 1 \\cdot \\sqrt{3} = \\sqrt{3}$\nThe area of $D$ is $4 \\cdot 1-\\frac{1}{4}\\pi \\cdot 2^2=4-\\pi$\nHence, finding $A+B-C+D$ , the desired area is $\\frac{7\\pi}{3}-\\sqrt{3}+4$ , so the answer is $7+3+3+4=\\boxed{17}$", "First we have to solve the area of the non-shaded area(the semicircles) that are in Circle $F$ .The middle semicircle has area $\\frac12\\pi$ and the other two have about half of their are inside the circle = $\\frac14\\pi\\ + \\frac14\\pi\\ + \\frac12\\pi\\ = \\pi$ . Then we subtract the part of the quartercircle that isn't in Circle $F$ . This is an area equal to that of a triangle minus an minor segment. The height of the triangle is the radius of the semicircles, which is $1$ . The length is the radius of Circle $F$ minus the length from the center of the middle semicircle up to until it is on the edge of the circle. Using the Pythagorean Theorem we can figure out that the length is: \\[\\sqrt{2^2 - 1^2} = \\sqrt{3}.\\] This means that the length of the triangle is $2 - \\sqrt{3}$ and so the area of the triangle is $\\frac{2 - \\sqrt{3}}{2}$ . For the area of the segment, it's the area of the sector minus the area of the triangle. The triangle's length is the radius of $F$ $2$ , while its height is the radius of the semicircles: $1$ , so the area is 1. The angle is $30^{\\circ}$ as the hypotenuse is the radius of $F$ and the opposite side is the radius of the semicircles, which means the area is $\\frac{1}{12}$ of the whole area, which is $4\\pi$ so the area of the sector is $\\frac{\\pi}{3}$ and the area of the segment is $\\frac{\\pi}{3} - 1$ and so the area of the part of the quartercircles that stick out of Circle $F$ is: \\[(\\frac{2 - \\sqrt{3}}{2})-(\\frac{\\pi}{3} - 1) = \\frac{2 - \\sqrt{3}}{2} + 1 - \\frac{\\pi}{3} = \\frac{4 - \\sqrt{3}}{2} - \\frac{\\pi}{3}.\\]\nSince there are two, one for each side, we have to multiply it by 2, so we have ${4 - \\sqrt{3}} - \\frac{2\\pi}{3}$ , which we subtract from $\\pi$ which gets us $\\frac{5\\pi}{3} - 4 + \\sqrt{3}$ which we subtract from $4\\pi$ $=$ $\\frac{12\\pi}{3}$ , which is $\\frac{7\\pi}{3} + 4 - \\sqrt{3}$ so we get $7+3+4+3=\\boxed{17}$" ]

[ "Divide the circle into four parts: the top semicircle by connecting E, F, and G( $A$ ); the bottom sector ( $B$ ), whose arc angle is $120^{\\circ}$ because the large circle's radius is $2$ and the short length (the radius of the smaller semicircles) is $1$ , giving a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle; the triangle formed by the radii of $A$ and the chord ( $C$ ); and the four parts which are the corners of a circle inscribed in a square ( $D$ ). Then the area is $A + B - C + D$ (in $B-C$ , we find the area of the bottom shaded region, and in $D$ we find the area of the shaded region above the semicircles but below the diameter).\nThe area of $A$ is $\\frac{1}{2} \\pi \\cdot 2^2 = 2\\pi$\nThe area of $B$ is $\\frac{120^{\\circ}}{360^{\\circ}} \\pi \\cdot 2^2 = \\frac{4\\pi}{3}$\nFor the area of $C$ , the radius of $2$ , and the distance of $1$ (the smaller semicircles' radius) to $BC$ , creates two $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangles, so $C$ 's area is $2 \\cdot \\frac{1}{2} \\cdot 1 \\cdot \\sqrt{3} = \\sqrt{3}$\nThe area of $D$ is $4 \\cdot 1-\\frac{1}{4}\\pi \\cdot 2^2=4-\\pi$\nHence, finding $A+B-C+D$ , the desired area is $\\frac{7\\pi}{3}-\\sqrt{3}+4$ , so the answer is $7+3+3+4=\\boxed{17}$", "First we have to solve the area of the non-shaded area(the semicircles) that are in Circle $F$ .The middle semicircle has area $\\frac12\\pi$ and the other two have about half of their are inside the circle = $\\frac14\\pi\\ + \\frac14\\pi\\ + \\frac12\\pi\\ = \\pi$ . Then we subtract the part of the quartercircle that isn't in Circle $F$ . This is an area equal to that of a triangle minus an minor segment. The height of the triangle is the radius of the semicircles, which is $1$ . The length is the radius of Circle $F$ minus the length from the center of the middle semicircle up to until it is on the edge of the circle. Using the Pythagorean Theorem we can figure out that the length is: \\[\\sqrt{2^2 - 1^2} = \\sqrt{3}.\\] This means that the length of the triangle is $2 - \\sqrt{3}$ and so the area of the triangle is $\\frac{2 - \\sqrt{3}}{2}$ . For the area of the segment, it's the area of the sector minus the area of the triangle. The triangle's length is the radius of $F$ $2$ , while its height is the radius of the semicircles: $1$ , so the area is 1. The angle is $30^{\\circ}$ as the hypotenuse is the radius of $F$ and the opposite side is the radius of the semicircles, which means the area is $\\frac{1}{12}$ of the whole area, which is $4\\pi$ so the area of the sector is $\\frac{\\pi}{3}$ and the area of the segment is $\\frac{\\pi}{3} - 1$ and so the area of the part of the quartercircles that stick out of Circle $F$ is: \\[(\\frac{2 - \\sqrt{3}}{2})-(\\frac{\\pi}{3} - 1) = \\frac{2 - \\sqrt{3}}{2} + 1 - \\frac{\\pi}{3} = \\frac{4 - \\sqrt{3}}{2} - \\frac{\\pi}{3}.\\]\nSince there are two, one for each side, we have to multiply it by 2, so we have ${4 - \\sqrt{3}} - \\frac{2\\pi}{3}$ , which we subtract from $\\pi$ which gets us $\\frac{5\\pi}{3} - 4 + \\sqrt{3}$ which we subtract from $4\\pi$ $=$ $\\frac{12\\pi}{3}$ , which is $\\frac{7\\pi}{3} + 4 - \\sqrt{3}$ so we get $7+3+4+3=\\boxed{17}$" ]

[ "Let the interior point be $P$ , let the points on $\\overline{BC}$ $\\overline{CA}$ and $\\overline{AB}$ be $D$ $E$ and $F$ , respectively. Let $x$ be the area of $\\triangle APE$ and $y$ be the area of $\\triangle CPD$ . Note that $\\triangle APF$ and $\\triangle BPF$ share the same altitude from $P$ , so the ratio of their areas is the same as the ratio of their bases. Similarly, $\\triangle ACF$ and $\\triangle BCF$ share the same altitude from $C$ , so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are actually the same, and thus in the same ratio. As a result, we have: $\\frac{40}{30} = \\frac{124 + x}{65 + y}$ or equivalently $372 + 3x = 260 + 4y$ and so $4y = 3x+ 112$\nApplying identical reasoning to the triangles with bases $\\overline{CD}$ and $\\overline{BD}$ , we get $\\frac{y}{35} = \\frac{x+y+84}{105}$ so that $3y = x + y + 84$ and $2y = x + 84$ . Substituting from this equation into the previous one gives $x = 56$ , from which we get $y = 70$ and so the area of $\\triangle ABC$ is $56 + 40 + 30 + 35 + 70 + 84 = \\Rightarrow \\boxed{315}$", "This problem can be done using mass points. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas that state (The sum of the two weights) = (The middle weight), and (The weight $\\times$ side) = (Other weight) $\\times$ (The other side), the problem yields the answer $\\boxed{315}$", "Let the interior point be $P$ and let the points on $\\overline{BC}$ $\\overline{CA}$ and $\\overline{AB}$ be $D$ $E$ and $F$ , respectively. Also, let $[APE]=x,[CPD]=y.$ Then notice that by Ceva's, $\\frac{FB\\cdot DC\\cdot EA}{DB\\cdot CE\\cdot AF}=1.$ However, we can deduce $\\frac{FB}{AF}=\\frac{3}{4}$ from the fact that $[AFP]$ and $[BPF]$ share the same height. Similarly, $x=\\frac{84CE}{EA}$ and $y=\\frac{35DC}{BD}.$ Plug and chug and you get $xy=84\\cdot 35\\cdot \\frac{3}{4}=2205.$ Then notice by the same height reasoning, $\\frac{84}{x}=\\frac{119+y}{x+70}.$ Clear the fractions and combine like terms to get $35x=5880-xy.$ We know $xy=2205$ so subtraction yields $35x=3675,$ or $x=105.$ Plugging this in to our previous ratio statement yields $\\frac{84}{105}=\\frac{4}{5}=\\frac{119+y}{175},$ so $y=21.$ Basic addition gives us $105+84+21+35+30+40=\\boxed{315}.$", "Let the interior point be $P$ and let the points on $\\overline{BC}$ $\\overline{CA}$ and $\\overline{AB}$ be $D$ $E$ and $F$ , respectively. Then the cevians $AD,BF,CE$ are concurrent, so we can use Ceva's Theorem, letting $\\frac{BD}{DC}=\\frac{a}{b}$ and $\\frac{CF}{FA}=\\frac{c}{d}$ . Notice that $\\frac{AE}{EB}=\\frac{[\\Delta APE]}{[\\Delta EPB]}=\\frac43.$ \\[\\frac{4}{3}\\cdot \\frac{a}{b}\\cdot \\frac{c}{d}=1\\implies \\frac{d}{c}=\\frac{a}{b}\\cdot \\frac{4}{3}.\\]\nWe know that $[\\Delta CPD]=35\\cdot \\frac ba$ and $[\\Delta APF]=84\\cdot \\frac dc,$ so \\[[\\Delta ABC] = 84+84\\cdot \\frac dc + 35\\cdot \\frac ba + 40+30+35 = \\left(1+\\frac ba\\right)(40+30+35).\\] We will now solve for $\\frac ba$\n\\[84+84\\cdot \\frac ab\\cdot \\frac 43 + 35\\cdot \\frac ba = \\frac ba\\cdot 105.\\] \\[12+16\\cdot \\frac ab + 5\\cdot \\frac ba = \\frac ba\\cdot 15\\] \\[10\\left(\\frac ba\\right)^2-12\\cdot \\frac ba-16=0\\] Factoring this gives $\\left(\\frac ba-2\\right)\\left(10\\cdot \\frac ba - 8\\right)=0,$ so the area of $\\triangle ABC$ is \\[\\left(1+\\frac ba\\right)(40+30+35)=3\\cdot 105=\\boxed{315.}\\]" ]

[ "Working backwards, if $6 \\le E \\le 10$ , then $6 \\pm 11 \\le A \\le 10 \\pm 11$ . Since $A$ is a positive integer, $17 \\le A \\le 21$\nSince $17 \\le A \\le 21$ , we know that $17 \\pm 19 \\le B \\le 21 \\pm 19$ . But if $B=36$ , which is the smallest possible \"plus\" value, then $E + A + B = 6 + 17 + 36 = 59$ , which is too much money.\nHence, $17 - 19 \\le B \\le 21 - 19$ . But since $B$ must be a positive integer, that leaves only two possibilities: $B = 1$ or $B=2$ , which correspond with $E = 9$ and $E = 10$\nConcentrating only on $E=9$ , we have $E=9$ leading to $A = 9 + 11 = 20$ , which leads to $B = 20 - 19 = 1$ , which leads to $C = 1 + 7 = 8$ . Thus far we have given out $9 + 20 + 1 + 8 = 38$ dollars. This means that Dick must have $56 - 38 = 18$ dollars. However, the difference between Carlos and Dick is not $5$ dollars.\nThus, the right answer must be $\\boxed{10}$ . Verifying, if $E = 10$ , then $A = 10 + 11 = 21$ $B = 21 - 19 = 2$ , which leads to $C = 2 + 7 = 9$ . Thus far, we have given out $10 + 21 + 2 + 9 = 42$ dollars, leaving $56 - 42 = 14$ dollars for Dick. Dick does indeed have $5$ dollars more than Carlos, and $4$ dollars more than Elgin." ]

[ "It follows from the givens that $a$ is a perfect fourth power $b$ is a perfect fifth power, $c$ is a perfect square and $d$ is a perfect cube . Thus, there exist integers $s$ and $t$ such that $a = t^4$ $b = t^5$ $c = s^2$ and $d = s^3$ . So $s^2 - t^4 = 19$ . We can factor the left-hand side of this equation as a difference of two squares, $(s - t^2)(s + t^2) = 19$ . 19 is a prime number and $s + t^2 > s - t^2$ so we must have $s + t^2 = 19$ and $s - t^2 = 1$ . Then $s = 10, t = 3$ and so $d = s^3 = 1000$ $b = t^5 = 243$ and $d-b=\\boxed{757}$" ]

[ "Note that each given equation is of the form \\[f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7\\] for some $k\\in\\{1,2,3\\}.$\nWhen we expand $f(k)$ and combine like terms, we obtain a quadratic function of $k:$ \\[f(k)=ak^2+bk+c,\\] where $a,b,$ and $c$ are linear combinations of $x_1,x_2,x_3,x_4,x_5,x_6,$ and $x_7.$\nWe are given that \\begin{alignat*}{10} f(1)&=\\phantom{42}a+b+c&&=1, \\\\ f(2)&=4a+2b+c&&=12, \\\\ f(3)&=9a+3b+c&&=123, \\end{alignat*} and we wish to find $f(4).$\nWe eliminate $c$ by subtracting the first equation from the second, then subtracting the second equation from the third: \\begin{align*} 3a+b&=11, \\\\ 5a+b&=111. \\end{align*} By either substitution or elimination, we get $a=50$ and $b=-139.$ Substituting these back produces $c=90.$\nFinally, the answer is \\[f(4)=16a+4b+c=\\boxed{334}.\\]", "For simplicity purposes, we number the given equations $(1),(2),$ and $(3),$ in that order. Let \\[16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=S. \\hspace{29.5mm}(4)\\] Subtracting $(1)$ from $(2),$ subtracting $(2)$ from $(3),$ and subtracting $(3)$ from $(4),$ we obtain the following equations, respectively: \\begin{align*} 3x_1 + 5x_2 + 7x_3 + 9x_4 + 11x_5 + 13x_6 + 15x_7 &=11, \\hspace{20mm}&(5) \\\\ 5x_1 + 7x_2 + 9x_3 + 11x_4 + 13x_5 + 15x_6 + 17x_7 &=111, &(6) \\\\ 7x_1 + 9x_2 + 11x_3 + 13x_4 + 15x_5 + 17x_6 + 19x_7 &=S-123. &(7) \\\\ \\end{align*} Subtracting $(5)$ from $(6)$ and subtracting $(6)$ from $(7),$ we obtain the following equations, respectively: \\begin{align*} 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=100, &(8) \\\\ 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=S-234. \\hspace{20mm}&(9) \\end{align*} Finally, applying the Transitive Property to $(8)$ and $(9)$ gives $S-234=100,$ from which $S=\\boxed{334}.$", "Note that the second differences of all quadratic sequences must be constant (but nonzero). One example is the following sequence of perfect squares:\n\nLabel equations $(1),(2),(3),$ and $(4)$ as Solution 2 does. Since the coefficients of $x_1,x_2,x_3,x_4,x_5,x_6,x_7,$ or $(1,4,9,16),(4,9,16,25),(9,16,25,36),(16,25,36,49),(25,36,49,64),(36,49,64,81),(49,64,81,100),$ respectively, all form quadratic sequences with second differences $2,$ we conclude that the second differences of equations $(1),(2),(3),(4)$ must be constant.\nIt follows that the second differences of $(1,12,123,S)$ must be constant, as shown below:\n\nFinally, we have $d_2=100,$ from which \\begin{align*} S&=123+d_1 \\\\ &=123+(111+d_2) \\\\ &=\\boxed{334} ~MRENTHUSIASM", "Notice that we may rewrite the equations in the more compact form as: \\begin{align*} \\sum_{i=1}^{7}i^2x_i&=c_1, \\\\ \\sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\\\ \\sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\\\ \\sum_{i=1}^{7}(i+3)^2x_i&=c_4, \\end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find.\nNow consider the polynomial given by $f(z) = \\sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients).\nNotice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$ . Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=\\boxed{334}$", "The idea is to multiply the first, second and third equations by $a,b,$ and $c,$ respectively.\nWe can only consider the coefficients of $x_1,x_2,$ and $x_3:$ \\begin{align} a+4b+9c&=16, \\\\ 4a+9b+16c&=25, \\\\ 9a+16b+25c&=36. \\end{align} Subtracting $(1)$ from $(2),$ we get \\[3a+5b+7c=9. \\hspace{15mm}(4)\\] Subtracting $3\\cdot(4)$ from $(3),$ we get \\[b+4c=9. \\hspace{25.5mm}(5)\\] Subtracting $(1)$ from $4\\cdot(5),$ we get \\[7c-a=20. \\hspace{23mm}(6)\\] From $(5)$ and $(6),$ we have $(a,b,c)=(7c-20,9-4c,c).$ Substituting this into $(2)$ gives $(a,b,c)=(1,-3,3).$\nTherefore, the answer is $1\\cdot1+12\\cdot(-3) + 123\\cdot3 = \\boxed{334}.$", "We let $(x_4,x_5,x_6,x_7)=(0,0,0,0)$ . Thus, we have \\begin{align*} x_1+4x_2+9x_3&=1,\\\\ 4x_1+9x_2+16x_3&=12,\\\\ 9x_1+16x_2+25x_3&=123.\\\\ \\end{align*} Grinding this out, we have $(x_1,x_2,x_3)=\\left(\\frac{797}{4},-229,\\frac{319}{4}\\right)$ which gives $\\boxed{334}$ as our final answer.", "Let $s_n = n^2$ be the sequence of perfect squares.\nBy either expanding or via finite differences, one can prove the miraculous recursion \\[s_n = 3s_{n-1} - 3s_{n-2} + s_{n-3}.\\] Hence, the answer is simply \\[3 \\cdot 123 - 3 \\cdot 12 + 1 = \\boxed{334}.\\]" ]

[ "There are $5+5+15+2=27$ million Blacks living in the U.S. Out of these, $15$ of them live in the South, so the percentage is $\\frac{15}{27}\\approx \\frac{60}{108}\\approx 56\\%$\n$\\boxed{56}$" ]

[ "If $x\\neq y,$ then $\\frac{x-y}{y-x}=-1.$ We use this fact to simplify the original expression: \\[\\frac{\\color{red}\\overset{-1}{\\cancel{a-3}}}{\\color{blue}\\underset{1}{\\cancel{5-c}}} \\cdot \\frac{\\color{green}\\overset{-1}{\\cancel{b-4}}}{\\color{red}\\underset{1}{\\cancel{3-a}}} \\cdot \\frac{\\color{blue}\\overset{-1}{\\cancel{c-5}}}{\\color{green}\\underset{1}{\\cancel{4-b}}}=(-1)(-1)(-1)=\\boxed{1}.\\] ~CoolJupiter ~MRENTHUSIASM", "At $(a,b,c)=(4,5,6),$ the answer choices become\n$\\textbf{(A) } {-}1 \\qquad \\textbf{(B) } 1 \\qquad \\textbf{(C) } 2 \\qquad \\textbf{(D) } {-}\\frac{1}{120} \\qquad \\textbf{(E) } \\frac{1}{120}$\nand the original expression becomes \\[\\frac{-1}{1}\\cdot\\frac{-1}{1}\\cdot\\frac{-1}{1}=\\boxed{1}.\\] ~MRENTHUSIASM" ]

[ "If $108$ students eat $2$ cookies on average, there will need to be $108\\cdot 2 = 216$ cookies. But with the smaller attendance, you will only need $100\\% - 25\\% = 75\\%$ of these cookies, or $75\\% \\cdot 216 = 0.75\\cdot 216 = 162$ cookies.\n$162$ cookies requires $\\frac{162}{15} = 10.8$ batches. However, since half-batches are forbidden, we must round up to get $\\left\\lceil \\frac{162}{15} \\right\\rceil = 11$ batches, and the correct answer is $\\boxed{11}$", "If there were $108$ students before, with the $25\\%$ of students missing, there will be $75\\%$ of $108$ students left. This is $75\\% \\cdot 108 = 0.75 \\cdot 108 = 81$ students. These students eat $81 \\cdot 2 = 162$ cookies. Follow the logic of the second paragraph above to find that there needs to be $11$ batches, and the correct answer is $\\boxed{11}$" ]

[ "If $108$ students eat $2$ cookies on average, there will need to be $108\\cdot 2 = 216$ cookies. There are $15$ cookies per pan, meaning there needs to be $\\frac{216}{15} = 14.4$ pans. However, since half-recipes are forbidden, we need to round up and make $\\lceil \\frac{216}{15}\\rceil = 15$ pans.\n$1$ pan requires $2$ eggs, so $15$ pans require $2\\cdot 15 = 30$ eggs. Since there are $6$ eggs in a half dozen, we need $\\frac{30}{6} = 5$ half-dozens of eggs, and the answer is $\\boxed{5}$" ]

[ "For $216$ cookies, you need to make $\\frac{216}{15} = 14.4$ pans. Since fractional pans are forbidden, round up to make $\\lceil \\frac{216}{15} \\rceil = 15$ pans.\nThere are $3$ tablespoons of butter per pan, meaning $3 \\cdot 15 = 45$ tablespoons of butter are required for $15$ pans.\nEach stick of butter has $8$ tablespoons, so we need $\\frac{45}{8} = 5.625$ sticks of butter. However, we must round up again because partial sticks of butter are forbidden! Thus, we need $\\lceil \\frac{45}{8} \\rceil = 6$ sticks of butter, and the answer is $\\boxed{6}$" ]

[ "Given that these are the only math teachers at Euclid Middle School and we are told how many from each class are taking the AMC 8, we simply add the three numbers to find the total. $11+8+9=\\boxed{28}$" ]

[ "We can see that this is a Venn Diagram Problem.\nFirst, we analyze the information given. There are $198$ students. Let's use A as the first issue and B as the second issue.\n$149$ students were for A, and $119$ students were for B. There were also $29$ students against both A and B.\nSolving this without a Venn Diagram, we subtract $29$ away from the total, $198$ . Out of the remaining $169$ , we have $149$ people for A and\n$119$ people for B. We add this up to get $268$ . Since that is more than what we need, we subtract $169$ from $268$ to get\n$\\boxed{99}$" ]

[ "Out of the soccer players, $40\\%$ swim. As the soccer players are $60\\%$ of the whole, the swimming soccer players are $0.4 \\cdot 0.6 = 0.24 = 24\\%$ of all children.\nThe non-swimming soccer players then form $60\\% - 24\\% = 36\\%$ of all the children.\nOut of all the children, $30\\%$ swim. We know that $24\\%$ of all the children swim and play soccer, hence $30\\%-24\\% = 6\\%$ of all the children swim and don't play soccer.\nFinally, we know that $70\\%$ of all the children are non-swimmers. And as $36\\%$ of all the children do not swim but play soccer, $70\\% - 36\\% = 34\\%$ of all the children do not engage in any activity.\nA quick summary of what we found out:\nNow we can compute the answer. Out of all children, $70\\%$ are non-swimmers, and again out of all children $36\\%$ are non-swimmers that play soccer. Hence the percent of non-swimmers that play soccer is $\\frac{36}{70} \\approx 51\\% \\Rightarrow \\boxed{51}$", "Let us set the total number of children as $100$ . So $60$ children play soccer, $30$ swim, and $0.4\\times60=24$ play soccer and swim.\nThus, $60-24=36$ children only play soccer.\nSo our numerator is $36$\nOur denominator is simply $100-\\text{Swimmers}=100-30=70$\nAnd so we get $\\frac{36}{70}$ which is roughly $51.4\\% \\Rightarrow \\boxed{51}$", "WLOG, let the total number of students be $100$ . Draw a venn diagram with 2 circles encompassing these 4 regions:\nNon-soccer players, non-swimmers: 34 people\nSoccer players, non-swimmers: 36 people\nSoccer players, swimmers: 24 people\nNon-soccer players, swimmers: 6 people.\nHence the answer is $\\frac{36}{70}=\\frac{18}{35}$ . We know this is a little bit larger than $\\frac 12$ because $\\frac{17.5}{35}=\\frac 12$ $\\boxed{51}$" ]

[ "We can set up a system of equations where $a$ is the sets of twins, $b$ is the sets of triplets, and $c$ is the sets of quadruplets. \\[\\begin{split} 2a + 3b + 4c & = 1000 \\\\ b & = 4c \\\\ a & = 3b \\end{split}\\]\nSolving for $c$ and $a$ in the second and third equations and substituting into the first equation yields \\[\\begin{split} 2 (3b) + 3b + 4 (0.25b) & = 1000 \\\\ 6b + 3b + b & = 1000 \\\\ b & = 100 \\end{split}\\]\nSince we are trying to find the number of babies and NOT the number of sets of quadruplets, the solution is not $c$ , but rather $4c$ . Therefore, we strategically use the second initial equation to realize that $b$ $=$ $4c$ , leaving us with the number of babies born as quadruplets equal to $\\boxed{100}$", "Say there are $12x$ sets of twins, $4x$ sets of triplets, and $x$ sets of quadruplets. That's $12x\\cdot2=24x$ twins, $4x\\cdot3=12x$ triplets, and $x\\cdot4=4x$ quadruplets. A tenth of the babies are quadruplets and that's $\\frac{1}{10}(1000)=\\boxed{100}$" ]

[ "$60\\% \\cdot 20\\% = 12\\%$ of the people that claim that they like dancing actually dislike it, and $40\\% \\cdot 90\\% = 36\\%$ of the people that claim that they dislike dancing actually dislike it. Therefore, the answer is $\\frac{12\\%}{12\\%+36\\%} = \\boxed{25}$", "Assume WLOG that there are 100 people. Then 60 of them like dancing, 40 dislike dancing. Of the ones that like dancing, 48 say they like dancing and 12 say they dislike it. Of the ones who dislike dancing, 36 say they dislike dancing and 4 say they like it. We want the ratio of students like it but say they dislike it to the total amount of students that say they dislike it. This is $\\frac{12}{12+36}=\\frac{12}{48}=\\frac{1}{4}$ . We choose $\\boxed{25}$", "WLOG, assume that there are a total of $100$ students at Typico High School. We make a chart:\n\\[\\begin{tabular}[t]{|c|c|c|c|}\\hline & \\text{Likes dancing} & \\text{Doesn't like dancing} & \\text{Total} \\\\\\hline \\text{Says they like dancing} & & & \\\\\\hline \\text{Says they don't like dancing} & & & \\\\\\hline \\text{Total} & & & 100 \\\\\\hline \\end{tabular}\\]\nWe know that $60$ of the students like dancing (since $60\\%$ of $100$ is $60$ ), so we fill that in:\n\\[\\begin{tabular}[t]{|c|c|c|c|}\\hline & \\text{Likes dancing} & \\text{Doesn't like dancing} & \\text{Total} \\\\\\hline \\text{Says they like dancing} & & & \\\\\\hline \\text{Says they don't like dancing} & & & \\\\\\hline \\text{Total} & 60 & & 100 \\\\\\hline \\end{tabular}\\]\n$80\\%$ of those $60$ kids say that they like dancing, so that's $48$ kids who like dancing and say that they like dancing. The other $12$ kids like dancing and say that they do not.\n\\[\\begin{tabular}[t]{|c|c|c|c|}\\hline & \\text{Likes dancing} & \\text{Doesn't like dancing} & \\text{Total} \\\\\\hline \\text{Says they like dancing} & 48 & & \\\\\\hline \\text{Says they don't like dancing} & 12 & & \\\\\\hline \\text{Total} & 60 & & 100 \\\\\\hline \\end{tabular}\\]\n$40$ students do not like dancing. $90\\%$ of those $40$ students say that they do not like it, which is $36$ of them.\n\\[\\begin{tabular}[t]{|c|c|c|c|}\\hline & \\text{Likes dancing} & \\text{Doesn't like dancing} & \\text{Total} \\\\\\hline \\text{Says they like dancing} & 48 & & \\\\\\hline \\text{Says they don't like dancing} & 12 & 36 & \\\\\\hline \\text{Total} & 60 & 40 & 100 \\\\\\hline \\end{tabular}\\]\nAt this point, one can see that there are $12+36=48$ total students who say that they do not like dancing. $12$ of those actually like it, so that is $\\dfrac{12}{48}=\\dfrac14=\\boxed{25}.$" ]

[ "There are two cases:\nCase 1: One man and one woman is chosen from each department.\nCase 2: Two men are chosen from one department, two women are chosen from another department, and one man and one woman are chosen from the third department.\nFor the first case, in each department there are ${{2}\\choose{1}} \\times {{2}\\choose{1}} = 4$ ways to choose one man and one woman. Thus there are $4^3 = 64$ total possibilities conforming to case 1.\nFor the second case, there is only ${{2}\\choose{2}} = 1$ way to choose two professors of the same gender from a department, and again there are $4$ ways to choose one man and one woman. Thus there are $1 \\cdot 1 \\cdot 4 = 4$ ways to choose two men from one department, two women from another department, and one man and one woman from the third department. However, there are $3! = 6$ different department orders, so the total number of possibilities conforming to case 2 is $4 \\cdot 6 = 24$\nSumming these two values yields the final answer: $64 + 24 = \\boxed{088}$", "Use generating functions . For each department, there is 1 way to pick 2 males, 4 ways to pick one of each, and 1 way to pick 2 females. Since there are three departments in total, and we wish for three males and three females, the answer will be equal to the coefficient of $x^3y^3$ in the expansion of $(x^2+4xy+y^2)^3$ . The requested coefficient is $\\boxed{088}$" ]

[ "We start with $2^{1+\\lfloor\\log_{2}(N-1)\\rfloor}-N = 19$ . After rearranging, we get $\\lfloor\\log_{2}(N-1)\\rfloor = \\log_{2} \\left(\\frac{N+19}{2}\\right)$\nSince $\\lfloor\\log_{2}(N-1)\\rfloor$ is a positive integer, $\\frac{N+19}{2}$ must be in the form of $2^{m}$ for some positive integer $m$ . From this fact, we get $N=2^{m+1}-19$\nIf we now check integer values of N that satisfy this condition, starting from $N=19$ , we quickly see that the first values that work for $N$ are $45$ and $109$ , that is, $2^6-19$ and $2^7 -19$ , giving values of $5$ and $6$ for $m$ , respectively. Adding up these two values for $N$ , we get $45 + 109 = 154 \\rightarrow \\boxed{154}$", "We examine the value that $2^{1+\\lfloor\\log_{2}(N-1)\\rfloor}$ takes over various intervals. The $\\lfloor\\log_{2}(N-1)\\rfloor$ means it changes on each multiple of 2, like so:\n2 --> 1\n3 - 4 --> 2\n5 - 8 --> 3\n9 - 16 --> 4\nFrom this, we see that $2^{1+\\lfloor\\log_{2}(N-1)\\rfloor} - N$ is the difference between the next power of 2 above $2^{\\lfloor\\log_{2}(N-1)\\rfloor}$ and $N$ . We are looking for $N$ such that this difference is 19. The first two $N$ that satisfy this are $45 = 64-19$ and $109=128-19$ for a final answer of $45 + 109 = 154 \\rightarrow \\boxed{154}$", "Note that each $N$ is $19$ less than a power of $2$ . So, the answer will be $38$ less than the sum of $2$ powers of $2$ . Adding $38$ to each answer, we get $76$ $128$ $192$ $444$ , and $1062$ . Obviously we can take out $76$ and $1062$ . Also, $128$ will not work because two powers of two will never sum to another power of $2$ (unless they are equal, which is a contradiction to the question). So, we have $192$ and $444$ . Note that $444 = 1 + 443 = 2 + 442 = 4 + 440 = 8 + 436 = 16 + 428 = 32 + 412$ , etc. We quickly see that $444$ will not work, leaving $192$ which corresponds to $\\boxed{154}$ . We can also confirm that this works because $192 = 128 + 64 = 2^7 + 2^6$", "In order to fix the exponent and get rid of the logarithm term, let $N = 2^m + k + 1$ , with $0 \\leq k < 2^m$ . Doing so, we see that $\\lfloor \\log_2{N - 1} \\rfloor = m$ , which turns our given relation into \\[2^m = 20 + k,\\] for which the solutions of the form $(m, k)$ $(5, 12)$ and $(6, 44)$ , follow trivially. Adding up the two values of $N$ gives us $32 + 12 + 1 + 64 + 44 + 1 = 154$ , so the answer is $\\boxed{154}$" ]

[ "Each one of the ten people has to shake hands with all the $20$ other people they don’t know. So $10\\cdot20 = 200$ . From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands from $10$ , or $\\binom{10}{2} = 45$ . Thus the answer is $200 + 45 = \\boxed{245}$", "We can also use complementary counting. First of all, $\\dbinom{30}{2}=435$ handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from $435$ to find the handshakes. Hugs only happen between the $20$ people who know each other, so there are $\\dbinom{20}{2}=190$ hugs. $435-190= \\boxed{245}$", "We can focus on how many handshakes the $10$ people who don't know anybody get.\nThe first person gets $29$ handshakes with other people not him/herself, the second person gets $28$ handshakes with other people not him/herself and not the first person, ..., and the tenth receives $20$ handshakes with other people not him/herself and not the first, second, ..., ninth person. We can write this as the sum of an arithmetic sequence:\n$\\frac{10(20+29)}{2}\\implies 5(49)\\implies 245.$ Therefore, the answer is $\\boxed{245}$", "First, we can find out the number of handshakes that the $10$ people who don't know anybody share with the $20$ other people. This is simply $10 \\cdot 20 = 200$ . Next, we need to find out the number of handshakes that are shared within the $10$ people who don't know anybody. Here, we can use the formula $\\frac{n(n-1)}{2}$ , where $n$ is the number of people being counted. The reason we divide by $2$ is because $n(n-1)$ counts the case where the $1^{st}$ person shakes hands with the $2^{nd}$ person $and$ the case where the $2^{nd}$ shakes hands with the $1^{st}$ (and these 2 cases are the same). Thus, plugging $n=10$ gives us $\\frac{10 \\cdot 9}{2} \\implies 45$ . Adding up the 2 cases gives us $200+45=\\boxed{245}$" ]

[ "All of the handshakes will involve at least one person from the $10$ who knows no one. Label these ten people $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ $I$ $J$\nPerson $A$ from the group of $10$ will initiate a handshake with everyone else ( $29$ people). Person $B$ initiates $28$ handshakes plus the one already counted from person $A$ . Person $C$ initiates $27$ new handshakes plus the two we already counted. This continues until person $J$ initiates $20$ handshakes plus the nine we already counted from $A$ ... $I$\n$29+28+27+26+25+24+23+22+21+20 = \\boxed{245}$", "Let the group of people who all know each other be $A$ , and let the group of people who know no one be $B$ . Handshakes occur between each pair $(a,b)$ such that $a\\in A$ and $b\\in B$ , and between each pair of members in $B$ . Thus, the answer is\n$|A||B|+{|B|\\choose 2} = 20\\cdot 10+{10\\choose 2} = 200+45 = \\boxed{245}$", "The number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are ${30\\choose 2}$ and ${20\\choose 2}$ , respectively. Thus, the total amount of handshakes is ${30\\choose 2} - {20\\choose 2} = 435 - 190= \\boxed{245}$", "Each of the $10$ people who do not know anybody will shake hands with all $20$ of the people who do know each other. This means there will be at least $20 * 10 = 200$ handshakes. In addition, those $10$ people will also shake hands with each other, giving us another $9+8+7+6+5+4+3+2+1 = 45$ handshakes. Therefore, there is a total of $200+45 = \\boxed{245}$ handshakes.", "Every one of the $20$ people who know each other will shake hands with each of the $10$ people who know no one, so there are $20\\cdot 10 = 200$ handshakes here. Each of the $10$ people will also shake hands with each other, so there will be ${10 \\choose 2}$ $=45$ handshakes for this case. In total, there are $200+45 = \\boxed{245}$ handshakes." ]

[ "There are $46$ students paired with a blue partner. The other $11$ students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are $64$ students remaining. Therefore the requested number of pairs is $\\tfrac{64}{2}=\\boxed{32}$ ~Punxsutawney Phil" ]

[ "There are $46$ students paired with a blue partner. The other $11$ students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are $64$ students remaining. Therefore the requested number of pairs is $\\tfrac{64}{2}=\\boxed{32}$ ~Punxsutawney Phil" ]

[ "Assume arbitrarily (and WLOG) there are $5$ women in the room, of which $5 \\cdot \\frac25 = 2$ are single and $5-2=3$ are married. Each married woman came with her husband, so there are $3$ married men in the room as well for a total of $5+3=8$ people. The fraction of the people that are married men is $\\boxed{38}$" ]

[ "If each man danced with $3$ women, then there will be a total of $3\\times12=36$ pairs of men and women. However, each woman only danced with $2$ men, so there must have been $\\frac{36}2 \\Longrightarrow \\boxed{18}$ women.", "Consider drawing out a diagram. Let a circle represent a man, and let a shaded circle represent a woman.\nThen, we know that for every 2 men, there will be 3 woman using our diagram. Therefore, the ratio between the number of men and women is 2:3.\nHence, we know that:\n$\\frac{2}{3} = \\frac{12}{x} \\implies x = 18 \\implies \\boxed{18}.$" ]

[ "The minimum dimensions of the rectangle are $1.5$ inches by $2.5$ inches. The minimum area is $1.5\\times2.5=\\boxed{3.75}$ square inches." ]

[ "There are $18$ total twins and $18$ total triplets. Each of the twins shakes hands with the $16$ twins not in their family and $9$ of the triplets, a total of $25$ people. Each of the triplets shakes hands with the $15$ triplets not in their family and $9$ of the twins, for a total of $24$ people. Dividing by two to accommodate the fact that each handshake was counted twice, we get a total of $\\frac{1}{2}\\times 18 \\times (25+24) = 9 \\times 49 = 441 \\rightarrow \\boxed{441}$" ]

[ "At the fourth practice she made $48/2=24$ throws, at the third one it was $24/2=12$ , then we get $12/2=6$ throws for the second practice, and finally $6/2=3\\Rightarrow\\boxed{3}$ throws at the first one." ]

[ "Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made $\\frac{1}{2} \\cdot 48 = 24$ free throws, on the third $12$ , on the second $6$ , and on the first $3 \\Rightarrow \\mathrm{(A)}$\nBecause there are five days, or four transformations between days (day 1 $\\rightarrow$ day 2 $\\rightarrow$ day 3 $\\rightarrow$ day 4 $\\rightarrow$ day 5), she makes $48 \\cdot \\frac{1}{2^4} = \\boxed{3}$" ]

[ "Say the student in the center starts out with $a$ coins, the students neighboring the center student each start with $b$ coins, and all other students start out with $c$ coins. Then the $a$ -coin student has five neighbors, all the $b$ -coin students have three neighbors, and all the $c$ -coin students have four neighbors.\nNow in order for each student's number of coins to remain equal after the trade, the number of coins given by each student must be equal to the number received, and thus\n\\begin{align*} a &= 5 \\cdot \\frac{b}{3}\\\\ b &= \\frac{a}{5} + 2 \\cdot \\frac{c}{4}\\\\ c &= 2 \\cdot \\frac{c}{4} + 2 \\cdot \\frac{b}{3}. \\end{align*}\nSolving these equations, we see that $\\frac{a}{5} = \\frac{b}{3} = \\frac{c}{4}.$ Also, the total number of coins is $a + 5b + 10c = 3360,$ so $a + 5 \\cdot \\frac{3a}{5} + 10 \\cdot \\frac{4a}{5} = 3360 \\rightarrow a = \\frac{3360}{12} = \\boxed{280}.$", "Since the students give the same number of gifts of coins as they receive and still end up the same number of coins, we can assume that every gift of coins has the same number of coins. Let $x$ be the number of coins in each gift of coins. There $10$ people who give $4$ gifts of coins, $5$ people who give $3$ gifts of coins, and $1$ person who gives $5$ gifts of coins. Thus,\n\\begin{align*} 10(4x)+5(3x)+5x &= 3360\\\\ 40x+15x+5x &= 3360\\\\ 60x &= 3360\\\\ x &= 56 \\end{align*} Therefore the answer is $5(56) = \\boxed{280}.$", "Mark the number of coins from inside to outside as $a$ $b_1$ $b_2$ $b_3$ $b_4$ $b_5$ $c_1$ $c_2$ $c_3$ $c_4$ $c_5$ $d_1$ $d_2$ $d_3$ $d_4$ $d_5$ .\nThen, we obtain \\begin{align*} d_1 &= \\frac{d_5}{4} + \\frac{d_2}{4} + \\frac{c_1}{4} + \\frac{c_2}{4}\\\\ d_2 &= \\frac{d_1}{4} + \\frac{d_3}{4} + \\frac{c_2}{4} + \\frac{c_3}{4}\\\\ d_3 &= \\frac{d_2}{4} + \\frac{d_4}{4} + \\frac{c_3}{4} + \\frac{c_4}{4}\\\\ d_4 &= \\frac{d_3}{4} + \\frac{d_5}{4} + \\frac{c_4}{4} + \\frac{c_5}{4}\\\\ d_5 &= \\frac{d_4}{4} + \\frac{d_1}{4} + \\frac{c_5}{4} + \\frac{c_1}{4}\\\\ \\end{align*} Letting $D = d_1 + d_2 + d_3 + d_4 + d_5$ $C = c_1 + c_2 + c_3 + c_4 + c_5$ gets us $D = \\frac{D}{4} + \\frac{D}{4} + \\frac{C}{4} + \\frac{C}{4}$ and $D = C$ .\nIn the same way, $C = \\frac{D}{4} + \\frac{D}{4} + \\frac{B}{3} + \\frac{B}{3}$ $B = \\frac{3D}{4}$ $B = \\frac{C}{4} + \\frac{C}{4} + a$ $a = \\frac{D}{4}$ .\nThen, with $a + B + C + D = 3360$ $D = 1120$ $a = \\boxed{280}$", "Define $x$ as the number of coins the student in the middle has. Since this student connects to $5$ other students, each of those students must have passed $\\dfrac15 x$ coins to the center to maintain the same number of coins.\nEach of these students connect to $3$ other students, passing $\\dfrac15 x$ coins to each, so they must have $\\dfrac35 x$ coins. These students must then recieve $\\dfrac35 x$ coins, $\\dfrac 15 x$ of which were given to by the center student. Thus, they must also have received $\\dfrac25 x$ coins from the outer layer, and since this figure has symmetry, these must be the same.\nEach of the next layer of students must have given $\\dfrac15 x$ coins. Since they gave $4$ people coins, they must have started with $\\dfrac45 x$ coins. They received $\\dfrac25 x$ of them from the inner layer, making the two other connections have given them the same amount. By a similar argument, they recieved $\\dfrac15 x$ from each of them.\nBy a similar argument, the outermost hexagon of students must have had $\\dfrac45 x$ coins each. Summing this all up, we get the number of total coins passed out as a function of $x$ . This ends up to be \\begin{align*} \\dfrac45 x + \\dfrac45 x + \\dfrac45 x + \\dfrac45 x + \\dfrac45 x + \\dfrac45 x + \\dfrac45 x + \\dfrac45 x + \\dfrac45 x + \\dfrac45 x + \\dfrac35 x + \\dfrac35 x + \\dfrac35 x + \\dfrac35 x + \\dfrac35 x + x &= 10 \\cdot \\dfrac45 x + 5 \\cdot \\dfrac35 x + x \\\\ &= 8x + 3x + x \\\\ &= 12 x. \\end{align*}\nSince this all sums up to $3360$ , which is given, we find that \\begin{align*} 12x &= 3360 \\\\ x &= 280 \\end{align*}\nWe defined $x$ to be the number of coins that the center person has, so the answer is $x$ , which is $\\boxed{280}$" ]

[ "At noon on a certain day, let $M$ and $L$ be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that $M=L+N.$\nAt $4{:}00,$ we get \\begin{align*} |(M-5)-(L+3)| &= 2 \\\\ |M-L-8| &= 2 \\\\ |N-8| &= 2. \\end{align*} We have two cases:\nTogether, the product of all possible values of $N$ is $10\\cdot6=\\boxed{60}.$", "At noon on a certain day, the difference of temperatures in Minneapolis and St. Louis is $N$ degrees.\nAt $4{:}00,$ the difference of temperatures in Minneapolis and St. Louis is $N-8$ degrees.\nIt follows that \\[|N-8|=2.\\] We continue with the casework in Solution 1 to get the answer $\\boxed{60}.$" ]

[ "At noon on a certain day, let $M$ and $L$ be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that $M=L+N.$\nAt $4{:}00,$ we get \\begin{align*} |(M-5)-(L+3)| &= 2 \\\\ |M-L-8| &= 2 \\\\ |N-8| &= 2. \\end{align*} We have two cases:\nTogether, the product of all possible values of $N$ is $10\\cdot6=\\boxed{60}.$", "At noon on a certain day, the difference of temperatures in Minneapolis and St. Louis is $N$ degrees.\nAt $4{:}00,$ the difference of temperatures in Minneapolis and St. Louis is $N-8$ degrees.\nIt follows that \\[|N-8|=2.\\] We continue with the casework in Solution 1 to get the answer $\\boxed{60}.$" ]

[ "First, find the amount of money one will pay for three sandals without the discount. We have $\\textdollar 50\\times 3 \\text{ sandals} = \\textdollar 150$\nThen, find the amount of money using the discount: $50 + 0.6 \\times 50 + \\frac{1}{2} \\times 50 = \\textdollar 105$\nFinding the percentage yields $\\frac{105}{150} = 70 \\%$\nTo find the percent saved, we have $100 \\% -70 \\%= \\boxed{30}$" ]

[ "The first six gallons are irrelevant. We start with the odometer at $56,200$ miles, and a full gas tank. The total gas consumed by the car during the trip is equal to the total gas the driver had to buy to make the tank full again, i.e., $12+20=32$ gallons. The distance covered is $57,060 - 56,200 = 860$ miles. Hence the average MPG ratio is $860 / 32 \\approx 26.9 \\rightarrow \\boxed{26.9}$" ]

[ "Clearly, the minimum possible value would be $70 - 50 = 20\\%$ . The maximum possible value would be $30 + 50 = 80\\%$ . The difference is $80 - 20 = \\boxed{60}$" ]

[ "Clearly, the minimum possible value would be $70 - 50 = 20\\%$ . The maximum possible value would be $30 + 50 = 80\\%$ . The difference is $80 - 20 = \\boxed{60}$" ]

[ "Lisa's goal was to get an $A$ on $80\\% \\cdot 50 = 40$ quizzes. She already has $A$ 's on $22$ quizzes, so she needs to get $A$ 's on $40-22=18$ more. There are $50-30=20$ quizzes left, so she can afford to get less than an $A$ on $20-18=\\boxed{2}$ of them." ]

[ "Lisa's goal was to get an $A$ on $80\\% \\cdot 50 = 40$ quizzes. She already has $A$ 's on $22$ quizzes, so she needs to get $A$ 's on $40-22=18$ more. There are $50-30=20$ quizzes left, so she can afford to get less than an $A$ on $20-18=\\boxed{2}$ of them." ]

[ "In $1994$ , if Water is $x$ years old, then Walter's grandmother is $2x$ years old.\nThis means that Walter was born in $1994 - x$ , and Walter's grandmother was born in $1994 - 2x$\nThe sum of those years is $3838$ , so we have:\n$1994 - x + 1994 - 2x = 3838$\n$3988 - 3x = 3838$\n$x = 50$\nIf Walter is $50$ years old in $1994$ , then he will be $55$ years old in $1999$ , thus giving answer $\\boxed{55}$" ]

[ "Last week, each box was $\\frac{5}{4} = 1.25$\nThis week, each box is $\\frac{4}{5} = 0.80$\nPercent decrease is given by $\\frac{X_{old} - X_{new}}{X_{old}} \\cdot 100\\%$\nThis, the percent decrease is $\\frac{1.25 - 0.8}{1.25}\\cdot 100\\% = \\frac{45}{125} \\cdot 100\\% = 36\\%$ , which is closest to $\\boxed{35}$" ]

[ "Suppose $x$ is the price of an adult ticket. The price of a child ticket would be $\\frac{x}{2}$\n\\begin{eqnarray*} 5x + 4(x/2) = 7x &=& 24.50\\\\ x &=& 3.50\\\\ \\end{eqnarray*}\nPlug in for 8 adult tickets and 6 child tickets.\n\\begin{eqnarray*} 8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\\\ &=&\\boxed{38.50}" ]

[ "If Brianna is half as old as Aunt Anna, then Brianna is $\\frac{42}{2}$ years old, or $21$ years old.\nIf Caitlin is $5$ years younger than Brianna, she is $21-5$ years old, or $16$\nSo, the answer is $\\boxed{16}$", "Since Brianna is half of Aunt Anna's age this means that Brianna is $21$ years old.\nNow we just find Caitlin's age by doing $21-5$ . This makes $16$ or $\\boxed{16}$" ]

[ "The way to Temple took $\\frac{50}{60}=\\frac56$ hours, and the way back took $\\frac{50}{40}=\\frac54$ for a total of $\\frac56 + \\frac54 = \\frac{25}{12}$ hours. The trip is $50\\cdot2=100$ miles. The average speed is $\\frac{100}{25/12} = \\boxed{48}$ miles per hour.", "We calculate the harmonic mean of Austin and Temple. \nPlugging in, we have: $\\frac{2ab}{a+b} = \\frac{2 \\cdot 60 \\cdot 40}{60 + 40} = \\frac{4800}{100} = \\boxed{48}$ miles per hour.", "The 50 miles part of this question is redundant. We find that the average speed is skewed towards the lower speed, since you spend more time in the lower speed. So, we take the LCM of 40 and 60, getting 120. So, we do (40 + 40 + 40 + 60 + 60)/5, calculation the ratio of the parts and taking the mean. Solving, we get 240/5 = $\\boxed{48} \\qquad$ -themathgood" ]

[ "We need to find out the number of configurations with 3 $O$ and 3 $X$ with 3 $O$ in a row, and 3 $X$ not in a row.\n$\\textbf{Case 1}$ : 3 $O$ are in a horizontal row or a vertical row.\nStep 1: We determine the row that 3 $O$ occupy.\nThe number of ways is 6.\nStep 2: We determine the configuration of 3 $X$\nThe number of ways is $\\binom{6}{3} - 2 = 18$\nIn this case, following from the rule of product, the number of ways is $6 \\cdot 18 = 108$\n$\\textbf{Case 2}$ : 3 $O$ are in a diagonal row.\nStep 1: We determine the row that 3 $O$ occupy.\nThe number of ways is 2.\nStep 2: We determine the configuration of 3 $X$\nThe number of ways is $\\binom{6}{3} = 20$\nIn this case, following from the rule of product, the number of ways is $2 \\cdot 20 = 40$\nPutting all cases together, the total number of ways is $108 + 40 = 148$\nTherefore, the answer is $\\boxed{148}$" ]

[ "Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label.\n2022AIMEIIP2.png\nThis problem can be solved by using $2$ cases.\n$\\textbf{Case 1:}$ $C$ 's opponent for the semifinal is $A$\nThe probability $C$ 's opponent is $A$ is $\\frac13$ . Therefore the probability $C$ wins the semifinal in this case is $\\frac13 \\cdot \\frac13$ . The other semifinal game is played between $J$ and $S$ , it doesn't matter who wins because $C$ has the same probability of winning either one. The probability of $C$ winning in the final is $\\frac34$ , so the probability of $C$ winning the tournament in case 1 is $\\frac13 \\cdot \\frac13 \\cdot \\frac34$\n$\\textbf{Case 2:}$ $C$ 's opponent for the semifinal is $J$ or $S$\nIt doesn't matter if $C$ 's opponent is $J$ or $S$ because $C$ has the same probability of winning either one. The probability $C$ 's opponent is $J$ or $S$ is $\\frac23$ . Therefore the probability $C$ wins the semifinal in this case is $\\frac23 \\cdot \\frac34$ . The other semifinal game is played between $A$ and $J$ or $S$ . In this case it matters who wins in the other semifinal game because the probability of $C$ winning $A$ and $J$ or $S$ is different.\n$\\textbf{Case 2.1:}$ $C$ 's opponent for the final is $A$\nFor this to happen, $A$ must have won $J$ or $S$ in the semifinal, the probability is $\\frac34$ . Therefore, the probability that $C$ won $A$ in the final is $\\frac34 \\cdot \\frac13$\n$\\textbf{Case 2.2:}$ $C$ 's opponent for the final is $J$ or $S$\nFor this to happen, $J$ or $S$ must have won $A$ in the semifinal, the probability is $\\frac14$ . Therefore, the probability that $C$ won $J$ or $S$ in the final is $\\frac14 \\cdot \\frac34$\nIn Case 2 the probability of $C$ winning the tournament is $\\frac23 \\cdot \\frac34 \\cdot (\\frac34 \\cdot \\frac13 + \\frac14 \\cdot \\frac34)$\nAdding case 1 and case 2 together we get $\\frac13 \\cdot \\frac13 \\cdot \\frac34 + \\frac23 \\cdot \\frac34 \\cdot (\\frac34 \\cdot \\frac13 + \\frac14 \\cdot \\frac34) = \\frac{29}{96},$ so the answer is $29 + 96 = \\boxed{125}$" ]

[ "There are a total of $(12+1) \\times (12+1) = 169$ products, and a product is odd if and only if both its factors are odd. There are $6$ odd numbers between $0$ and $12$ , namely $1, 3, 5, 7, 9, 11,$ hence the number of odd products is $6 \\times 6 = 36$ . Therefore the answer is $36/169 \\doteq \\boxed{0.21}$", "Note that if we had an $11$ by $11$ multiplication table, the fraction of odd products becomes $0.25$ . If we add the $12$ by $12$ , the fraction of odd products decreases. Because $0.21$ is the only option less than $0.25$ , our answer is $\\boxed{0.21}$" ]

[ "By adding a number from Bag A and a number from Bag B together, the values we can get are $3, 5, 7, 5, 7, 9, 7, 9, 11.$ Therefore the number of different values is $\\boxed{5}$", "The sum of an even number added to an odd number is always odd. The smallest possible sum is $3$ , and the largest possible sum is $11$ . The odd numbers in between can be achieved by replacing chips with ${\\displaystyle \\pm }2$ within the same bag. Therefore, we can conclude that there are $(11-3)/2(*)+1=\\boxed{5}$ possible sums." ]

[ "To solve this question, you can use $y = mx + b$ where the $x$ is Fahrenheit and the $y$ is Breadus. We have $(110,0)$ and $(350,100)$ . We want to find the value of $y$ in $(200,y)$ that falls on this line. The slope for these two points is $\\frac{5}{12}$ $y = \\frac{5}{12}x + b$ . Solving for $b$ using $(110, 0)$ $\\frac{550}{12} = -b$ . We get $b = \\frac{-275}{6}$ . Plugging in $(200, y), \\frac{1000}{12}-\\frac{550}{12}=y$ . Simplifying, $\\frac{450}{12} = \\boxed{37.5}$", "Let $^\\circ B$ denote degrees Breadus. We notice that $200^\\circ F$ is $90^\\circ F$ degrees to $0^\\circ B$ , and $150^\\circ F$ to $100^\\circ B$ . This ratio is $90:150=3:5$ ; therefore, $200^\\circ F$ will be $\\dfrac3{3+5}=\\dfrac38$ of the way from $0$ to $100$ , which is $\\boxed{37.5.}$", "From $110$ to $350$ degrees Fahrenheit, the Breadus scale goes from $1$ to $100$ $110$ to $350$ degrees is a span of $240$ , and we can use this to determine how many Fahrenheit each Breadus unit is worth. $240$ divided by $100$ is $2.4$ , so each Breadus unit is $2.4$ Fahrenheit, starting at $110$ Fahrenheit. For example, $1$ degree on the Breadus scale is $110 + 2.4$ , or $112.4$ Fahrenheit. Using this information, we can figure out how many Breadus degrees $200$ Fahrenheit is. $200-110$ is $90$ , so we divide $90$ by $2.4$ to find the answer, which is $\\boxed{37.5}$", "We note that the range of F temperatures that $0-100$ $\\text{Br}^\\circ$ represents is $350-110 = 240$ $\\text{F}^\\circ$ $200$ $\\text{F}^\\circ$ is $(200-110) = 90$ $\\text{F}^\\circ$ along the way to getting to $240$ $\\text{F}^\\circ$ , the end of this range, or $90/240 = 9/24 = 3/8 = 0.375$ of the way. Therefore if we switch to the Br scale, we are $0.375$ of the way to $100$ from $0$ , or at $\\boxed{37.5}^\\circ$", "We have the points $(0, 110)$ and $(100, 350)$ . We want to find $(x, 200)$ . The equation of the line is $y=\\frac{12}{5}x+110$ . We use this to find $x=\\frac{75}{2}=37.5$ , or $\\boxed{37.5}$ .\n~MC413551" ]

[ "Barney travels $1661-1441=220$ miles in $4+6=10$ hours for an average of $220/10=\\boxed{22}$ miles per hour." ]

[ "Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number(common sum) would be to add another even number to 44, and a different one to 38. Since there is only one even prime (2), the middle card's hidden number cannot be an odd prime, and so must be even. Therefore, the middle card's hidden number must be 2, so the constant sum is $59+2=61$ . Thus, the first card's hidden number is $61-44=17$ , and the last card's hidden number is $61-38=23$\nSince the sum of the hidden primes is $2+17+23=42$ , the average of the primes is $\\dfrac{42}{3}=\\boxed{14}$" ]

[ "We casework to find the number of ways to get each possible score. Note that the lowest possible score is $2$ and the highest possible score is $7$ . Let the bijective function $f(x)=\\{1,2,3,4,5,6\\} \\to \\{1,2,3,4,5,6\\}$ denote the row number of the rook for the corresponding column number.\nThus, the expected sum is $\\dfrac{120 \\cdot 2 + 216 \\cdot 3 + 222 \\cdot 4 + 130 \\cdot 5 + 31 \\cdot 6 + 1 \\cdot 7}{720}= \\dfrac{2619}{720}=\\dfrac{291}{80}$ , so the desired answer is $291+80=\\boxed{371}$", "If the score is $n+1$ , then one of the rooks must appear in the $n$ th antidiagonal, and this is the first antidiagonal in which a rook can appear. To demonstrate this, we draw the following diagram when $n=4$\n We first count the number of arrangements that avoid the squares above the $n$ th diagonal, and then we subtract from these the number of arrangements that avoid all squares above the $(n+1)$ th diagonal. In the first column, there are $7-n$ rows in which to place the rook. In the second column, there is one more possible row, but one of the rows is used up by the rook in the first column, hence there are still $7-n$ places to place the rook. This pattern continues through the $n$ th column, so there are $(7-n)^n$ ways to place the first $n$ rooks while avoiding the crossed out squares. We can similarly compute that there are $(6-n)^n$ ways to place the rooks in the first $n$ columns that avoid both the crossed out and shaded squares. Therefore, there are $(7-n)^n-(6-n)^n$ ways to place the first $n$ rooks such that at least one of them appears in a shaded square.\nAfter this, there are $(6-n)$ rows and $(6-n)$ columns in which to place the remaining rooks, and we can do this in $(6-n)!$ ways. Hence the number of arrangements with a score of $n$ is $((7-n)^n-(6-n)^n)\\cdot(6-n)!$ . We also know that $n$ can range from from $1$ to $6$ , so the average score is given by\n\\[\\frac{2\\cdot(6^1-5^1)\\cdot5!+3\\cdot(5^2-4^2)\\cdot4!+4\\cdot(4^3-3^3)\\cdot3!+5\\cdot(3^4-2^4)\\cdot 2!+6\\cdot(2^5-1^5)\\cdot 1!+7\\cdot(1^6-0^6)\\cdot 0!}{6!}=\\frac{291}{80}.\\] Thus the answer is $\\boxed{371}$", "So we first count the number of permutations with score $\\ge 2$ . This is obviously $6!=720$ . Then, the number of permutations with score $\\ge 3$ can also be computed: in the first column, there are five ways to place a rook- anywhere but the place with score $1$ . In the next column, there are $5$ ways to place a rook- anywhere but the one in the same row as the previous row. We can continue this to obtain that the number of permutations with score $\\ge 3$ is $600$ . Doing the same for scores $\\ge 4$ $\\ge 5$ $\\ge 6$ , and $\\ge 7$ we obtain that these respective numbers are $384$ $162$ $32$ $1$\nNow, note that if $a_k$ is the number of permutations with score $\\ge k$ , then $a_k-a_{k-1}=b_{k}$ , where $b_k$ is the number of permutations with score exactly $k$ . Thus, we can compute the number of permutations with scores $2$ $3$ , etc as $120,216,222,130,31,1$ . We then compute \\[\\frac{120(2)+216(3)+222(4)+130(5)+31(6)+1(7)}{720}=\\frac{291}{80}\\] leading us to the answer of $291+80=\\boxed{371}$ $\\blacksquare$", "The problem is asking us to compute $\\mathbb{E}[S]$ , where $S$ is the random variable that takes an arrangement of rooks and outputs its score, which is a non-negative integer quantity. For any random variable $S$ with non-negative integer values, we have the tail sum formula \\[\\mathbb{E}[S] = \\sum_{n = 1}^{\\infty}\\mathbb{P}(S\\geq n).\\] These probabilities can be computed as in Solution 3, giving us the following table.\nHence \\begin{align*} \\mathbb{E}[S] &= \\frac{720 + 720 + 600 + 384 + 162 + 32 + 1}{720} \\\\ &= 2 + \\frac{1179}{720} = 2 + \\frac{131}{80} = \\frac{291}{80}, \\end{align*} and the final answer is $291 + 80 = \\boxed{371}$ as above." ]

[ "Let $x$ be the amount of paint in a strip. The total amount of paint is $130+164+188=482$ . After painting the wall, the total amount of paint is $482-4x$ . Because the total amount in each color is the same after finishing, we have\n\\[\\frac{482-4x}{3} = 130-x\\] \\[482-4x=390-3x\\] \\[x=92\\]\nOur answer is $482-4\\cdot 92 = 482 - 368 = \\boxed{114}$", "Let the stripes be $b, r, w,$ and $p$ , respectively. Let the red part of the pink be $\\frac{r_p}{p}$ and the white part be $\\frac{w_p}{p}$ for $\\frac{r_p+w_p}{p}=p$\nSince the stripes are of equal size, we have $b=r=w=p$ . Since the amounts of paint end equal, we have $130-b=164-r-\\frac{r_p}{p}=188-w-\\frac{w_p}{p}$ . Thus, we know that \\[130-p=164-p-\\frac{r_p}{p}=188-p-\\frac{w_p}{p}\\] \\[130=164-\\frac{r_p}{p}=188-\\frac{w_p}{p}\\] \\[r_p=34p, w_p=58p\\] \\[\\frac{r_p+w_p}{p}=92=p=b.\\] Each paint must end with $130-92=38$ oz left, for a total of $3 \\cdot 38 = \\boxed{114}$ oz.", "After the pink stripe is drawn, all three colors will be used equally so the pink stripe must bring the amount of red and white paint down to $130$ ounces each. Say $a$ is the fraction of the pink paint that is red paint and $x$ is the size of each stripe. Then equations can be written: $ax = 164 - 130 = 34$ and $(1-a)x = 188 - 130 = 58$ . The second equation becomes $x - ax = 58$ and substituting the first equation into this one we get $x - 34 = 58$ so $x = 92$ . The amount of each color left over at the end is thus $130 - 92 = 38$ and $38 * 3 = \\boxed{114}$", "We know that all the stripes are of equal size. We can then say that $r$ is the amount of paint per stripe. Then $130 - r$ will be the amount of blue paint left. Now for the other two stripes. The amount of white paint left after the white stripe and the amount of red paint left after the blue stripe are $188 - r$ and $164 - r$ respectively. The pink stripe is also r ounces of paint, but let there be $k$ ounces of red paint in the mixture and $r - k$ ounces of white paint. We now have two equations: $164 - r - k = 188 - r - (r-k)$ and $164 - r - k = 130 - r$ . Solving yields k = 34 and r = 92. We now see that there will be $130 - 92 = 38$ ounces of paint left in each can. $38 * 3 = \\boxed{114}$", "Let the amount of paint each stripe painted used be $x$ . Also, let the amount of paint of each color left be $y$ . 1 stripe is drawn with the blue paint, and 3 stripes are drawn with the red and white paints. Add together the amount of red and white paint, $164 + 188 = 352$ and obtain the following equations : $352 - 3x = 2y$ and $130 - x = y$ . Solve to obtain $x = 92$ . Therefore $y$ is $130 - 92 = 38$ , with three cans of equal amount of paint, the answer is $38 * 3 = \\boxed{114}$", "Let $x$ be the number of ounces of paint needed for a single stripe. We know that in the end, the total amount of red and white paint will equal double the amount of blue paint.\nAfter painting, the amount of red and white paint remaining is equal to $164+188-2x$ , and then minus another $x$ for the pink stripe. The amount of blue paint remaining is equal to $130-x$ . \nSo, we get the equation $2\\cdot(130-x)=164+188-3x$ . Simplifying, we get $x=38$ and $3x=\\boxed{114}$", "Just like in solution 1, we note that all colors will be used equally, except for the pink stripe. This must bring red and white down to $130$ each, so $34$ red and $58$ white are used, making for a total of $92$ for the pink stripe. Thus, the other stripes also use $92$ . The answer is $130+164+188-4(92)=\\boxed{114}$" ]

[ "At the beginning of the problem, the Unicorns had played $y$ games and they had won $x$ of these games. From the information given in the problem, we can say that $\\frac{x}{y}=0.45.$ Next, the Unicorns win 6 more games and lose 2 more, for a total of $6+2=8$ games played during district play. We are told that they end the season having won half of their games, or $0.5$ of their games. We can write another equation: $\\frac{x+6}{y+8}=0.5.$ This gives us a system of equations: $\\frac{x}{y}=0.45$ and $\\frac{x+6}{y+8}=0.5.$ We first multiply both sides of the first equation by $y$ to get $x=0.45y.$ Then, we multiply both sides of the second equation by $(y+8)$ to get $x+6=0.5(y+8).$ Applying the Distributive Property gives yields $x+6=0.5y+4.$ Now we substitute $0.45y$ for $x$ to get $0.45y+6=0.5y+4.$ Solving gives us $y=40.$ Since the problem asks for the total number of games, we add on the last 8 games to get the solution $\\boxed{48}$", "Because 45% can be simplified to $9/20$ , and we know that we cannot play a fraction amount of games, we know that the amount of games before district play is divisible by 20. After district play, there was $8$ games, so in total there must be $20x+8$ . The only answer in this format is $\\boxed{48}$", "Let $n$ be the number of pre-district games. Therefore, we can write the percentage of total games won as a weighted average, namely $.45(n)+.75(8)=(n+8)(.5)$ . Solving this equation for $n$ gives $40$ , but since the problem asked for all games, the answer is $n+8=40+8=\\boxed{48}$", "Let the number of games they won be $x$ and the number of games they lost be $y$ . We are told that $\\frac{x}{x+y}=\\frac{9}{20}$ , which can be manipulated to $20x=9x+9y$ which simplifies down to $11x=9y$ . Then, after district games, we are told $\\frac{x+6}{x+y+8}=\\frac12$ , which can be changed into $2x+12=x+y+8$ which simplifies down to $x+4=y$ . Then we can solve for $x$ using substitution: \\[11x=9x+36\\] \\[2x=36\\] \\[x=18\\] Now that we know $x=18$ , we can figure out that $y=22$ $18+22=40$ . Now we need to add on the district games: $40+8=\\boxed{48}$", "Let $x$ be the total number of games before the district play. The Unicorns have $0.45x$ wins, therefore the rest $0.55x$ are losses. But after the district play, they won 6 and lost 2 more games. We can solve for x by forming the equation $0.45x+6=0.55x+2$ . Subtracting $2$ and $0.45x$ from both sides gives us $0.10x=4$ , and from here we multiply both sides by 10 to get $x=40$ . We are not finished yet as the problem is asking for the total games (which includes the games after the district play), so we add 8 to our value of x to get our answer which is $\\boxed{48}$", "We can check each answer choice from left to right to see which one is correct. Suppose the Unicorns played $48$ games in total. Then, after district play, they would have won $24$ games. Now, consider the situation before district play. The Unicorns would have won $18$ games out of $40$ . Converting to a percentage, $18/40 = 45$ %. Thus, the answer is $\\boxed{48}$" ]

[ "Simplifying 45% to $\\frac{9}{20}$ , we see that the numbers of games before district play are a multiple of 20. After that the Aces played 8 more games to the total number of games is in the form of 20x+8 where x is any positive integer. The only answer choice is $\\boxed{48}$ , which is 20(2)+8.", "First we simplify $45$ % to $\\frac{9}{20}$ . Ratio of won to total is $\\frac{9}{20}$ , but ratio of total number won to total number played is $\\frac{9x}{20x}$ for some $x$ . After they won 6 more games and lost 2 more games the number of games they won is $9x+6$ , and the total number of games is $20x+8$ . Turning it into a fraction we get $\\frac{9x+6}{20x+8}=\\frac{1}{2}$ , so solving for $x$ we get $x=2.$ Plugging in 2 for $x$ we get $20(2)+8=\\boxed{48}$" ]

[ "Every 10 feet Bella goes, Ella goes 50 feet, which means a total of 60 feet. They need to travel that 60 feet $10560\\div60=176$ times to travel the entire 2 miles. Since Bella goes 10 feet 176 times, this means that she travels a total of 1760 feet. And since she walks 2.5 feet each step, $1760\\div2.5=\\boxed{704}$", "We know that Bella goes 2.5 feet per step and since Ella rides 5 times faster than Bella she must go 12.5 feet on her bike for every step of Bella's. For Bella, it takes 4,224 steps, and for Ella, it takes 1/5th those steps since Ella goes 5 times faster than Bella, taking her 844.8 steps. The number of steps where they meet therefore must be less than 844.8. The only answer choice less than it is $\\boxed{704}$", "We can turn $2 \\tfrac{1}{2}$ into an improper fraction. It will then become 5/2. Since Ella bikes 5 times faster, we multiply 5/2 by 5 to get 25/2. Then we add 5/2 to it in order to find the distance they walk and bike together in total. After adding, you should get 30/2 which is equal to 15. This means that after 15 times, they will meet. So you have to divide 10,560 by 15. The answer should be $\\boxed{704}$" ]

[ "Work backwards. The last number Bernardo produces must be in the range $[950,999]$ . That means that before this, Silvia must produce a number in the range $[475,499]$ . Before this, Bernardo must produce a number in the range $[425,449]$ . Before this, Silvia must produce a number in the range $[213,224]$ . Before this, Bernardo must produce a number in the range $[163,174]$ . Before this, Silvia must produce a number in the range $[82,87]$ . Before this, Bernardo must produce a number in the range $[32,37]$ . Before this, Silvia must produce a number in the range $[16,18]$ . Silvia could not have added 50 to any number before this to obtain a number in the range $[16,18]$ , hence the minimum $N$ is 16 with the sum of digits being $\\boxed{7}$", "If our first number is $N,$ then the sequence of numbers will be \\[2N,~2N+50,~4N+100,~4N+150,~8N+300,~8N+350,~16N+700,~16N+750\\] Note that we cannot go any further because doubling gives an extra $1500$ at the end, which is already greater than $1000.$ The smallest $N$ will be given if $16N+750>1000>16N+700 \\implies 15<N<19.$ Since the problem asks for the smallest possible value of $N,$ we get $16,$ and the sum of its digits is $1+6=\\boxed{7}$" ]

[ "The last number that Bernardo says has to be between 950 and 999. Note that $1\\rightarrow 2\\rightarrow 52\\rightarrow 104\\rightarrow 154\\rightarrow 308\\rightarrow 358\\rightarrow 716\\rightarrow 766$ contains 4 doubling actions. Thus, we have $x \\rightarrow 2x \\rightarrow 2x+50 \\rightarrow 4x+100 \\rightarrow 4x+150 \\rightarrow 8x+300 \\rightarrow 8x+350 \\rightarrow 16x+700$\nThus, $950<16x+700<1000$ . Then, $16x>250 \\implies x \\geq 16$\nBecause we are looking for the smallest integer $x$ $x=16$ . Our answer is $1+6=\\boxed{7}$ , which is A." ]

[ "Work backwards. The last number Bernardo produces must be in the range $[950,999]$ . That means that before this, Silvia must produce a number in the range $[475,499]$ . Before this, Bernardo must produce a number in the range $[425,449]$ . Before this, Silvia must produce a number in the range $[213,224]$ . Before this, Bernardo must produce a number in the range $[163,174]$ . Before this, Silvia must produce a number in the range $[82,87]$ . Before this, Bernardo must produce a number in the range $[32,37]$ . Before this, Silvia must produce a number in the range $[16,18]$ . Silvia could not have added 50 to any number before this to obtain a number in the range $[16,18]$ , hence the minimum $N$ is 16 with the sum of digits being $\\boxed{7}$", "If our first number is $N,$ then the sequence of numbers will be \\[2N,~2N+50,~4N+100,~4N+150,~8N+300,~8N+350,~16N+700,~16N+750\\] Note that we cannot go any further because doubling gives an extra $1500$ at the end, which is already greater than $1000.$ The smallest $N$ will be given if $16N+750>1000>16N+700 \\implies 15<N<19.$ Since the problem asks for the smallest possible value of $N,$ we get $16,$ and the sum of its digits is $1+6=\\boxed{7}$" ]

[ "The last number that Bernardo says has to be between 950 and 999. Note that $1\\rightarrow 2\\rightarrow 52\\rightarrow 104\\rightarrow 154\\rightarrow 308\\rightarrow 358\\rightarrow 716\\rightarrow 766$ contains 4 doubling actions. Thus, we have $x \\rightarrow 2x \\rightarrow 2x+50 \\rightarrow 4x+100 \\rightarrow 4x+150 \\rightarrow 8x+300 \\rightarrow 8x+350 \\rightarrow 16x+700$\nThus, $950<16x+700<1000$ . Then, $16x>250 \\implies x \\geq 16$\nBecause we are looking for the smallest integer $x$ $x=16$ . Our answer is $1+6=\\boxed{7}$ , which is A." ]

[ "First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.\nSay that $N \\equiv a \\pmod{6}$\nalso that $N \\equiv b \\pmod{5}$\nSubstituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b < 5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$ $N \\equiv a \\pmod{6}$ $N \\equiv a \\pmod{5}$ $\\implies N=a \\pmod{30}$ $0 \\le a \\le 4$\nTherefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$\nJust keep in mind that $y$ can be one of five choices: $0, 1, 2, 3,$ or $4$ , ;\nAlso, we have already found which digits of $y$ will add up into the units digits of $2N$\nNow, examine the tens digit, $x$ by using $\\mod{25}$ and $\\mod{36}$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work)\nThen we take $N=30x+y$ $\\mod{25}$ and $\\mod{36}$ to find the last two digits in the base $5$ and $6$ representation. \\[N \\equiv 30x \\pmod{36}\\] \\[N \\equiv 30x \\equiv 5x \\pmod{25}\\] Both of those must add up to \\[2N\\equiv60x \\pmod{100}\\]\n$33 \\ge x \\ge 4$\nNow, since $y=0,1,2,3,4$ will always work if $x$ works, then we can treat $x$ as a units digit instead of a tens digit in the respective bases and decrease the mods so that $x$ is now the units digit. \\[N \\equiv 5x \\pmod{6}\\] \\[N \\equiv 6x \\equiv x \\pmod{5}\\] \\[2N\\equiv 6x \\pmod{10}\\]\nSay that $x=5m+n$ (m is between 0-6, n is 0-4 because of constraints on x)\nThen\n\\[N \\equiv 5m+n \\pmod{5}\\] \\[N \\equiv 25m+5n \\pmod{6}\\] \\[2N\\equiv30m + 6n \\pmod{10}\\]\nand this simplifies to\n\\[N \\equiv n \\pmod{5}\\] \\[N \\equiv m+5n \\pmod{6}\\] \\[2N\\equiv 6n \\pmod{10}\\]\nFrom careful inspection, this is true when\n$n=0, m=6$\n$n=1, m=6$\n$n=2, m=2$\n$n=3, m=2$\n$n=4, m=4$\nThis gives you $5$ choices for $x$ , and $5$ choices for $y$ , so the answer is $5* 5 = \\boxed{25}$", "Notice that there are exactly $1000-100=900=5^2\\cdot 6^2$ possible values of $N$ . This means, in $100\\le N\\le 999$ , every possible combination of $2$ digits will happen exactly once. We know that $N=900,901,902,903,904$ works because $900\\equiv\\dots00_5\\equiv\\dots00_6$\nWe know for sure that the units digit will add perfectly every $30$ added or subtracted, because $\\text{lcm }5,6=30$ . So we only have to care about cases of $N$ every $30$ subtracted. In each case, $2N$ subtracts $6$ /adds $4$ $N_5$ subtracts $1$ and $N_6$ adds $1$ for the $10$ 's digit.\n\\[\\textbf{5 }\\textcolor{red}{\\text{ 0}}\\text{ 4 3 2 1 0 }\\textcolor{red}{\\text{4}}\\text{ 3 2 1 0 4 3 2 1 0 4 }\\textcolor{red}{\\text{3 2}}\\text{ 1 0 4 3 2 1 0 4 3 2 }\\textcolor{red}{\\text{1}}\\]\n\\[\\textbf{6 }\\textcolor{red}{\\text{ 0}}\\text{ 1 2 3 4 5 }\\textcolor{red}{\\text{0}}\\text{ 1 2 3 4 5 0 1 2 3 4 }\\textcolor{red}{\\text{5 0}}\\text{ 1 2 3 4 5 0 1 2 3 4 }\\textcolor{red}{\\text{5}}\\]\n\\[\\textbf{10}\\textcolor{red}{\\text{ 0}}\\text{ 4 8 2 6 0 }\\textcolor{red}{\\text{4}}\\text{ 8 2 6 0 4 8 2 6 0 4 }\\textcolor{red}{\\text{8 2}}\\text{ 6 0 4 8 2 6 0 4 8 2 }\\textcolor{red}{\\text{6}}\\]\nAs we can see, there are $5$ cases, including the original, that work. These are highlighted in $\\textcolor{red}{\\text{red}}$ . So, thus, there are $5$ possibilities for each case, and $5\\cdot 5=\\boxed{25}$", "Notice that $N_5$ ranges from $3$ to $5$ digits and $N_6$ ranges from $3$ to $4$ digits.\nThen let $a_i$ $b_i$ denotes the digits of $N_5$ $N_6$ , respectively such that \\[0\\le a_i<5,0\\le b_i<6\\] Thus we have \\[N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4\\] \\[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\\] Now we are given \\[2N \\equiv S \\equiv N_5+N_6\\pmod{100}\\] \\[2(625a_1+125a_2+25a_3+5a_4+a_5) \\equiv (10000a_1+1000a_2+100a_3+10a_4+a_5)+(1000b_1+100b_2+10b_3+b_4)\\pmod{100}\\] \\[1250a_1+250a_2+50a_3+10a_4+2a_5 \\equiv 10000a_1+1000a_2+1000b_1+100a_3+100b_2+10a_4+10b_3+a_5+b_4\\pmod{100}\\] \\[50a_1+50a_2+50a_3+10a_4+2a_5 \\equiv 10a_4+10b_3+a_5+b_4\\pmod{100}\\] Canceling out $a_5$ left with \\[50a_1+50a_2+50a_3+10a_4+a_5 \\equiv 10a_4+10b_3+b_4\\pmod{100}\\]\nSince $a_5$ $b_4$ determine the unit digits of the two sides of the congruence equation, we have $a_5=b_4=0,1,2,3,4$ . Thus,\n\\[50a_1+50a_2+50a_3+10a_4 \\equiv 10a_4+10b_3\\pmod{100}\\] canceling out $10a_4$ , we have \\[50a_1+50a_2+50a_3 \\equiv 10b_3\\pmod{100}\\] \\[5a_1+5a_2+5a_3 \\equiv b_3\\pmod{10}\\] \\[5(a_1+a_2+a_3) \\equiv b_3\\pmod{10}\\]\nThus $b_3$ is a multiple of $5$\nNow going back to our original equation \\[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\\] Since $a_5=b_4$ \\[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\\] \\[5(125a_1+25a_2+5a_3+a_4)=6(36b_1+6b_2+b_3)\\] \\[5(125a_1+25a_2+5a_3+a_4)=6[6(6b_1+b_2)+b_3]\\]\nSince the left side is a multiple of $5$ , then so does the right side. Thus $5\\mid6(6b_1+b_2)+b_3$\nSince we already know that $5\\mid b_3$ , then $5\\mid6(6b_1+b_2)$ , from where we also know that $5\\mid6b_1+b_2$\nFor $b_1,b_2<6$ , there is a total of 7 ordered pairs that satisfy the condition. Namely,\n\\[(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)\\]\nSince $N_6$ has at least $3$ digits, $(b_1,b_2)=(0,0)$ doesn't work. Furthermore, when $b_1=5$ $216b_1$ exceeds $1000$ which is not possible as $N$ is a three digit number, thus $(b_1,b_2)=(5,0)$ won't work as well.\nSince we know that $a_i<5$ , for each of the ordered pairs $(b_1,b_2)$ , there is respectively one and only one solution $(a_1,a_2,a_3,a_4)$ that satisfies the equation\n\\[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\\]\nThus there are five solutions to the equation. Also since we have 5 possibilities for $a_5=b_4$ , we have a total of $5\\cdot5=25$ values for $N$ $\\boxed{25}$", "Observe that the maximum possible value of the sum of the last two digits of the base $5$ number and the base $6$ number is $44+55=99$ . \nLet $N \\equiv a \\pmod {25}$ and $N \\equiv b \\pmod {36}$\nIf $a < \\frac{25}{2}$ $2N \\equiv 2a \\pmod {25}$ and if $a > \\frac{25}{2}$ $2N \\equiv 2a - 25 \\pmod {25}$\nUsing the same logic for $b$ , if $b < 18$ $2N \\equiv 2b \\pmod {36}$ , and in the other case $2N \\equiv 2b - 36 \\pmod {36}$\nWe can do four cases:\nCase 1: $a + b = 2a - 25 + 2b - 36 \\implies a + b = 61$\nFor this case, there is trivially only one possible solution, $(a, b) = (25, 36)$ , which is equivalent to $(a, b) = (0, 0)$\nCase 2: $a + b = 2a - 25 + 2b \\implies a + b = 25$\nNote that in this case, $a \\geq 13$ must hold, and $b < 18$ must hold. \nWe find the possible ordered pairs to be: $(13, 12), (14, 11), (15, 10), ..., (24, 1)$ for a total of $12$ ordered pairs.\nCase 3: $a + b = 2a + 2b - 36 \\implies a + b = 36$\nNote that in this case, $b \\geq 18$ must hold, and $a < 13$ must hold.\nWe find the possible ordered pairs to be: $(24, 12), (25, 11), (26, 10), ..., (35, 1)$ for a total of $12$ ordered pairs.\nCase 4: $a + b = 2a + 2b$\nTrivially no solutions except $(a, b) = (0, 0)$ , which matches the solution in Case 1, which makes this an overcount.\nBy CRT, each solution $(a, b)$ corresponds exactly one positive integer in a set of exactly $\\text{lcm} (25, 36) = 900$ consecutive positive integers, and since there are $900$ positive integers between $100$ and $999$ , our induction is complete, and our answer is $1 + 12 + 12 = \\boxed{25}$" ]