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[ "We have \\[\\sqrt{1} + \\sqrt{1+3} + \\sqrt{1+3+5} + \\sqrt{1+3+5+7} = \\sqrt{1} + \\sqrt{4} + \\sqrt{9} + \\sqrt{16}\\ = 1 + 2 + 3 + 4 = \\boxed{10}.\\] Note: This comes from the fact that the sum of the first $n$ odds is $n^2$" ]

[ "Notice this is the term $a_6$ in a recursive sequence, defined recursively as $a_1 = 3, a_n = 2a_{n-1} + 1.$ Thus: \\[\\begin{split} a_2 = 3 \\cdot 2 + 1 = 7.\\\\ a_3 = 7 \\cdot 2 + 1 = 15.\\\\ a_4 = 15 \\cdot 2 + 1 = 31.\\\\ a_5 = 31 \\cdot 2 + 1 = 63.\\\\ a_6 = 63 \\cdot 2 + 1 = \\boxed{127}\\]", "Starting to compute the inner expressions, we see the results are $1, 3, 7, 15, \\ldots$ . This is always $1$ less than a power of $2$ . The only admissible answer choice by this rule is thus $\\boxed{127}$", "Working our way from the innermost parenthesis outwards and directly computing, we have $\\boxed{127}$", "If you distribute this you get a sum of the powers of $2$ . The largest power of $2$ in the series is $64$ , so the sum is $\\boxed{127}$", "$(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$ $=(2(2(2(2(2(3)+1)+1)+1)+1)+1)$ $=(2(2(2(2(6+1)+1)+1)+1)+1)$ $=(2(2(2(2(7)+1)+1)+1)+1)$ $=(2(2(2(14+1)+1)+1)+1)$ $=(2(2(2(15)+1)+1)+1)$ $=(2(2(30+1)+1)+1)$ $=(2(2(31)+1)+1)$ $=(2(62+1)+1)$ $=(2(63)+1)$ $=(126+1)$ $=127 \\Longrightarrow \\boxed{127}$", "Notice that $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = x (x (x (x (x (x + 1) + 1) + 1) + 1) + 1) + 1$ . Substituting $2$ for $x$ , we get \\[2(2(2(2(2(2+1)+1)+1)+1)+1)+1 = 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2 + 1 = 2^7 - 1 \\Longrightarrow \\boxed{127}\\]" ]

[ "By the distributive property,\n\\[(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \\boxed{11}\\]", "Inputting 4 into $x$ in the original equation,\n\\[[3(4)-2][4(4)+1]-[3(4)-2][4(4)]+1 = 170-160+1 = \\boxed{11}\\]" ]

[ "By the distributive property,\n\\[(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \\boxed{11}\\]", "Inputting 4 into $x$ in the original equation,\n\\[[3(4)-2][4(4)+1]-[3(4)-2][4(4)]+1 = 170-160+1 = \\boxed{11}\\]" ]

[ "$(625^{\\log_5 2015})^\\frac{1}{4} = ((5^4)^{\\log_5 2015})^\\frac{1}{4} = (5^{4 \\cdot \\log_5 2015})^\\frac{1}{4} = (5^{\\log_5 2015 \\cdot 4})^\\frac{1}{4} = ((5^{\\log_5 2015})^4)^\\frac{1}{4} = (2015^4)^\\frac{1}{4} = \\boxed{2015}$", "$(625^{\\log_5 2015})^{\\frac{1}{4}} = (625^{\\frac{1}{4}})^{\\log_5 2015} = 5^{\\log_5 2015} = \\boxed{2015}$", "We note that the year number is just $2015$ , so just guess $\\boxed{2015}$" ]

[ "We can rewrite $\\log_5 2015$ as as $5^x = 2015$ . Thus, $625^{x \\cdot \\frac{1}{4}} = 5^x = \\boxed{2015}.$" ]

[ "By the order of operations , we have \\[(8 \\times 4 + 2) - (8 + 4 \\times 2) = (32+2) - (8+8) = 34 - 16 = \\boxed{18}.\\] ~apex304, TaeKim, peelybonehead, MRENTHUSIASM", "We can simplify the expression above in another way: \\[(8 \\times 4 + 2) - (8 + 4 \\times 2)=8\\times4+2-8-4\\times2=32+2-8-8=34-16=\\boxed{18}.\\]" ]

[ "Rearranging the terms, we get $(1-2)+(3-4)+(5-6)+...(2017-2018)+2019$ , and our answer is $-1009+2019=\\boxed{1010}$", "We can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of two. So, do the second last ones and so on. Now, all we need to find is the number of integers in any of the sets (I chose even) to get $\\boxed{1010}$", "It is similar to the Solution 1: \nRearranging the terms, we get $1+(3-2)+(5-4)+(6-5)...(2017-2016)+(2019-2018)$ , and our answer is $1+1009=\\boxed{1010}$" ]

[ "Note that the sum of consecutive odd numbers can be expressed as a square, namely $1+3+5+7+...+2017+2019 = 1010^2$ . We can modify the negative numbers in the same way by adding 1 to each negative term, factoring a negative sign, and accounting for the extra 1's by subtracting 1009. We then have $1010^2-1009^2-1009$ . Using difference of squares, we obtain $(1010+1009)(1010-1009)-1009 = 2019-1009 = \\boxed{1010}$" ]

[ "We see that $1, 2, 3,$ and $4$ each appear in the ones, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$ , we find that the sum is equal to \\[10\\cdot(1+10+100+1000)=\\boxed{11110}.\\] Note that it is equally valid to manually add all four numbers together to get the answer.", "We have \\[1234 + 2341 + 3412 + 4123 = 1111 \\left( 1 + 2 + 3 + 4 \\right) = \\boxed{11110}.\\] ~Steven Chen (www.professorchenedu.com)", "We see that the units digit must be $0$ , since $4+3+2+1$ is $0$ . But every digit from there, will be a $1$ since we have that each time afterwards, we must carry the $1$ from the previous sum. The answer choice that satisfies these conditions is $\\boxed{11110}$", "We can simply add the numbers. $1234 + 2341 + 3412 + 4123 = 11110 \\implies \\boxed{11110}$" ]

[ "We see that $1, 2, 3,$ and $4$ each appear in the ones, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$ , we find that the sum is equal to \\[10\\cdot(1+10+100+1000)=\\boxed{11110}.\\] Note that it is equally valid to manually add all four numbers together to get the answer.", "We have \\[1234 + 2341 + 3412 + 4123 = 1111 \\left( 1 + 2 + 3 + 4 \\right) = \\boxed{11110}.\\] ~Steven Chen (www.professorchenedu.com)", "We see that the units digit must be $0$ , since $4+3+2+1$ is $0$ . But every digit from there, will be a $1$ since we have that each time afterwards, we must carry the $1$ from the previous sum. The answer choice that satisfies these conditions is $\\boxed{11110}$", "We can simply add the numbers. $1234 + 2341 + 3412 + 4123 = 11110 \\implies \\boxed{11110}$" ]

[ "We group the addends inside the parentheses two at a time: \\begin{align*} -1 + 2 - 3 + 4 - 5 + 6 - 7 + \\ldots + 1000 &= (-1 + 2) + (-3 + 4) + (-5 + 6) + \\ldots + (-999 + 1000) \\\\ &= \\underbrace{1+1+1+\\ldots + 1}_{\\text{500 1's}} \\\\ &= 500. \\end{align*} Then the desired answer is $4 \\times 500 = \\boxed{2000}$" ]

[ "We can use subtraction of fractions to get \\[\\frac{11!-10!}{9!} = \\frac{11!}{9!} - \\frac{10!}{9!} = 110 -10 = \\boxed{100}.\\]", "Factoring out $9!$ gives $\\frac{11!-10!}{9!} = \\frac{9!(11 \\cdot 10 - 10)}{9!} = 110-10=\\boxed{100}$", "We are given the equation $\\frac{11!-10!}{9!}$\nThis is equivalent to $\\frac{11(10!) - 1(10!)}{9!}$ Simplifying, we get $\\frac{(11-1)(10!)}{9!}$ , which equals $10 \\cdot 10$\nTherefore, the answer is $10^2$ $\\boxed{100}$" ]

[ "We can use subtraction of fractions to get \\[\\frac{11!-10!}{9!} = \\frac{11!}{9!} - \\frac{10!}{9!} = 110 -10 = \\boxed{100}.\\]", "Factoring out $9!$ gives $\\frac{11!-10!}{9!} = \\frac{9!(11 \\cdot 10 - 10)}{9!} = 110-10=\\boxed{100}$", "We are given the equation $\\frac{11!-10!}{9!}$\nThis is equivalent to $\\frac{11(10!) - 1(10!)}{9!}$ Simplifying, we get $\\frac{(11-1)(10!)}{9!}$ , which equals $10 \\cdot 10$\nTherefore, the answer is $10^2$ $\\boxed{100}$" ]

[ "We have \\[\\frac{(2112-2021)^2}{169}=\\frac{91^2}{169}=\\frac{91^2}{13^2}=\\left(\\frac{91}{13}\\right)^2=7^2=\\boxed{49}.\\] ~MRENTHUSIASM", "We have \\[\\frac{(2112-2021)^2}{169}=\\frac{91^2}{169}=\\frac{(10^2-3^2)^2}{13^2}=\\frac{((10+3)(10-3))^2}{13^2}=\\frac{(13\\cdot7)^2}{13^2}=\\frac{13^2 \\cdot 7^2}{13^2}=7^2=\\boxed{49}.\\]", "We know that $2112-2021 = 91$ . Approximate this as $100$ as it is pretty close to it. Also, approximate $169$ to $170$ . We then have \\[\\frac{(2112 - 2021)^2}{169} \\approx \\frac{100^2}{170} \\approx \\frac{1000}{17} \\approx 58.\\] Now check the answer choices. The two closest answers are $49$ and $64$ . As the numerator is actually bigger than it should be, it should be the smaller answer, or $\\boxed{49}$" ]

[ "We have \\[\\frac{(2112-2021)^2}{169}=\\frac{91^2}{169}=\\frac{91^2}{13^2}=\\left(\\frac{91}{13}\\right)^2=7^2=\\boxed{49}.\\] ~MRENTHUSIASM", "We have \\[\\frac{(2112-2021)^2}{169}=\\frac{91^2}{169}=\\frac{(10^2-3^2)^2}{13^2}=\\frac{((10+3)(10-3))^2}{13^2}=\\frac{(13\\cdot7)^2}{13^2}=\\frac{13^2 \\cdot 7^2}{13^2}=7^2=\\boxed{49}.\\]", "We know that $2112-2021 = 91$ . Approximate this as $100$ as it is pretty close to it. Also, approximate $169$ to $170$ . We then have \\[\\frac{(2112 - 2021)^2}{169} \\approx \\frac{100^2}{170} \\approx \\frac{1000}{17} \\approx 58.\\] Now check the answer choices. The two closest answers are $49$ and $64$ . As the numerator is actually bigger than it should be, it should be the smaller answer, or $\\boxed{49}$" ]

[ "By: Dragonfly\nWe find that $a^{-1}$ is the same as $2$ , since a number to the power of $-1$ is just the reciprocal of that number. We then get the equation to be\n\\[\\frac{2\\times2+\\frac{2}{2}}{\\frac{1}{2}}\\]\nWe can then simplify the equation to get $\\boxed{10}$" ]

[ "Factorizing the numerator, $\\frac{\\frac{1}{a}\\cdot(2+\\frac{1}{2})}{a}$ then becomes $\\frac{\\frac{5}{2}}{a^{2}}$ which is equal to $\\frac{5}{2}\\cdot 2^2$ which is $\\boxed{10}$", "Substituting $\\frac{1}{2}$ for $a$ in $\\frac{\\frac{1}{a}\\cdot(2+\\frac{1}{2})}{a}$ gives us $\\boxed{10}$ . ~peelybonehead" ]

[ "$|x-1|$ is the distance between $x$ and $1$ $|x-2|$ is the distance between $x$ and $2$\nTherefore, the given equation says $x$ is equidistant from $1$ and $2$ , so $x=\\frac{1+2}2=\\frac{3}{2}\\Rightarrow\\boxed{32}$", "We know that either $x-1=x-2$ or $x-1=-(x-2)$ . The first equation simplifies to $-1=-2$ , which is false, so $x-1=-(x-2)$ . Solving, we get $x=\\frac{3}{2}\\Rightarrow\\boxed{32}$" ]

[ "$(4-2)-(9-3)+(16-4)=2-6+12=8.$ This corresponds to $\\boxed{8}.$", "We have \\begin{align*} \\left(2^2-2\\right)-\\left(3^2-3\\right)+\\left(4^2-4\\right) &= 2(2-1)-3(3-1)+4(4-1) \\\\ &= 2(1)-3(2)+4(3) \\\\ &= 2-6+12 \\\\ &= \\boxed{8} ~MRENTHUSIASM", "We have \\begin{align*} (2^2-2)-(3^2-3)+(4^2-4) &= 2^2+4^2-3^2-2+3-4 \\\\ &=2^2+(4-3)(4+3)-3 \\\\ &=2^2+7-3=2^2+4 \\\\ &=4\\cdot 2 \\\\ &=\\boxed{8} PureSwag", "Using the difference of squares, we have \\[(2-\\sqrt{2})(2+\\sqrt{2}) - (3-\\sqrt{3})(3+\\sqrt{3}) + (4-2)(4+2).\\] Knowing that $\\sqrt{2} \\approx 1.41$ and $\\sqrt{3} \\approx 1.73,$ we get \\[(2-\\sqrt{2})(2+\\sqrt{2}) - (3-\\sqrt{3})(3+\\sqrt{3}) + (4-2)(4+2) \\approx 0.59\\cdot 3.41 -1.26\\cdot 4.73 + 2 \\cdot 6 =8.0521,\\] which is closest to $\\boxed{8}.$" ]

[ "Let $\\text{log } 2 = x$ . The expression then becomes \\[(1+x)^3+(1-x)^3+(3x)(-2x)=\\boxed{2}.\\]", "Using sum of cubes \\[(\\log 5)^{3}+(\\log 20)^{3}\\] \\[= (\\log 5 + \\log 20)((\\log 5)^{2}-(\\log 5)(\\log 20) + (\\log 20)^{2})\\] \\[= 2((\\log 5)^{2}-(\\log 5)(2\\log 2 + \\log 5) + (2\\log 2 + \\log 5)^{2})\\] Let x = $\\log 5$ and y = $\\log 2$ , so $x+y=1$\nThe entire expression becomes \\[2(x^2-x(2y+x)+(2y+x)^2)-6y^2\\] \\[=2(x^2+2xy+4y^2-3y^2)\\] \\[=2(x+y)^2 = \\boxed{2}\\]", "We can estimate the solution. Using $\\log(2) \\approx 0.3, \\log(20) = \\log(10)+\\log(2) = 1 + 0.3 \\approx 1.3, \\log(8) \\approx 0.9$ and $\\log(.25) = \\log(1)-\\log(4)= 0 - 0.6\\approx -0.6,$ we have\n\\[0.7^3 + 1.7^3 + .9\\cdot(-0.6) = \\boxed{2}\\] ~kxiang", "Using log properties, we combine the terms to make our expression equal to $\\log {( (5^{\\log^2{5}}) \\cdot (20^{\\log^2{20}}) \\cdot 8 ^ {\\log{\\frac{1}{4}}} ) }$ . \nBy exponent properties, we separate the part with base $20$ to become $20^{\\log^2{5}} \\cdot 20^{\\log^2{20}-\\log^2{5}}$ . Then, we substitute this into the original expression to get $\\log {( (5^{\\log^2{5}}) \\cdot 20^{\\log^2{5}} \\cdot 20^{\\log^2{20}-\\log^2{5}} \\cdot 8 ^ {\\log{\\frac{1}{4}}} ) } = \\log {( (100^{\\log^2{5}}) \\cdot 20^{\\log^2{20}-\\log^2{5}} \\cdot 8 ^ {\\log{\\frac{1}{4}}} ) }$ . \nBecause $100^{\\log^2{5}} = 25^{\\log{5}}$ , and $\\log^2{20}-\\log^2{5} = (\\log{20}+\\log{5})(\\log{20}-\\log{5}) = \\log{100}\\cdot\\log{4} = 2\\log{4}$ , this expression is equal to $\\log {( 25 ^ {\\log{5}} \\cdot 400^{\\log{4}} \\cdot 8 ^ {\\log{\\frac{1}{4}}} ) }$ . \nWe perform the step with the base combining on $25$ and $400$ to get $25 ^ {\\log{5}} \\cdot 400^{\\log{4}} = 25 ^ {\\log{5}-\\log{4}} \\cdot 10000^{\\log{4}} = 25^{\\log{\\frac{5}{4}}}\\cdot 256$ . Putting this back into the whole equation gives $\\log{( 25^{\\log{\\frac{5}{4}}}\\cdot 256 \\cdot 8^{\\log{\\frac{1}{4}}})}$ . \nOne last base merge remains - but $25\\cdot 8$ isn't a power of 10. We can rectify this by converting $8^{\\log{\\frac{1}{4}}}$ to $(4^\\frac{3}{2})^{\\log{\\frac{1}{4}}} = 4^{\\log{ \\frac{1}{8} }}$ . \nFinally, we complete this arduous process by performing the base merge on $\\log{( 25^{\\log{\\frac{5}{4}}}\\cdot 256 \\cdot 4^{\\log{\\frac{1}{8}}})}$ . \nWe get $25^{\\log{\\frac{5}{4}}} \\cdot 4^{\\log{\\frac{1}{8}}} = 25^{\\log{\\frac{5}{4}}-\\log{\\frac{1}{8}}} \\cdot 100^{\\log{\\frac{1}{8}}} = 25^{\\log{10}} \\cdot \\frac{1}{64} = \\frac{25}{64}$ . \nPutting this back into that original equation one last time, we get $\\log(256 \\cdot \\frac{25}{64}) = \\log{100} = \\boxed{2}$ .\n~aop2014" ]

[ "Looking at the numbers, you see that every set of $4$ has $3$ positive numbers and 1 negative number. Calculating the sum of the first couple sets gives us $2+10+18...+394$ . Clearly, this pattern is an arithmetic sequence. By using the formula we get $\\frac{2+394}{2}\\cdot 50=\\boxed{9900}$", "Split the even numbers and the odd numbers apart. If we group every 2 even numbers together and add them, we get a total of $50\\cdot (-2)=-100$ . Summing the odd numbers is equivalent to summing the first 100 odd numbers, which is equal to $100^2=10000$ . Adding these two, we obtain the answer of $\\boxed{9900}$", "We can break this entire sum down into $4$ integer bits, in which the sum is $2x$ , where $x$ is the first integer in this bit. We can find that the first sum of every sequence is $4x-3$ , which we plug in for the $50$ bits in the entire sequence is $1+2+3+\\cdots+50=1275$ , so then we can plug it into the first term of every sequence equation we got above $4(1275)-3(50)=4950$ , and so the sum of every bit is $2x$ , and we only found the value of $x$ , the sum of the sequence is $4950\\cdot2=\\boxed{9900}$ .-middletonkids", "Another solution involves adding everything and subtracting out what is not needed. The first step involves solving $1+2+3+4+5+6+7+8+\\cdots+197+198+199+200$ . To do this, we can simply multiply $200$ and $201$ and divide by $2$ to get us $20100$ . The next step involves subtracting out the numbers with minus signs. We actually have to do this twice, because we need to take out the numbers we weren’t supposed to add and then subtract them from the problem. Then, we can see that from $4$ to $200$ , incrementing by $4$ , there are $50$ numbers that we have to subtract. To do this we can do $50$ times $51$ divided by $2$ , and then we can multiply by $4$ , because we are counting by fours, not ones. Our answer will be $5100$ , but remember, we have to do this twice. Once we do that, we will get $10200$ . Finally, we just have to do $20100-10200$ , and our answer is $\\boxed{9900}$", "In this solution, we group every 4 terms. Our groups should be: $1 + 2 + 3 - 4 = 2$ $5 + 6 + 7 - 8 = 10$ $9 + 10 + 11 - 12 = 18$ , ... $197 + 198 + 199 - 200 = 394$ . We add them together to get this expression: $2 + 10 + 18 + ... + 394$ . This can be rewritten as $8 * (0 + 1 + 2 + ... + 49) + 100$ . We add this to get $\\boxed{9900}$ . ~Baolan", "We can split up this long sum into groups of four integers. Finding the first few sums, we have that $1 + 2 + 3 - 4 = 2$ $5 + 6 + 7 - 8 = 10$ , and $9 + 10 + 11 - 12 = 18$ . Notice that this is an increasing arithmetic sequence, with a common difference of $8$ . We can find the sum of the arithmetic sequence by finding the average of the first and last terms, and then multiplying by the number of terms in the sequence. The first term is $1 + 2 + 3 - 4$ , or $2$ , the last term is $197 + 198 + 199 - 200$ , or $394$ , and there are $200\\div 4$ or $50$ terms. So, we have that the sum of the sequence is $\\frac{(394+2)\\cdot 50}{2}$ , or $\\boxed{9900}$ . ~Arctic_Bunny", "Note that the original expression is equal to \\[(1+2+3+\\dots+199+200)-2(4(1+2+3+\\dots+49+50)).\\] Since the sum of the first $n$ positive integers is $\\frac{n(n+1)}{2}$ , this is equal to \\[\\frac{200(201)}{2}-2\\left(4\\left(\\frac{50(51)}{2}\\right)\\right),\\] which can be simplified as \\[20100-4(50)(51)=20100-10200.\\] Doing the subtraction yields $\\boxed{9900}$ . -BorealBear", "We can split the sum into 4 groups so that the terms in these groups have the same common difference of 4 $(1+5+9+...+197)+(2+6+10+...+198)+(3+7+11+...+199)-(4+8+12+...+200)$ .\nThen, we can use the sum formula. It's equal to $\\frac{50\\cdot(1+197)}{2} + \\frac{50\\cdot(2+198)}{2} + \\frac{50\\cdot(3+199)}{2} - \\frac{50\\cdot(4+200)}{2}$ .\nThis can also be expressed as $(25\\cdot198) + (25\\cdot200) + (25\\cdot202) - (25\\cdot204)$ .\nAnd it is equal to $25\\cdot(198+200+202-204) = 25\\cdot396 = 9900$ .\nTherefore, the answer is $\\boxed{9900}$ . ~kurbanlik_21" ]

[ "We know that when we subtract negative numbers, $a-(-b)=a+b$\nThe equation becomes $1+2-3+4-5+6 = \\boxed{5}$", "Like Solution 1, we know that when we subtract $a-(-b)$ , that will equal $a+b$ as the opposite/negative of a negative is a positive. Thus, $1-(-2)-3-(-4)-5-(-6)=1+2-3+4-5+6$ . We can group together a few terms to make our computation a bit simpler. $1+(2-3)+4+(-5+6)= 1+(-1)+4+1=\\boxed{5}$" ]

[ "To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas.\n\\[2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3\\]\n$=(2-1)(2^2+1 \\cdot 2+1^2)+(4-3)(4^2+4 \\cdot 3+3^2)+(6-5)(6^2+6 \\cdot 5+5^2)+...+(18-17)(18^2+18 \\cdot 17+17^2)$\n$=(2^2+1 \\cdot 2+1^2)+(4^2+4 \\cdot 3+3^2)+(6^2+6 \\cdot 5+5^2)+...+(18^2+18 \\cdot 17+17^2)$\n$=1^2+2^2+3^2+4^2+5^2+6^2...+17^2+18^2+1 \\cdot 2+3 \\cdot 4+5 \\cdot 6+...+17 \\cdot 18$\n$=\\frac{18(18+1)(36+1)}{6}+1 \\cdot 2+3 \\cdot 4+5 \\cdot 6+...+17 \\cdot 18$\nwe could rewrite the second part as $\\sum_{n=1}^{9}(2n-1)(2n)$\n$(2n-1)(2n)=4n^2-2n$\n$\\sum_{n=1}^{9}4n^2=4(\\frac{9(9+1)(18+1)}{6})$\n$\\sum_{n=1}^{9}-2n=-2(\\frac{9(9+1)}{2})$\nHence,\n$1 \\cdot 2+3 \\cdot 4+5 \\cdot 6+...+17 \\cdot 18 = 4(\\frac{9(9+1)(18+1)}{6})-2(\\frac{9(9+1)}{2})$\nAdding everything up:\n$2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$\n$=\\frac{18(18+1)(36+1)}{6}+4(\\frac{9(9+1)(18+1)}{6})-2(\\frac{9(9+1)}{2})$\n$=3(19)(37)+6(10)(19)-9(10)$\n$=2109+1140-90$\n$=\\boxed{3159}$", "Think about $2^3+4^3+6^3+...+18^3$ . Once we factor out $2^3=8$ , we get $1^3+2^3+...+9^3$ , which can be found using the sum of cubes formula, $(\\frac{n(n+1)}{2})^2$ . Now think about $1^3+3^3+...+17^3$ . This is just the previous sum subtracted from the total sum of $18$ cubes. So now we have the two things we need to add. The sum of all the even cubes is $8\\cdot (\\frac{90}{2})^2\\rightarrow 8\\cdot 2025 = 16200$ . The sum of all cubes from $1^3$ to $18^3$ is $(\\frac{18\\cdot 19}{2})^2=29241$ . The sum of the odd cubes is then $29241-16200=13041$ . Thus we get $16200-13041=\\boxed{3159}$ ~amcrunner", "Using the same sum of cubes formula, we can rewrite as $2(2^3 + 4^3 + ... + 18^3) - (1^3 + 2^3 + ... + 18^3)$\n$= 2(2^3)(1^3 + ... + 9^3) - (1^3 + ... + 18^3)$\n$= 16(5 \\cdot 9)^2 - (9 \\cdot 19)^2 = 9^2(20^2 - 19^2) = 81 \\cdot 39 = \\boxed{3159}$ ~AoPSuser216", "For any real numbers $x$ and $y$ $x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)((x-y)^2+3xy)=(x-y)^3+3xy(x-y)$\nWhen $x = y + 1$ , with the above formula, we will get $x^3-y^3=1^3+3xy=1 + 3xy$\nTherefore,\n$2^3 - 1^3 + 4^3 - 3^3 + \\dots + 18^3 - 17^3$\n$= (1 + 3\\cdot 1\\cdot 2) + (1 + 3\\cdot 3\\cdot 4) + \\dots + (1 + 3\\cdot 17\\cdot 18)$\n$= 9 + 3\\cdot (1\\cdot 2 + 3\\cdot 4 + \\dots + 17\\cdot 18)$\n$= 9 + 3 \\cdot (2 + 12 + 30 + 56 + 90 + 132 + 182 + 240 + 306)$\n$= 9 + 3 \\cdot 1050$\n$= \\boxed{3159}$", "We rewrite the sum as\n\\[\\sum_{k=1}^{9}(2k)^3-(2k-1)^3\\] \\[=\\sum_{k=1}^{9} 12k^2 - 6k + 1\\] \\[=12\\sum_{k=1}^{9}k^2 - 6\\sum_{k=1}^{9}k + 9\\] \\[=12\\cdot\\frac{9\\cdot 10\\cdot 19}{6} -6\\cdot \\frac{9\\cdot 10}{2} + 9\\] \\[=3420 - 270 + 9 = \\boxed{3159}\\]", "We see $=12\\cdot\\frac{9\\cdot 10\\cdot 19}{6} -6\\cdot \\frac{9\\cdot 10}{2} + 9 = 9\\cdot (12\\cdot\\frac{10\\cdot 19}{6} -6\\cdot \\frac{10}{2} + 1)$ which is clearly a multiple of 9. The only answer choice which is a multiple of 9 is $\\boxed{3159}$ ~Ilaggo2432", "$2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \\dots + 18^3 - 17^3$\n$=8-1+64-27+216-125+512-343+1000-729+1728-1331+2744-2197+4096-3375+5832-4913$\n$= \\boxed{3159}$", "Reduce all terms mod 9. This yields:\n$2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \\dots + 18^3 - 17^3$ $\\equiv -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 + 0 - -1 + -1 - 1 + 1 - 0 +0 - -1 (\\mod 9)$ $\\equiv 0 (\\mod 9)$\nThe only answer choice which is also ≡0 mod 9 is $= \\boxed{3159}$" ]

[ "We evaluate the given expression to get that \\[2^{1+2+3}-(2^1+2^2+2^3)=2^6-(2^1+2^2+2^3)=64-2-4-8=50 \\implies \\boxed{50}\\]" ]

[ "We apply the difference of squares to the denominator, and then regroup factors: \\begin{align*} \\frac{\\left(1+\\frac13\\right)\\left(1+\\frac15\\right)\\left(1+\\frac17\\right)}{\\sqrt{\\left(1-\\frac{1}{3^2}\\right)\\left(1-\\frac{1}{5^2}\\right)\\left(1-\\frac{1}{7^2}\\right)}} &= \\frac{\\left(1+\\frac13\\right)\\left(1+\\frac15\\right)\\left(1+\\frac17\\right)}{\\sqrt{\\left(1+\\frac13\\right)\\left(1+\\frac15\\right)\\left(1+\\frac17\\right)}\\cdot\\sqrt{\\left(1-\\frac13\\right)\\left(1-\\frac15\\right)\\left(1-\\frac17\\right)}} \\\\ &= \\frac{\\sqrt{\\left(1+\\frac13\\right)\\left(1+\\frac15\\right)\\left(1+\\frac17\\right)}}{\\sqrt{\\left(1-\\frac13\\right)\\left(1-\\frac15\\right)\\left(1-\\frac17\\right)}} \\\\ &= \\frac{\\sqrt{\\frac43\\cdot\\frac65\\cdot\\frac87}}{\\sqrt{\\frac23\\cdot\\frac45\\cdot\\frac67}} \\\\ &= \\frac{\\sqrt{4\\cdot6\\cdot8}}{\\sqrt{2\\cdot4\\cdot6}} \\\\ &= \\frac{\\sqrt8}{\\sqrt2} \\\\ &= \\boxed{2} ~MRENTHUSIASM", "Since these numbers are fairly small, we can use brute force as follows: \\[\\frac{(1+\\frac{1}{3})(1+\\frac{1}{5})(1+\\frac{1}{7})}{\\sqrt{(1-\\frac{1}{3^2})(1-\\frac{1}{5^2})(1-\\frac{1}{7^2})}} =\\frac{\\frac{4}{3}\\cdot\\frac{6}{5}\\cdot\\frac{8}{7}}{\\sqrt{\\frac{8}{9}\\cdot\\frac{24}{25}\\cdot\\frac{48}{49}}} =\\frac{\\frac{4\\cdot6\\cdot8}{3\\cdot5\\cdot7}}{\\sqrt{\\frac{(2^3)(2^3\\cdot3^1)(2^4\\cdot3^1)}{(3^2)(5^2)(7^2)}}} =\\frac{\\frac{64}{35}}{\\frac{96}{105}}=\\frac{64}{35}\\cdot\\frac{105}{96}=\\boxed{2}.\\] ~not_slay", "This solution starts precisely like the one above. We simplify to get the following:\n\\[\\frac{(1+\\frac{1}{3})(1+\\frac{1}{5})(1+\\frac{1}{7})}{\\sqrt{(1-\\frac{1}{3^2})(1-\\frac{1}{5^2})(1-\\frac{1}{7^2})}} = \\frac{\\frac{4\\cdot6\\cdot8}{3\\cdot5\\cdot7}}{\\sqrt{\\frac{(2^3)(2^3\\cdot3^1)(2^4\\cdot3^1)}{(3^2)(5^2)(7^2)}}}\\]\nBut now, we can get a nice simplification as shown: \\[\\frac{\\frac{4\\cdot6\\cdot8}{3\\cdot5\\cdot7}}{\\sqrt{\\frac{(2^3)(2^3\\cdot3^1)(2^4\\cdot3^1)}{(3^2)(5^2)(7^2)}}} = \\dfrac{\\frac{4\\cdot6\\cdot8}{3\\cdot5\\cdot7}}{\\sqrt{\\frac{2^{10} \\cdot 3^{2}}{3^2\\cdot 5^2\\cdot 7^2}}} = \\dfrac{\\frac{4\\cdot6\\cdot8}{3\\cdot5\\cdot7}}{\\frac{2^5 \\cdot 3}{3\\cdot5\\cdot 7}} =\\dfrac{4\\cdot6\\cdot8}{3\\cdot5\\cdot7} \\hspace{0.05 in} \\cdot \\hspace{0.05 in}\\dfrac{3\\cdot5\\cdot 7}{2^5 \\cdot 3} =\\dfrac{2^6\\cdot 3}{2^5\\cdot 3} = \\boxed{2}.\\]" ]

[ "We will apply the following logarithmic identity: \\[\\log_{p^n}{q^n}=\\log_{p}{q},\\] which can be proven by the Change of Base Formula: \\[\\log_{p^n}{q^n}=\\frac{\\log_{p}{q^n}}{\\log_{p}{p^n}}=\\frac{n\\log_{p}{q}}{n}=\\log_{p}{q}.\\] Now, we simplify the expressions inside the summations: \\begin{align*} \\log_{5^k}{{3^k}^2}&=\\log_{5^k}{(3^k)^k} \\\\ &=k\\log_{5^k}{3^k} \\\\ &=k\\log_{5}{3}, \\end{align*} and \\begin{align*} \\log_{9^k}{25^k}&=\\log_{3^{2k}}{5^{2k}} \\\\ &=\\log_{3}{5}. \\end{align*} Using these results, we evaluate the original expression: \\begin{align*} \\left(\\sum_{k=1}^{20} \\log_{5^k} 3^{k^2}\\right)\\cdot\\left(\\sum_{k=1}^{100} \\log_{9^k} 25^k\\right)&=\\left(\\sum_{k=1}^{20} k\\log_{5}{3}\\right)\\cdot\\left(\\sum_{k=1}^{100} \\log_{3}{5}\\right) \\\\ &= \\left(\\log_{5}{3}\\cdot\\sum_{k=1}^{20} k\\right)\\cdot\\left(\\log_{3}{5}\\cdot\\sum_{k=1}^{100} 1\\right) \\\\ &= \\left(\\sum_{k=1}^{20} k\\right)\\cdot\\left(\\sum_{k=1}^{100} 1\\right) \\\\ &= \\frac{21\\cdot20}{2}\\cdot100 \\\\ &= \\boxed{21000} ~MRENTHUSIASM (Solution)", "First, we can get rid of the $k$ exponents using properties of logarithms: \\[\\log_{5^k} 3^{k^2} = k^2 \\cdot \\frac{1}{k} \\cdot \\log_{5} 3 = k\\log_{5} 3 = \\log_{5} 3^k.\\] (Leaving the single $k$ in the exponent will come in handy later). Similarly, \\[\\log_{9^k} 25^{k} = k \\cdot \\frac{1}{k} \\cdot \\log_{9} 25 = \\log_{9} 5^2.\\] Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms: \\begin{align*} \\sum_{k=1}^{20} \\log_{5} 3^k &= \\log_{5} 3^1 + \\log_{5} 3^2 + \\dots + \\log_{5} 3^{20} \\\\ &= \\log_{5} 3^{(1 + 2 + \\dots + 20)} \\\\ &= \\log_{5} 3^{\\frac{20(20+1)}{2}} &&\\hspace{15mm}(*) \\\\ &= \\log_{5} 3^{210}, \\\\ \\sum_{k=1}^{100} \\log_{9} 5^2 &= \\log_{9} 5^2 + \\log_{9} 5^2 + \\dots + \\log_{9} 5^2 \\\\ &= \\log_{9} 5^{2(100)} \\\\ &= \\log_{9} 5^{200}. \\end{align*} In $(*),$ we use the triangular numbers equation: \\[1 + 2 + \\dots + n = \\frac{n(n+1)}{2} = \\frac{20(20+1)}{2} = 210.\\] Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify: \\[\\log_{a} b\\log_{x} y = \\log_{a} y\\log_{x} b.\\] Thus, \\begin{align*} \\left(\\log_{5} 3^{210}\\right)\\left(\\log_{3^2} 5^{200}\\right) &= \\left(\\log_{5} 5^{200}\\right)\\left(\\log_{3^2} 3^{210}\\right) \\\\ &= \\left(\\log_{5} 5^{200}\\right)\\left(\\log_{3} 3^{105}\\right) \\\\ &= (200)(105) \\\\ &= \\boxed{21000} ~Joeya (Solution)", "In $\\sum_{k=1}^{20} \\log_{5^k} 3^{k^2},$ note that the addends are greater than $1$ for all $k\\geq2.$\nIn $\\sum_{k=1}^{100} \\log_{9^k} 25^k,$ note that the addends are greater than $1$ for all $k\\geq1.$\nWe have the inequality \\[\\left(\\sum_{k=1}^{20} \\log_{5^k} 3^{k^2}\\right)\\cdot\\left(\\sum_{k=1}^{100} \\log_{9^k} 25^k\\right)>\\left(\\sum_{k=2}^{20} 1\\right)\\cdot\\left(\\sum_{k=1}^{100} 1\\right)=19\\cdot100=1900,\\] which eliminates choices $\\textbf{(A)}, \\textbf{(B)},$ and $\\textbf{(C)}.$ We get the answer $\\boxed{21000}$ by either an educated guess or a continued approximation:", "Using the identity \\[\\log_{p^n}{q^n}=\\log_{p}{q},\\] simplify \\begin{align*} \\log_{5^k}{{3^k}^2}&=\\log_{5^k}{(3^k)^k} \\\\ &=\\log_{5}{3^k} \\\\ \\end{align*} and \\begin{align*} \\log_{9^k}{25^k}&=\\log_{3^{2k}}{5^{2k}} \\\\ &=\\log_{3}{5}. \\end{align*} . Now we have the product: \\[\\left(\\sum_{k=1}^{20} \\log_{5} 3^{k}\\right)\\cdot\\left(\\sum_{k=1}^{100} \\log_{3} 5\\right)\\] \\begin{align*} \\sum_{k=1}^{20} \\log_{5} 3^k &= \\log_{5} 3^1 + \\log_{5} 3^2 + \\dots + \\log_{5} 3^{20} \\\\ &= \\log_{5} 3^{(1 + 2 + \\dots + 20)} \\\\ &= \\log_{5} 3^{\\frac{20(20+1)}{2}} \\\\ &= \\log_{5} 3^{210} \\\\ &= {210}\\cdot\\log_{5} {3}, \\\\ \\sum_{k=1}^{100}\\log_{3} {5} &= {100}\\cdot\\log_{3} {5}. \\end{align*} With the reciprocal rule the logarithms cancel out leaving: $\\boxed{21000}.$" ]

[ "From the Change of Base Formula, we have \\[\\frac{\\prod_{i=3}^{13} \\log (2i+1)}{\\prod_{i=1}^{11}\\log (2i+1)} = \\frac{\\log 25 \\cdot \\log 27}{\\log 3 \\cdot \\log 5} = \\frac{(2\\log 5)\\cdot(3\\log 3)}{\\log 3 \\cdot \\log 5} = \\boxed{6}.\\]", "Using the chain rule of logarithms $\\log _{a} b \\cdot \\log _{b} c = \\log _{a} c,$ we get \\begin{align*} \\log_37\\cdot\\log_59\\cdot\\log_711\\cdot\\log_913\\cdots\\log_{21}25\\cdot\\log_{23}27 &= (\\log _{3} 7 \\cdot \\log _{7} 11 \\cdots \\log _{23} 27) \\cdot (\\log _{5} 9 \\cdot \\log _{9} 13 \\cdots \\log _{21} 25) \\\\ &= \\log _{3} 27 \\cdot \\log _{5} 25 \\\\ &= 3 \\cdot 2 \\\\ &= \\boxed{6}" ]

[ "Directly calculating:\nWe evaluate both the top and bottom: $\\frac{40320}{36}$ . This simplifies to $\\boxed{1120}$", "It is well known that the sum of all numbers from $1$ to $n$ is $\\frac{n(n+1)}{2}$ . Therefore, the denominator is equal to $\\frac{8 \\cdot 9}{2} = 4 \\cdot 9 = 2 \\cdot 3 \\cdot 6$ . Now, we can cancel the factors of $2$ $3$ , and $6$ from both the numerator and denominator, only leaving $8 \\cdot 7 \\cdot 5 \\cdot 4 \\cdot 1$ . This evaluates to $\\boxed{1120}$", "First, we evaluate $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8$ to get 36. We notice that 36 is 6 squared, so we can factor the denominator like $\\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 \\cdot 8}{6 \\cdot 6}$ then cancel the 6s out,\\ to get $\\frac{4 \\cdot 5 \\cdot 7 \\cdot 8}{1}$ . Now that we have escaped fraction form, we multiply $4 \\cdot 5 \\cdot 7 \\cdot 8$ . Multiplying these, we get $\\boxed{1120}$" ]

[ "$\\sqrt{16\\sqrt{8\\sqrt{4}}}$ $\\sqrt{16\\sqrt{8\\cdot 2}}$ $\\sqrt{16\\sqrt{16}}$ $\\sqrt{16\\cdot 4}$ $\\sqrt{64}$ $\\boxed{8}$" ]

[ "Using difference of squares to factor the left term, we get \\[\\frac{100^2-7^2}{70^2-11^2} \\cdot \\frac{(70-11)(70+11)}{(100-7)(100+7)} = \\frac{(100-7)(100+7)}{(70-11)(70+11)} \\cdot \\frac{(70-11)(70+11)}{(100-7)(100+7)}.\\] Cancelling all the terms, we get $\\boxed{1}$ as the answer." ]

[ "By adding up the numbers in each of the $6$ parentheses, we get:\n$\\frac{2}{1} \\cdot \\frac{3}{2} \\cdot \\frac{4}{3} \\cdot \\frac{5}{4} \\cdot \\frac{6}{5} \\cdot \\frac{7}{6}$\nUsing telescoping, most of the terms cancel out diagonally. We are left with $\\frac{7}{1}$ which is equivalent to $7$ . Thus, the answer would be $\\boxed{7}$" ]

[ "We see that $\\frac{44}{11}$ is $4$ $\\frac{110}{44}$ simplifies to $\\frac{5}{2}$ , which is $2.5$\nand $\\frac{44}{1100}$ simplifies to $\\frac{1}{25}$ , which is $0.04$\n$4+2.5+0.04$ reveals \\[\\frac{44}{11} + \\frac{110}{44} + \\frac{44}{1100}\\] is $\\boxed{6.54}$ . \n~ le_petit_chouetteur from TSMV" ]

[ "Drawing the tetrahedron out and testing side lengths, we realize that the $\\triangle ACD, \\triangle ABC,$ and $\\triangle ABD$ are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take $\\triangle ADC$ as the base, then $\\overline{AB}$ must be the altitude. The volume of tetrahedron $ABCD$ is $\\dfrac{1}{3} \\cdot \\dfrac{3 \\cdot 4}{2} \\cdot 2=\\boxed{4}.$", "We will place tetrahedron $ABCD$ in the $xyz$ -plane. By the Converse of the Pythagorean Theorem, we know that $\\triangle ACD$ is a right triangle. Without the loss of generality, let $A=(0,0,0), C=(3,0,0), D=(0,4,0),$ and $B=(x,y,z).$\nWe apply the Distance Formula to $\\overline{BA},\\overline{BC},$ and $\\overline{BD},$ respectively: \\begin{align*} x^2+y^2+z^2&=2^2, &(1) \\\\ (x-3)^2+y^2+z^2&=\\sqrt{13}^2, &(2) \\\\ x^2+(y-4)^2+z^2&=\\left(2\\sqrt5\\right)^2. &\\hspace{1mm} (3) \\end{align*} Subtracting $(1)$ from $(2)$ gives $-6x+9=9,$ from which $x=0.$\nSubtracting $(1)$ from $(3)$ gives $-8y+16=16,$ from which $y=0.$\nSubstituting $(x,y)=(0,0)$ into $(1)$ produces $z^2=4,$ or $|z|=2.$\nLet the brackets denote areas. Finally, we find the volume of tetrahedron $ABCD$ using $\\triangle ACD$ as the base: \\begin{align*} V_{ABCD}&=\\frac13\\cdot[ACD]\\cdot h_B \\\\ &=\\frac13\\cdot\\left(\\frac12\\cdot AC \\cdot AD\\right)\\cdot |z| \\\\ &=\\boxed{4} ~MRENTHUSIASM" ]

[ "Taking the seventh root of both sides, we get $(7x)^2=14x$\nSimplifying the LHS gives $49x^2=14x$ , which then simplifies to $7x=2$\nThus, $x=\\boxed{27}$" ]

[ "Notice that a number in row $k$ is $2k$ less than the number directly below it. For example, $5$ , which is in row $3$ , is $(2)(3)=6$ less than the number below it, $11$\nFrom row 1 to row $k$ , there are $k \\left(\\frac{1+(-1+2k)}{2} \\right) = k^2$ numbers in those $k$ rows. Because there are $12^2=144$ numbers up to the 12th row, $142$ is in the $k^{th}$ row. The number directly above is in the 11th row, and is $22$ less than $142$ . Thus the number directly above $142$ is $142-22=\\boxed{120}$", "Writing a couple more rows, the last number in each row ends in a perfect square. Thus $142$ is two left from the last number in its row, $144$ . One left and one up from $144$ is the last number of its row, also a perfect square, and is $121$ . This is one right and one up from $142$ , so the number directly above $142$ is one less than $121$ , or $\\boxed{120}$", "We can notice that even numbers are above even numbers so we can narrow our answers down to 2 solutions, $120$ and $122$ . Notice each number on the right end of each row is a perfect square. The perfect square closest to these is $121$ so $122$ is the first number on the next row. Notice $142$ is the third to last number on its row so it is too far away from 122, so our answer is $\\boxed{120}$" ]

[ "Adding all of the numbers gives us $\\frac{11\\cdot12}{2}=66$ as the current total. Since there are $11$ numbers, the current average is $\\frac{66}{11}=6$ . We need to take away a number from the total and then divide the result by $10$ because there will only be $10$ numbers left to give an average of $6.1$ . Setting up the equation:\n$\\frac{66-x}{10}=6.1$\n$66 - x = 61$\n$x = 5$\nThus, the answer is $\\boxed{5}$", "Similar to the first solution, the current total is $66$ . Since there are $11$ numbers on the list, taking $1$ number away will leave $10$ numbers. If those $10$ numbers have an average of $6.1$ , then those $10$ numbers must have a sum of $10 \\times 6.1 = 61$ . Thus, the number that was removed must be $66 - 61 = 5$ , and the answer is $\\boxed{5}$" ]

[ "We can approximate $7,928,564$ to $8,000,000$ and $0.000315$ to $0.0003.$ Multiplying the two yields $2400.$ Thus, it shows our answer is $\\boxed{2400}.$" ]

[ "\\begin{align*} 10^2-93&=7\\\\ 10^3-93&=907\\\\ 10^4-93&=9907\\\\ \\end{align*}\nThis can be generalized into $10^n-93$ is equal is $n-2$ nines followed by the digits $07$ . Then $10^{93}-93$ is equal to $91$ nines followed by $07$ . The sum of the digits is equal to $9(91)+7=819+7=\\boxed{826}$" ]

[ "Let $x$ be the sum of the integers and $y$ be the number of elements in the list. Then we get the equations $\\dfrac{x+15}{y+1}=\\dfrac{x}{y}+2$ and $\\dfrac{x+15+1}{y+1+1}=\\dfrac{x+16}{y+2}=\\frac{x}{y}+2-1=\\frac{x}{y}+1$ . With a lot of algebra, the solution is found to be $y= \\boxed{4}$", "We let $n$ be the original number of elements in the set and we let $m$ be the original average of the terms of the original list. Then we have $mn$ is the sum of all the elements of the list. So we have two equations: \\[mn+15=(m+2)(n+1)=mn+m+2n+2\\] and \\[mn+16=(m+1)(n+2)=mn+2m+n+2.\\] Simplifying both equations and we get, \\[13=m+2n\\] \\[14=2m+n\\] Solving for $m$ and $n$ , we get $m=5$ and $n=\\boxed{4}$", "Warning: This solution will rarely ever work in any other case. However, seeing that you can so easily plug and chug in problem 25 it is funny to see this.\nPlug and chug random numbers with the answer choices, starting with the choice of $4$ numbers. You see that if you have 4 5s and you add 15 to the set, the resulting mean will be 7; we can verify this with math \\[\\frac{5+5+5+5+15}{5}=7\\] adding in 1 to the set you result in the mean to be 6. \\[\\frac{5+5+5+5+15+1}{6}=6\\] Thus we conclude that 4 is the correct choice or $\\boxed{4}$" ]

[ "Note that the units digits of the powers of 9 have a pattern: $9^1 = {\\bf 9}$ $9^2 = 8{\\bf 1}$ $9^3 = 72{\\bf 9}$ $9^4 = 656{\\bf 1}$ , and so on. Since all natural numbers with the same last digit have the same remainder when divided by 5, the entire number doesn't matter, just the last digit. For even powers of $9$ , the number ends in a $1$ . Since the exponent is even, the final digit is $1$ . Note that all natural numbers that end in $1$ have a remainder of $1$ when divided by $5$ . So, our answer is $\\boxed{1}$", "Write $1999$ as $2000-1$ . We are taking $(2000-1)^{2000} \\mod{10}$ . Using the binomial theorem, we see that ALL terms in this expansion are divisible by $2000$ except for the very last term, which is just $(-1)^{2000}$ . This is clear because the binomial expansion is just choosing how many $2000$ s and how many $-1$ s there are for each term. Using this, we can take the entire polynomial $\\mod{10}$ , which leaves just $(-1)^{2000}=\\boxed{1}$", "As $1999 \\equiv -1 \\pmod{5}$ , we have $1999^{2000} \\equiv (-1)^{2000} \\equiv 1 \\pmod{5}$ . Thus, the answer is $\\boxed{1}$", "A sum/product of any two natural numbers has the same remainder, when divided by $n \\in \\{\\mathbb N}$ (Error compiling LaTeX. Unknown error_msg) , as the sum/product of their remainders. Thus, we can use the basic definition of an exponent and view the problem as $1999 \\cdot 1999 \\cdot \\cdot \\cdot 1999$\nUsing the fact stated in the first sentence, we see that the remainder of $1999$ , when divided by $5$ , is $4$ . The problem can now be written viewed as finding the remainder of $4^{2000}$ when it is divided by $5$ , which is already much simpler.\nNow, toying with the simplified problem a little, see notice that powers of $4$ alternate from ending in $4$ and $6$ . We notice that even powers of $4$ always end in $6$ , and also that $2000$ is even. Thus $4^{2000}$ must end in $6$ , which, when divided by $5$ gives a remainder of $\\boxed{1}$" ]

[ "We can use number separation for this problem. If we set each of the dice value to $D\\{a, b, c, d, e, f, g, h\\}$ , we can say $D = 10$ and each of $D$ 's elements are larger than $0$ . Using the positive number separation formula, which is $\\dbinom{n-1}{r-1}$ , we can make the following equations. \\begin{align*} D &= 10 \\\\ a+b+c+d+e+f+g &= 10 \\\\ \\dbinom{10-1}{7-1} &= \\\\ \\dbinom{9}{6} &= \\\\ \\dbinom{9}{3} &= \\\\ \\dfrac{9 \\cdot 8 \\cdot 7}{3 \\cdot 2 \\cdot 1} &= \\\\ 12 \\cdot 7 &= \\boxed{84}" ]

[ "Add possibilities. There are $3$ ways to sum to $10$ , listed below.\n\\[4,1,1,1,1,1,1: 7\\] \\[3,2,1,1,1,1,1: 42\\] \\[2,2,2,1,1,1,1: 35.\\]\nAdd up the possibilities: $35+42+7=\\boxed{84}$", "We can use generating functions, where $(x+x^2+...+x^6)$ is the function for each die. We want to find the coefficient of $x^{10}$ in $(x+x^2+...+x^6)^7$ , which is the coefficient of $x^3$ in $\\left(\\frac{1-x^7}{1-x}\\right)^7$ . This evaluates to $\\dbinom{-7}{3} \\cdot (-1)^3=\\boxed{84}$" ]

[ "If we let each number take its minimum value of 1, we will get 7 as the minimum sum. So we can do $10$ $7$ $3$ to find the number of balls we need to distribute to get three more added to the minimum to get 10, so the problem is asking how many ways can you put $3$ balls into $7$ boxes. From there we get $\\binom{7+3-1}{7-1}=\\binom{9}{6}=\\boxed{84}$" ]

[ "This is equivalent to $\\frac{(a-1)^6}{a^6}.$ Its expansion has 7 terms, whose coefficients are the same as those of $(a-1)^6$ .\nBy the Binomial Theorem, the sum of the last three coefficients is $\\binom{6}{2}-\\binom{6}{1}+\\binom{6}{0}=15-6+1=\\boxed{10}$" ]

[ "We start by trying to prove a function of $n$ , and then we can apply the function and equate it to $936$ to find the value of $n$\nIt is helpful to think of this problem in the format $(1+2+3+4+5+6) \\cdot (1+2+3+4+5+6) \\dots$ . Note that if we represent the scenario in this manner, we can think of picking a $1$ for one factor and then a $5$ for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for $n=2$ $4$ can be reached by picking $1$ and $4$ or $2$ and $2$ . However, this form gives us insights that will be useful later in the problem.\nNote that there are only $3$ primes in the set $\\{1,2,3,4,5,6\\}$ $2,3,$ and $5$ . Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form $2^h \\cdot 3^i \\cdot 5^j$ (the choice of variables will become clear later), for integer nonnegative values $h,i,j$ . So now the problem boils down to how many distinct triplets $(h,i,j)$ can be formed by taking the product of $n$ dice values.\nWe start our work on representing $j$ : the powers of $5$ , because it is the simplest in this scenario because there is only one factor of $5$ in the set. Because of this, having $j$ fives in our prime factorization of the product is equivalent to picking $j$ factors from the polynomial $(1+\\dots + 6) \\cdots$ and choosing each factor to be a $5$ . Now that we've selected $j$ factors, there are $n-j$ factors remaining to choose our powers of $3$ and $2$\nSuppose our prime factorization of this product contains $i$ powers of $3$ . These powers of $3$ can either come from a $3$ factor or a $6$ factor, but since both $3$ and $6$ contain only one power of $3$ , this means that a product with $i$ powers of $3$ corresponds directly to picking $i$ factors from the polynomial, each of which is either $3$ or $6$ (but this distinction doesn't matter when we consider only the powers of $3$\nNow we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair $(i,j)$ that match the requirements, corresponding to the number of $3$ 's and the number of $5$ 's our product will have. Then how many different $h$ values for the powers of $2$ are possible?\nIn the $i+j$ factors we have already chosen, we obviously can't have any factors of $2$ in the $j$ factors with $5$ . However, we can have a factor of $2$ pairing with factors of $3$ , if we choose a $6$ . The maximal possible power of $2$ in these $i$ factors is thus $2^i$ , which occurs when we pick every factor to be $6$\nWe now have $n-i-j$ factors remaining, and we want to allocate these to solely powers of $2$ . For each of these factors, we can choose either a $1,2,$ or $4$ . Therefore the maximal power of $2$ achieved in these factors is when we pick $4$ for all of them, which is equivalent to $2^{2\\cdot (n-i-j)}$\nNow if we multiply this across the total $n$ factors (or $n$ dice) we have a total of $2^{2n-2i-2j} \\cdot 2^i = 2^{2n-i-2j}$ , which is the maximal power of $2$ attainable in the product for a pair $(i,j)$ . Now note that every power of $2$ below this power is attainable: we can simply just take away a power of $2$ from an existing factor by dividing by $2$ . Therefore the powers of $2$ , and thus the $h$ value ranges from $h=0$ to $h=2n-i-2j$ , so there are a total of $2n+1-i-2j$ distinct values for $h$ for a given pair $(i,j)$\nNow to find the total number of distinct triplets, we must sum this across all possible $i$ s and $j$ s. Lets take note of our restrictions on $i,j$ : the only restriction is that $i+j \\leq n$ , since we're picking factors from $n$ dice.\n\\[\\sum_{i+j\\leq n}^{} 2n+1-i-2j = \\sum_{i+j \\leq n}^{} 2n+1 - \\sum_{i+j \\leq n}^{} i+2j\\]\nWe start by calculating the first term. $2n+1$ is constant, so we just need to find out how many pairs there are such that $i+j \\leq n$ . Set $i$ to $0$ $j$ can range from $0$ to $n$ , then set $i$ to $1$ $j$ can range from $0$ to $n-1$ , etc. The total number of pairs is thus $n+1+n+n-1+\\dots+1 = \\frac{(n+1)(n+2)}{2}$ . Therefore the left summation evaluates to \\[\\frac{(2n+1)(n+1)(n+2)}{2}\\]\nNow we calculate $\\sum_{i+j \\leq n}^{} i+2j$ . This simplifies to $\\sum_{i+j \\leq n}^{} i + 2 \\cdot \\sum_{i+j \\leq n}^{} j$ . Note that because $i+j = n$ is symmetric with respect to $i,j$ , the sum of $i$ in all of the pairs will be equal to the sum of $j$ in all of the pairs. Thus this is equal to calculating $3 \\cdot \\sum_{i+j \\leq n}^{} i$\nIn the pairs, $i=1$ appears for $j$ ranging between $0$ and $n-1$ so the sum here is $1 \\cdot (n)$ . Similarly $i=2$ appears for $j$ ranging from $0$ to $n-2$ , so the sum is $2 \\cdot (n-1)$ . If we continue the pattern, the sum overall is $(n)+2 \\cdot (n-1) + 3 \\cdot (n-2) + \\dots + (n) \\cdot 1$ . We can rearrange this as $((n)+(n-1)+ \\dots + 1) + ((n-1)+(n-2)+ \\dots + 1)+ ((n-2)+(n-3)+ \\dots + 1) + \\dots + 1)$\n\\[= \\frac{(n)(n+1)}{2} + \\frac{(n-1)(n)}{2}+ \\dots + 1\\]\nWe can write this in easier terms as $\\sum_{k=0}^{n} \\frac{(k)(k+1)}{2} = \\frac{1}{2} \\cdot \\sum_{k=0}^{n} k^2+k$ \\[=\\frac{1}{2} \\cdot( \\sum_{k=0}^{n} k^2 + \\sum_{k=0}^{n} k)\\] \\[= \\frac{1}{2} \\cdot ( \\frac{(n)(n+1)(2n+1)}{6} + \\frac{(n)(n+1)}{2})\\] \\[= \\frac{1}{2} \\cdot ( \\frac{(n)(n+1)(2n+1)}{6} + \\frac{3n(n+1)}{6}) = \\frac{1}{2} \\cdot \\frac{n(n+1)(2n+4)}{6}\\] \\[= \\frac{n(n+1)(n+2)}{6}\\]\nWe multiply this by $3$ to obtain that \\[\\sum_{i+j \\leq n}^{} i+2j = \\frac{n(n+1)(n+2)}{2}\\]\nThus our final answer for the number of distinct triplets $(h,i,j)$ is: \\[\\sum_{i+j\\leq n}^{} 2n+1-i-2j = \\frac{(2n+1)(n+1)(n+2)}{2} - \\frac{n(n+1)(n+2)}{2}\\] \\[= \\frac{(n+1)(n+2)}{2} \\cdot (2n+1-n) = \\frac{(n+1)(n+2)}{2} \\cdot (n+1)\\] \\[= \\frac{(n+1)^2(n+2)}{2}\\]\nNow most of the work is done. We set this equal to $936$ and prime factorize. $936 = 12 \\cdot 78 = 2^3 \\cdot 3^2 \\cdot 13$ , so $(n+1)^2(n+2) = 936 \\cdot 2 = 2^4 \\cdot 3^2 \\cdot 13$ . Clearly $13$ cannot be anything squared and $2^4 \\cdot 3^2$ is a perfect square, so $n+2 = 13$ and $n = 11 = \\boxed{11}$", "The product can be written as \\begin{align*} 2^a 3^b 4^c 5^d 6^e & = 2^{a + 2c + e} 3^{b + e} 5^d . \\end{align*}\nTherefore, we need to find the number of ordered tuples $\\left( a + 2c + e, b+e, d \\right)$ where $a$ $b$ $c$ $d$ $e$ are non-negative integers satisfying $a+b+c+d+e \\leq n$ .\nWe denote this number as $f(n)$\nDenote by $g \\left( k \\right)$ the number of ordered tuples $\\left( a + 2c + e, b+e \\right)$ where $\\left( a, b, c, e \\right) \\in \\Delta_k$ with $\\Delta_k \\triangleq \\left\\{ (a,b,c,e) \\in \\Bbb Z_+^4: a+b+c+e \\leq k \\right\\}$\nThus, \\begin{align*} f \\left( n \\right) & = \\sum_{d = 0}^n g \\left( n - d \\right) \\\\ & = \\sum_{k = 0}^n g \\left( k \\right) . \\end{align*}\nNext, we compute $g \\left( k \\right)$\nDenote $i = b + e$ . Thus, for each given $i$ , the range of $a + 2c + e$ is from 0 to $2 k - i$ .\nThus, the number of $\\left( a + 2c + e, b + e \\right)$ is \\begin{align*} g \\left( k \\right) & = \\sum_{i=0}^k \\left( 2 k - i + 1 \\right) \\\\ & = \\frac{1}{2} \\left( k + 1 \\right) \\left( 3 k + 2 \\right) . \\end{align*}\nTherefore, \\begin{align*} f \\left( n \\right) & = \\sum_{k = 0}^n g \\left( k \\right) \\\\ & = \\sum_{k=0}^n \\frac{1}{2} \\left( k + 1 \\right) \\left( 3 k + 2 \\right) \\\\ & = \\frac{3}{2} \\sum_{k=0}^n \\left( k + 1 \\right)^2 - \\frac{1}{2} \\sum_{k=0}^n \\left( k + 1 \\right) \\\\ & = \\frac{3}{2} \\cdot \\frac{1}{6} \\left( n+1 \\right) \\left( n+2 \\right) \\left( 2n + 3 \\right) - \\frac{1}{2} \\cdot \\frac{1}{2} \\left( n + 1 \\right) \\left( n + 2 \\right) \\\\ & = \\frac{1}{2} \\left( n + 1 \\right)^2 \\left( n + 2 \\right) . \\end{align*}\nBy solving $f \\left( n \\right) = 936$ , we get $n = \\boxed{11}$", "The product can be written as \\begin{align*} 2^x 3^y 5^z \\end{align*}\nLetting $n=1$ , we get $(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)$ , 6 possible values. But if the only restriction of the product if that $2x\\le n,y\\le n,z\\le n$ , we can get $(2+1)(1+1)(1+1)=12$ possible values. We calculate the ratio \\[r = \\frac{\\text{possible values of real situation}}{\\text{possible values of ideal situation}} = \\frac{6}{12}=0.5.\\]\nLetting $n=2$ , we get $(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),$ $(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)$ , 17 possible values.\nThe number of possibilities in the ideal situation is $5*3*3=45$ , making $r = 17/45 \\approx 0.378$\nNow we can predict the trend of $r$ : as $n$ increases, $r$ decreases.\nLetting $n=3$ , you get possible values of ideal situation= $7*4*4=112$ $n=4$ , the number= $9*5*5=225$ $n=5$ , the number= $11*6*6=396$ $n=6$ , the number= $13*7*7=637,637<936$ so 6 is not the answer. $n=7$ , the number= $15*8*8=960$ $n=8$ , the number= $17*9*9=1377$ ,but $1377*0.378$ $521$ still much smaller than 936. $n=9$ , the number= $19*10*10=1900$ ,but $1900*0.378$ $718$ still smaller than 936. $n=10$ , the number= $21*11*11=2541$ $2541*0.378$ $960$ is a little bigger 936, but the quotient of (possible values of real situation)/(possible values of ideal situation) is much smaller than 0.378 now, so 10 is probably not the answer,so the answer is $\\boxed{11}$" ]

[ "We have $x^5\\left(x+\\frac{1}{x}\\right)\\left(1+\\frac{2}{x}+\\frac{3}{x^2}\\right) = (x^6+x^4)\\left(1+\\frac{2}{x}+\\frac{3}{x^2}\\right) = x^6 + \\text{lower order terms}$ , where we know that the $x^6$ will not get cancelled out by e.g. a $-x^6$ term since all the terms inside the brackets are positive. Thus the degree is $6$ , which is choice $\\boxed{6}$" ]

[ "Obviously, we can factor out an $x$ first to get $x(x^8-1)$ .\nNext, we repeatedly factor differences of squares: \\[x(x^4+1)(x^4-1)\\] \\[x(x^4+1)(x^2+1)(x^2-1)\\] \\[x(x^4+1)(x^2+1)(x+1)(x-1)\\] None of these 5 factors can be factored further, so the answer is $\\boxed{5}$" ]

[ "Using synthetic division, we get that the remainder is $\\boxed{2}$", "By the remainder theorem, the remainder is equal to the expression $x^{13}+1$ when $x=1.$ This gives the answer of $\\boxed{2.}$", "Note that $x^{13} - 1 = (x - 1)(x^{12} + x^{11} \\cdots + 1)$ , so $x^{13} - 1$ is divisible by $x-1$ , meaning $(x^{13} - 1) + 2$ leaves a remainder of $\\boxed{2.}$" ]

[ "When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes $= 3\\times60 + 30$ minutes $= 210$ minutes, thus running $\\frac{210}{15} = 14$ minutes per mile. Now that he is an old man, he can walk $10$ miles in $4$ hours $= 4 \\times 60$ minutes $= 240$ minutes, thus walking $\\frac{240}{10} = 24$ minutes per mile. Therefore, it takes him $\\boxed{10}$ minutes longer to walk a mile now compared to when he was a boy." ]

[ "In total, there were $3+4=7$ marbles left from both Ringo and Paul.We know that $7 \\equiv 1 \\pmod{6}$ . This means that there would be $1$ marble leftover, or $\\boxed{1}$", "Let $r$ be the number of marbles Ringo has and let $p$ be the number of marbles Paul has. we have the following equations: \\[r \\equiv 4 \\mod{6}\\] \\[p \\equiv 3 \\mod{6}\\] Adding these equations we get: \\[p + r \\equiv 7 \\mod{6}\\] We know that $7 \\equiv 1 \\mod{6}$ so therefore: \\[p + r \\equiv 7 \\equiv 1 \\mod{6} \\implies p + r \\equiv 1 \\mod{6}\\] Thus when Ringo and Paul pool their marbles, they will have $\\boxed{1}$ marble left over." ]

[ "The tank started at $\\frac{1}{8}$ full, and ended at $\\frac{5}{8}$ full. Therefore, Walter filled $\\frac{5}{8} - \\frac{1}{8} = \\frac{4}{8} = \\frac{1}{2}$ of the tank.\nIf Walter fills half the tank with $7.5$ gallons, then Walter can fill two halves of the tank (or a whole tank) with $7.5 \\times 2 = 15$ gallons, giving an answer of $\\boxed{15}$" ]

[ "The sum of the digits from $1-9$ are $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$ . Note that one of the answer choices (that is equal to Yunju's digits) subtracted from her sum of $45$ must equal a square.\nNote that $6^2 = 36$ is a very close square to the sum of 45. Checking, we see that $45 - 9 = 36 = 6^2$ works.\nTherefore, the missing number is $\\boxed{9}$" ]

[ "Denote the probability of getting a heads in one flip of the biased coin as $h$ . Based upon the problem, note that ${5\\choose1}(h)^1(1-h)^4 = {5\\choose2}(h)^2(1-h)^3$ . After canceling out terms, we get $1 - h = 2h$ , so $h = \\frac{1}{3}$ . The answer we are looking for is ${5\\choose3}(h)^3(1-h)^2 = 10\\left(\\frac{1}{3}\\right)^3\\left(\\frac{2}{3}\\right)^2 = \\frac{40}{243}$ , so $i+j=40+243=\\boxed{283}$", "Denote the probability of getting a heads in one flip of the biased coins as $h$ and the probability of getting a tails as $t$ . Based upon the problem, note that ${5\\choose1}(h)^1(t)^4 = {5\\choose2}(h)^2(t)^3$ . After cancelling out terms, we end up with $t = 2h$ . To find the probability getting $3$ heads, we need to find ${5\\choose3}\\dfrac{(h)^3(t)^2}{(h + t)^5} =10\\cdot\\dfrac{(h)^3(2h)^2}{(h + 2h)^5}$ (recall that $h$ cannot be $0$ ). The result after simplifying is $\\frac{40}{243}$ , so $i + j = 40 + 243 = \\boxed{283}$" ]

[ "We start with final output of $1$ and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities each stage: \\[\\{1\\}\\rightarrow\\{2\\}\\rightarrow\\{4\\}\\rightarrow\\{1,8\\}\\rightarrow\\{2,16\\}\\rightarrow\\{4,5,32\\}\\rightarrow\\{1,8,10,64\\}\\] where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$ ), but $16$ could come from $32$ or $5$ (as $\\frac{32}{2} = 3 \\cdot 5 + 1 = 16$ , and $32$ is even while $5$ is odd). By construction, last set in this sequence contains all the numbers which will lead to number $1$ to end of the $6$ -step process, and sum is $1+8+10+64=\\boxed{83}$", "As in Solution 1, we work backwards from $1$ , this time showing the possible cases in a tree diagram:\n\nThe possible numbers are those at the \"leaves\" of the tree (the ends of the various branches), which are $1$ $8$ $64$ , and $10$ . Thus the answer is $1+8+64+10=\\boxed{83}$", "We begin by finding the inverse of the function that the machine uses. Call the input $I$ and the output $O$ . If $I$ is even, $O=\\frac{I}{2}$ , and if $I$ is odd, $O=3I+1$ . We can therefore see that $I=2O$ when $I$ is even and $I=\\frac{O-1}{3}$ when $I$ is odd. Therefore, starting with $1$ , if $I$ is even, $I=2$ , and if $I$ is odd, $I=0$ , but the latter is not valid since $0$ is not actually odd. This means that the 2nd-to-last number in the sequence has to be $2$ . Now, substituting $2$ into the inverse formulae, if $I$ is even, $I=4$ (which is indeed even), and if $I$ is odd, $I=\\frac{1}{3}$ , which is not an integer. This means the 3rd-to-last number in the sequence has to be $4$ . Substituting in $4$ , if $I$ is even, $I=8$ , but if $I$ is odd, $I=1$ . Both of these are valid solutions, so the 4th-to-last number can be either $1$ or $8$ . If it is $1$ , then by the argument we have just made, the 5th-to-last number has to be $2$ , the 6th-to-last number has to be $4$ , and the 7th-to-last number, which is the first number, must be either $1$ or $8$ . In this way, we have ultimately found two solutions: $N=1$ and $N=8$\nOn the other hand, if the 4th-to-last number is $8$ , substituting $8$ into the inverse formulae shows that the 5th-to-last number is either $16$ or $\\frac{7}{3}$ , but the latter is not an integer. Substituting $16$ shows that if $I$ is even, $I=32$ , but if I is odd, $I=5$ , and both of these are valid. If the 6th-to-last number is $32$ , then the first number must be $64$ , since $\\frac{31}{3}$ is not an integer; if the 6th-to-last number is $5,$ then the first number has to be $10$ , as $\\frac{4}{3}$ is not an integer. This means that, in total, there are $4$ solutions for $N$ , specifically, $1$ $8$ $10$ , and $64$ , which sum to $\\boxed{83}$" ]

[ "Let one leg of the triangle have length $a$ and let the other leg have length $b$ . When we rotate around the leg of length $a$ , the result is a cone of height $a$ and radius $b$ , and so of volume $\\frac 13 \\pi ab^2 = 800\\pi$ . Likewise, when we rotate around the leg of length $b$ we get a cone of height $b$ and radius $a$ and so of volume $\\frac13 \\pi b a^2 = 1920 \\pi$ . If we divide this equation by the previous one, we get $\\frac ab = \\frac{\\frac13 \\pi b a^2}{\\frac 13 \\pi ab^2} = \\frac{1920}{800} = \\frac{12}{5}$ , so $a = \\frac{12}{5}b$ . Then $\\frac{1}{3} \\pi \\left(\\frac{12}{5}b\\right)b^2 = 800\\pi$ so $b^3 = 1000$ and $b = 10$ so $a = 24$ . Then by the Pythagorean Theorem , the hypotenuse has length $\\sqrt{a^2 + b^2} = \\boxed{026}$", "Let $a$ $b$ be the $2$ legs, we have the $2$ equations \\[\\frac{a^2b\\pi}{3}=800\\pi,\\frac{ab^2\\pi}{3}=1920\\pi\\] Thus $a^2b=2400, ab^2=5760$ . Multiplying gets \\begin{align*} (a^2b)(ab^2)&=2400\\cdot5760\\\\ (ab)^3&=(2^5\\cdot3\\cdot5^2)(2^7\\cdot3^2\\cdot5)\\\\ ab&=\\sqrt[3]{2^{12}\\cdot3^3\\cdot5^3}=240 \\end{align*} Adding gets \\begin{align*} a^2b+ab^2=ab(a+b)&=2400+5760\\\\ 240(a+b)&=240\\cdot(10+24)\\\\ a+b&=34 \\end{align*} Let $h$ be the hypotenuse then \\begin{align*} h&=\\sqrt{a^2+b^2}\\\\ &=\\sqrt{(a+b)^2-2ab}\\\\ &=\\sqrt{34^2-2\\cdot240}\\\\ &=\\sqrt{676}\\\\ &=\\boxed{26}", "Let $a$ and $b$ be the two legs of the equation. We can find $\\frac{a}{b}$ by doing $\\frac{1920\\pi}{800\\pi}$ . This simplified is $\\frac{12}{5}$ . We can represent the two legs as $12x$ and $5x$ for $a$ and $b$ respectively.\nSince the volume of the first cone is $800\\pi$ , we use the formula for the volume of a cone and get $100\\pi x^3=800 \\pi$ . Solving for $x$ , we get $x=2$\nPlugging in the side lengths to the Pythagorean Theorem, we get an answer of $\\boxed{026}$" ]

[ "We are given that $66\\Bigl(\\underline{1}.\\overline{\\underline{a} \\ \\underline{b}}\\Bigr)-0.5=66\\Bigl(\\underline{1}.\\underline{a} \\ \\underline{b}\\Bigr),$ from which \\begin{align*} 66\\Bigl(\\underline{1}.\\overline{\\underline{a} \\ \\underline{b}}\\Bigr)-66\\Bigl(\\underline{1}.\\underline{a} \\ \\underline{b}\\Bigr)&=0.5 \\\\ 66\\Bigl(\\underline{1}.\\overline{\\underline{a} \\ \\underline{b}} - \\underline{1}.\\underline{a} \\ \\underline{b}\\Bigr)&=0.5 \\\\ 66\\Bigl(\\underline{0}.\\underline{0} \\ \\underline{0} \\ \\overline{\\underline{a} \\ \\underline{b}}\\Bigr)&=0.5 \\\\ 66\\left(\\frac{1}{100}\\cdot\\underline{0}.\\overline{\\underline{a} \\ \\underline{b}}\\right)&=\\frac12 \\\\ \\underline{0}.\\overline{\\underline{a} \\ \\underline{b}}&=\\frac{25}{33} \\\\ \\underline{0}.\\overline{\\underline{a} \\ \\underline{b}}&=0.\\overline{75} \\\\ \\underline{a} \\ \\underline{b}&=\\boxed{75} ~MRENTHUSIASM", "It is known that $\\underline{0}.\\overline{\\underline{a} \\ \\underline{b}}=\\frac{\\underline{a} \\ \\underline{b}}{99}$ and $\\underline{0}.\\underline{a} \\ \\underline{b}=\\frac{\\underline{a} \\ \\underline{b}}{100}.$\nLet $x=\\underline{a} \\ \\underline{b}.$ We have \\[66\\biggl(1+\\frac{x}{99}\\biggr)-66\\biggl(1+\\frac{x}{100}\\biggr)=0.5.\\] Expanding and simplifying give $\\frac{x}{150}=0.5,$ so $x=\\boxed{75}.$", "We have \\[66 \\cdot \\left(1 + \\frac{10a+b}{100}\\right) + \\frac{1}{2} = 66 \\cdot \\left(1+ \\frac{10a+b}{99}\\right).\\] Expanding both sides, we have \\[66 + \\frac{33(10a+b)}{50} + \\frac{1}{2} = 66 + \\frac{2(10a+b)}{3}.\\] Subtracting $66$ from both sides, we have \\[\\frac{33(10a+b)}{50} + \\frac{1}{2} = \\frac{2(10a+b)}{3}.\\] Multiplying both sides by $50 \\cdot 3 = 150,$ we have \\[99(10a+b) + 75 = 100(10a+b).\\] Thus, the answer is $10a+b = \\boxed{75}.$" ]

[ "We are given that $66\\Bigl(\\underline{1}.\\overline{\\underline{a} \\ \\underline{b}}\\Bigr)-0.5=66\\Bigl(\\underline{1}.\\underline{a} \\ \\underline{b}\\Bigr),$ from which \\begin{align*} 66\\Bigl(\\underline{1}.\\overline{\\underline{a} \\ \\underline{b}}\\Bigr)-66\\Bigl(\\underline{1}.\\underline{a} \\ \\underline{b}\\Bigr)&=0.5 \\\\ 66\\Bigl(\\underline{1}.\\overline{\\underline{a} \\ \\underline{b}} - \\underline{1}.\\underline{a} \\ \\underline{b}\\Bigr)&=0.5 \\\\ 66\\Bigl(\\underline{0}.\\underline{0} \\ \\underline{0} \\ \\overline{\\underline{a} \\ \\underline{b}}\\Bigr)&=0.5 \\\\ 66\\left(\\frac{1}{100}\\cdot\\underline{0}.\\overline{\\underline{a} \\ \\underline{b}}\\right)&=\\frac12 \\\\ \\underline{0}.\\overline{\\underline{a} \\ \\underline{b}}&=\\frac{25}{33} \\\\ \\underline{0}.\\overline{\\underline{a} \\ \\underline{b}}&=0.\\overline{75} \\\\ \\underline{a} \\ \\underline{b}&=\\boxed{75} ~MRENTHUSIASM", "It is known that $\\underline{0}.\\overline{\\underline{a} \\ \\underline{b}}=\\frac{\\underline{a} \\ \\underline{b}}{99}$ and $\\underline{0}.\\underline{a} \\ \\underline{b}=\\frac{\\underline{a} \\ \\underline{b}}{100}.$\nLet $x=\\underline{a} \\ \\underline{b}.$ We have \\[66\\biggl(1+\\frac{x}{99}\\biggr)-66\\biggl(1+\\frac{x}{100}\\biggr)=0.5.\\] Expanding and simplifying give $\\frac{x}{150}=0.5,$ so $x=\\boxed{75}.$", "We have \\[66 \\cdot \\left(1 + \\frac{10a+b}{100}\\right) + \\frac{1}{2} = 66 \\cdot \\left(1+ \\frac{10a+b}{99}\\right).\\] Expanding both sides, we have \\[66 + \\frac{33(10a+b)}{50} + \\frac{1}{2} = 66 + \\frac{2(10a+b)}{3}.\\] Subtracting $66$ from both sides, we have \\[\\frac{33(10a+b)}{50} + \\frac{1}{2} = \\frac{2(10a+b)}{3}.\\] Multiplying both sides by $50 \\cdot 3 = 150,$ we have \\[99(10a+b) + 75 = 100(10a+b).\\] Thus, the answer is $10a+b = \\boxed{75}.$" ]

[ "Note that $n$ is equal to the number of integers between $53$ and $201$ , inclusive. Thus, $n=201-53+1=\\boxed{149}$" ]

[ "Note that $n$ is equal to the number of integers between $53$ and $201$ , inclusive. Thus, $n=201-53+1=\\boxed{149}$" ]

[ "Let's work on both parts of the problem separately. First, \\[855 \\equiv 787 \\equiv 702 \\equiv r \\pmod{m}.\\] We take the difference of $855$ and $787$ , and also of $787$ and $702$ . We find that they are $85$ and $68$ , respectively. Since the greatest common divisor of the two differences is $17$ (and the only one besides one), it's safe to assume that $m = 17$\nThen, we divide $855$ by $17$ , and it's easy to see that $r = 5$ . Dividing $787$ and $702$ by $17$ also yields remainders of $5$ , which means our work up to here is correct.\nDoing the same thing with $815$ $722$ , and $412$ , the differences between $815$ and $722$ and $412$ are $310$ and $93$ , respectively. Since the only common divisor (besides $1$ , of course) is $31$ $n = 31$ . Dividing all $3$ numbers by $31$ yields a remainder of $9$ for each, so $s = 9$ . Thus, $m + n + r + s = 17 + 5 + 31 + 9 = \\boxed{062}$", "We know that $702 = am + r, 787 = bm + r,$ and $855 = cm+r$ where $a-c$ are integers.\nSubtracting the first two, the first and third, and the last two, we get $85 = (b-a)m, 153=(c-a)m,$ and $68=(c-b)m.$\nWe know that $b-a, c-a$ and $c-b$ must be integers, so all the numbers are divisible by $m.$\nFactorizing the numbers, we get $85 = 5 \\cdot 17, 153 = 3^2 \\cdot 17,$ and $68 = 2^2 \\cdot 17.$ We see that all these have a factor of 17, so $m=17.$\nFinding the remainder when $702$ is divided by $17,$ we get $n=5.$\nDoing the same thing for the next three numbers, we get $17 + 5 + 31 + 9 = \\boxed{062}$", "As in Solution 1, we are given $855\\equiv787\\equiv702\\equiv r\\pmod{m}$ and $815\\equiv722\\equiv412\\equiv s\\pmod{n}.$ Tackling the first equation, we can simply look at $855\\equiv787\\equiv702\\pmod{m}$ . We subtract $702$ from each component of the congruency to get $153\\equiv85\\equiv0\\pmod{m}$ . Thus, we know that $153$ and $85$ must both be divisible by $m$ . The only possible $m$ , in this case, become $17$ and $1$ ; obviously, $m\\neq1$ , so we know $m=17$ . We go back to the original equation, plug in $m$ , and we find that $r=5$\nSimilarly, we can subtract out the smallest value in the second congruency, $412$ . We end up with $403\\equiv310\\equiv0\\pmod{n}$ . Again, we find that $n=31$ or $1$ , so $n=31$ . We also find that $s=9$\nThus, our answer is $\\boxed{062}$" ]

[ "We want the least common multiple of $2,3,4,5,6,7$ , which is $420$ , or choice $\\boxed{420}$" ]

[ "Four gallons is $\\frac{1}{2}-\\frac{1}{3}= \\frac{1}{6}$ . Thus, capacity of the tank in gallons is $6 \\times 4= \\boxed{24}$" ]

[ "Let $a$ be the original number of ounces of acid and $w$ be the original number of ounces of water. We can write two equations since we know the percentage of acid after some water and acid. \\[\\frac{a}{a+w+1} = \\frac{1}{5}\\] \\[\\frac{a+1}{a+w+2} = \\frac{1}{3}\\] Cross-multiply to get rid of the fractions. \\[5a = a+w+1\\] \\[3a+3=a+w+1\\] Solve the system to get $a=1$ and $w=3$ . The percentage of acid in the original mixture is $\\tfrac{1}{1+3} = \\boxed{25}$" ]

[ "Without loss of generality , assume that face $F$ has a 6, so the opposite face has a 1. Let $A(n)$ be the probability of rolling a number $n$ on one die and let $B(n)$ be the probability of rolling a number $n$ on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of $\\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{1}{36}$ , totaling $4 \\cdot \\frac{1}{36} = \\frac{1}{9}$ . Subtracting all these probabilities from $\\frac{47}{288}$ leaves $\\frac{15}{288}=\\frac{5}{96}$ chance of getting a 1 on die $A$ and a 6 on die $B$ or a 6 on die $A$ and a 1 on die $B$\n\\[A(6)\\cdot B(1)+B(6)\\cdot A(1)=\\frac{5}{96}\\]\nSince the two dice are identical, $B(1)=A(1)$ and $B(6)=A(6)$ so\n\\begin{align*}A(1)\\cdot A(6)+A(1)\\cdot A(6)&=\\frac{5}{96}\\\\ A(1)\\cdot A(6)&=\\frac{5}{192}\\end{align*}\nAlso, we know that $A(2)=A(3)=A(4)=A(5)=\\frac{1}{6}$ and that the total probability must be $1$ , so:\n\\[A(1)+4 \\cdot \\frac{1}{6}+A(6)=\\frac{6}{6} \\Longrightarrow A(1)+A(6)=\\frac{1}{3}\\]\nCombining the equations:\n\\begin{align*}A(6)\\left(\\frac{1}{3}-A(6)\\right)&=\\frac{5}{192}\\\\ 0 &= 192 \\left(A(6)\\right)^2 - 64 \\left(A(6)\\right) + 5\\\\ A(6)&=\\frac{64\\pm\\sqrt{64^2 - 4 \\cdot 5 \\cdot 192}}{2\\cdot192} =\\frac{5}{24}, \\frac{1}{8}\\end{align*} We know that $A(6)>\\frac{1}{6}$ , so it can't be $\\frac{1}{8}$ . Therefore, the probability is $\\frac{5}{24}$ and the answer is $5+24=\\boxed{29}$", "We have that the cube probabilities to land on its faces are $\\frac{1}{6}$ $\\frac{1}{6}$ $\\frac{1}{6}$ $\\frac{1}{6}$ $\\frac{1}{6}+x$ $\\frac{1}{6}-x$ we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: \\[4 \\cdot \\left(\\frac{1}{6} \\right)^2+2 \\left(\\frac{1}{6}+x \\right) \\left(\\frac{1}{6}-x \\right)=\\frac{47}{288}\\] multiplying by 288 we get: \\[32+16(1-6x)(6x+1)=47 \\Longrightarrow 16(1-36x^2)=15\\] dividing by 16 and rearranging we get: \\[\\frac{1}{16}=36x^2 \\longrightarrow x=\\frac{1}{24}\\] so the probability F which is greater than $\\frac{1}{6}$ is equal $\\frac{1}{6}+\\frac{1}{24}=\\frac{5}{24}\\longrightarrow 24+5=\\boxed{29}$", "Let $p(a,b)$ denote the probability of obtaining $a$ on the first die and $b$ on the second. Then the probability of obtaining a sum of 7 is \\[p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).\\] Let the probability of obtaining face $F$ be $(1/6)+x$ . Then the probability of obtaining the face opposite face $F$ is $(1/6)-x$ . Therefore\n\\begin{align*} {{47}\\over{288}}&= 4\\left({1\\over6}\\right)^2+2\\left({1\\over6}+x\\right) \\left({1\\over6}-x\\right)\\cr&= {4\\over36}+2\\left({1\\over36}-x^2\\right)\\cr&= {1\\over6}-2x^2. \\end{align*}\nThen $2x^2=1/288$ , and so $x=1/24$ . The probability of obtaining face $F$ is therefore $(1/6)+(1/24)=5/24$ , and $m+n=\\boxed{29}$" ]

[ "Notice that the number of kilobits in this song is $4.2 \\cdot 8000 = 8 \\cdot 7 \\cdot 6 \\cdot 100.$\nWe must divide this by $56$ in order to find out how many seconds this song would take to download: $\\frac{\\cancel{8}\\cdot\\cancel{7}\\cdot6\\cdot100}{\\cancel{56}} = 600.$\nFinally, we divide this number by $60$ because this is the number of seconds to get the answer $\\frac{600}{60}=\\boxed{10}.$", "We seek a value of $x$ that makes the following equation true, since every other quantity equals $1$\n\\[\\frac{x\\ \\text{min}}{4.2\\ \\text{mb}} \\cdot \\frac{56\\ \\text{kb}}{1\\ \\text{sec}} \\cdot \\frac{1\\ \\text{mb}}{8000\\ \\text{kb}} \\cdot \\frac{60\\ \\text{sec}}{1\\ \\text{min}} = 1.\\] Solving yields $x=\\boxed{10}$" ]

[ "If $x$ is the number, then moving the decimal point four places to the right is the same as multiplying $x$ by $10000$ . This gives us: \\[10000x=4\\cdot\\frac{1}{x} \\implies x^2=\\frac{4}{10000}\\] Since \\[x>0\\implies x=\\frac{2}{100}=\\boxed{0.02}\\]", "Alternatively, we could try each solution and see if it fits the problems given statements.\nAfter testing, we find that $\\boxed{0.02}$ is the correct answer." ]

[ "\\begin{align*} \\dfrac{49}{84} &= \\dfrac{7^2}{2^2\\cdot 3\\cdot 7} \\\\ &= \\dfrac{7}{2^2\\cdot 3} \\\\ &= \\dfrac{7}{12}. \\end{align*}\nThe sum of the numerator and denominator is $7+12=19\\rightarrow \\boxed{19}$" ]

[ "By Fermat's Little Theorem , we know that $2^{100} \\equiv 2^{1000 \\pmod{12}}\\pmod{13}$ . However, we find that $1000 \\equiv 4 \\pmod{12}$ , so $2^{1000} \\equiv 2^4 = 16 \\equiv 3 \\pmod{13}$ , so the answer is $\\boxed{3}$" ]

[ "We let $x=0.\\overline{36}$ . Thus, $100x=36.\\overline{36}$ . We find that $100x-x=99x=36.\\overline{36}-0.\\overline{36}=36$ , or $x=\\frac{36}{99}=\\frac{4}{11}$ . Since $4+11=15$ , the answer is $\\boxed{15}$" ]

[ "The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about $(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9$ . We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1 through 2 and 2 through 3 intervals make four of the expressions negative. There are $\\boxed{6}$ intervals.", "The roots of the factorized polynomial are intervals from numbers 1 to 10. We take each interval as being defined as the number behind it. To make the function positive, we need to have an even number of negative expressions. Real numbers raised to even powers are always positive, so we only focus on $x-1$ $x-3$ $x-5$ $x-7$ , and $x-9$ . The intervals 1 and 2 leave 4 negative expressions, so they are counted. The same goes for intervals 5, 6, 9, and 10. Intervals 3 and 4 leave 3 negative expressions and intervals 7 and 8 leave 1 negative expression. The solution is the number of intervals which is $\\boxed{6}$", "We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive.\nFirst, we evaluate any value on the interval $(-\\infty, 1)$ . Since the degree of $P(x)$ is $1+2+...+9+10$ $\\frac{10\\times11}{2}$ $55$ , and every term in $P(x)$ is negative, multiplying $55$ negatives gives a negative value. So $(-\\infty, 0)$ is a negative interval.\nWe know that the roots of $P(x)$ are at $1,2,...,10$ . When the degree of the term of each root is odd, the graph of $P(x)$ will pass through the graph and change signs, and vice versa. So at $x=1$ , the graph will change signs; at $x=2$ , the graph will not, and so on.\nThis tells us that the interval $(1,2)$ is positive, $(2,3)$ is also positive, $(3,4)$ is negative, $(4,5)$ is also negative, and so on, with the pattern being $+,+,-,-,+,+,-,-,...$\nThe positive intervals are therefore $(1,2)$ $(2,3)$ $(5,6)$ $(6,7)$ $(9,10)$ , and $(10,\\infty)$ , for a total of $\\boxed{6}$", "Denote by $I_k$ the interval $\\left( k - 1 , k \\right)$ for $k \\in \\left\\{ 2, 3, \\cdots , 10 \\right\\}$ and $I_1$ the interval $\\left( - \\infty, 1 \\right)$\nTherefore, the number of intervals that $P(x)$ is positive is \\begin{align*} 1 + \\sum_{i=1}^{10} \\Bbb I \\left\\{ \\sum_{j=i}^{10} j \\mbox{ is even} \\right\\} & = 1 + \\sum_{i=1}^{10} \\Bbb I \\left\\{ \\frac{\\left( i + 10 \\right) \\left( 11 - i \\right)}{2} \\mbox{ is even} \\right\\} \\\\ & = 1 + \\sum_{i=1}^{10} \\Bbb I \\left\\{ \\frac{- i^2 + i + 110}{2} \\mbox{ is even} \\right\\} \\\\ & = 1 + \\sum_{i=1}^{10} \\Bbb I \\left\\{ \\frac{i^2 - i}{2} \\mbox{ is odd} \\right\\} \\\\ & = \\boxed{6}" ]

[ "The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about $(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9$ . We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1 through 2 and 2 through 3 intervals make four of the expressions negative. There are $\\boxed{6}$ intervals.", "The roots of the factorized polynomial are intervals from numbers 1 to 10. We take each interval as being defined as the number behind it. To make the function positive, we need to have an even number of negative expressions. Real numbers raised to even powers are always positive, so we only focus on $x-1$ $x-3$ $x-5$ $x-7$ , and $x-9$ . The intervals 1 and 2 leave 4 negative expressions, so they are counted. The same goes for intervals 5, 6, 9, and 10. Intervals 3 and 4 leave 3 negative expressions and intervals 7 and 8 leave 1 negative expression. The solution is the number of intervals which is $\\boxed{6}$", "We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive.\nFirst, we evaluate any value on the interval $(-\\infty, 1)$ . Since the degree of $P(x)$ is $1+2+...+9+10$ $\\frac{10\\times11}{2}$ $55$ , and every term in $P(x)$ is negative, multiplying $55$ negatives gives a negative value. So $(-\\infty, 0)$ is a negative interval.\nWe know that the roots of $P(x)$ are at $1,2,...,10$ . When the degree of the term of each root is odd, the graph of $P(x)$ will pass through the graph and change signs, and vice versa. So at $x=1$ , the graph will change signs; at $x=2$ , the graph will not, and so on.\nThis tells us that the interval $(1,2)$ is positive, $(2,3)$ is also positive, $(3,4)$ is negative, $(4,5)$ is also negative, and so on, with the pattern being $+,+,-,-,+,+,-,-,...$\nThe positive intervals are therefore $(1,2)$ $(2,3)$ $(5,6)$ $(6,7)$ $(9,10)$ , and $(10,\\infty)$ , for a total of $\\boxed{6}$", "Denote by $I_k$ the interval $\\left( k - 1 , k \\right)$ for $k \\in \\left\\{ 2, 3, \\cdots , 10 \\right\\}$ and $I_1$ the interval $\\left( - \\infty, 1 \\right)$\nTherefore, the number of intervals that $P(x)$ is positive is \\begin{align*} 1 + \\sum_{i=1}^{10} \\Bbb I \\left\\{ \\sum_{j=i}^{10} j \\mbox{ is even} \\right\\} & = 1 + \\sum_{i=1}^{10} \\Bbb I \\left\\{ \\frac{\\left( i + 10 \\right) \\left( 11 - i \\right)}{2} \\mbox{ is even} \\right\\} \\\\ & = 1 + \\sum_{i=1}^{10} \\Bbb I \\left\\{ \\frac{- i^2 + i + 110}{2} \\mbox{ is even} \\right\\} \\\\ & = 1 + \\sum_{i=1}^{10} \\Bbb I \\left\\{ \\frac{i^2 - i}{2} \\mbox{ is odd} \\right\\} \\\\ & = \\boxed{6}" ]

[ "First we try for a positive product, meaning we either pick three positive numbers or one positive number and two negative numbers.\nIt is clearly impossible to pick three positive numbers. If we try the second case, we want to pick the numbers with the largest absolute values, so we choose $5$ $-3$ and $-2$ . Their product is $30\\rightarrow \\boxed{30}$" ]

[ "The positive divisors of $100$ are \\[1,2,4,5,10,20,25,50,100.\\] It is clear that $10\\leq c\\leq50,$ so we apply casework to $c:$\nTogether, the numbers $a,b,$ and $c$ can be chosen in $\\boxed{4}$ ways.", "The positive divisors of $100$ are \\[1,2,4,5,10,20,25,50,100.\\] We apply casework to $a$\nIf $a=1$ , then there are $3$ cases:\nIf $a=2$ , then there is only $1$ case:\nIn total, there are $3+1=\\boxed{4}$ ways to choose distinct positive integer values of $a,b,c$" ]

[ "This simplifies to \\[(a^{\\frac{9}{6}/3})^4 \\cdot (a^{\\frac{9}{3}/6})^4 = (a^{\\frac{1}{2}})^4 \\cdot (a^{\\frac{1}{2}})^4 = (a^2)(a^2) = \\boxed{4}.$" ]

[ "When dealing with positive decimals, the leftmost digits affect the change in value more. Thus, to get the largest number, we change the $1$ to a $9 \\rightarrow \\boxed{1}$" ]

[ "The largest numbers in the first, second, third, fourth and fifth columns are $13,7,15,12,9$ respectively. Of these, only $7$ is the smallest in its row $\\rightarrow \\boxed{7}$" ]

[ "We see that $1^2=1$ $2^2=4$ $5^2=25$ , and $4^2=16$ , so already we know that either $\\text{E}$ is the answer or the problem has some issues.\nFor integers, only the units digit affects the units digit of the final result, so we only need to test the squares of the integers from $0$ through $9$ inclusive. Testing shows that $8$ is unachievable, so the answer is $\\boxed{8}$" ]

[ "On the interval $[0, \\pi]$ sine is nonnegative; thus $\\sin(x + y) = \\sin x \\cos y + \\sin y \\cos x \\le \\sin x + \\sin y$ for all $x, y \\in [0, \\pi]$ and equality only occurs when $\\cos x = \\cos y = 1$ , which is cosine's maximum value. The answer is $\\boxed{0}$ . (CantonMathGuy)", "Expanding, \\[\\cos y \\sin x + \\cos x \\sin y \\le \\sin x + \\sin y\\] Let $\\sin x =a \\ge 0$ $\\sin y = b \\ge 0$ . We have that \\[(\\cos y)a+(\\cos x)b \\le a+b\\] Comparing coefficients of $a$ and $b$ gives a clear solution: both $\\cos y$ and $\\cos x$ are less than or equal to one, so the coefficients of $a$ and $b$ on the left are less than on the right. Since $a, b \\ge 0$ , that means that this equality is always satisfied over this interval, or $\\boxed{0}$", "If we plug in $\\pi$ , we can see that $\\sin(x+\\pi) \\le \\sin(x)$ . Note that since $\\sin(x)$ is always nonnegative, $\\sin(x+\\pi)$ is always nonpositive. So, the inequality holds true when $y=\\pi$ . The only interval that contains $\\pi$ in the answer choices is $\\boxed{0}$" ]

[ "Rewrite it as \\[\\cos y \\sin x + \\cos x \\sin y \\le \\sin x + \\sin y\\] then move the terms such that it appears as this \\[\\sin x (\\cos y -1) \\le \\sin y (1- \\cos x)\\] We know that on the interval of $[0, \\pi]$ $0\\le (1-\\cos x)\\le 2$ , and $0\\le \\sin y$ , such that the whole thing on the right hand side is always greater than or equal to $0$ . On the left hand side, $0\\le \\sin x$ and we want $(\\cos y -1)\\le 0$ , which $\\boxed{0}$ satisfies. \n~azure123456" ]

[ "Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get \\[x^2+(x^2-a)^2=a^2 \\implies x^2+x^4-2ax^2=0 \\implies x^2(x^2-(2a-1))=0\\] Since this is a quartic, there are $4$ total roots (counting multiplicity). We see that $x=0$ always has at least one intersection at $(0,-a)$ (and is in fact a double root).\nThe other two intersection points have $x$ coordinates $\\pm\\sqrt{2a-1}$ . We must have $2a-1> 0;$ otherwise we are in the case where the parabola lies entirely above the circle (tangent at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see that $\\boxed{12}$", "Substituting $y = x^2 - a$ gives $x^2 + (x^2 - a)^2 = a^2$ , which simplifies to $x^2 + x^4 - 2x^2a + a^2 = a^2$ . This further simplifies to $x^2(1 + x^2 - 2a) = 0$ . Thus, either $x^2 = 0$ , or $x^2 - 2a + 1 = 0$\nSince we care about $a$ , we consider the second case. We solve in terms of $a$ , giving $a = \\frac{x^2}{2} + \\frac{1}{2}$ . We see that in order to find the range in which $a$ lies, we must find the vertex of this equation, which turns out to be $\\left(0, \\frac{1}{2}\\right)$ . Hence, we know that the minimum is $\\frac{1}{2}$ , which further implies that $\\boxed{12}$", "\nLooking at a graph, it is obvious that the two curves intersect at $(0, -a)$ . We also see that if the parabola goes \"in\" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.\nWe have $x^2 - a = -\\sqrt{a^2 - x^2}$ . Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$ . Since $x = 0$ is already accounted for, we only need to find one solution for $x^2 = 2a - 1$ , where the right hand side portion is obviously increasing. Since $a = \\frac{1}{2}$ begets $x = 0$ (an overcount), we have $\\boxed{12}$ as the right answer.", "This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is $2$ , the radius of the circle that matches it has a radius of $\\frac{1}{2}$ . This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only $1$ point. When a larger circle is used, it is tangent to $3$ points because the points on either side are now separated from the vertex. Therefore, $\\boxed{12}$ is correct.", "Now, let's graph these two equations. We want the blue parabola to be inside this red circle. Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$ . Expanding, we get $x^4-2ax^2+x^2=0$ . Factoring out the $x^2$ , we get $x^2(x^2-2a+1)=0$ . Then we find that $x=0$ or $x=\\pm\\sqrt{2a-1}$ . Therefore, $2a-1>0$ , which means $\\boxed{12}$", "In order to solve for the values of $a$ , we need to just count multiplicities of the roots when the equations are set equal to each other: in other words, take the derivative. We know that $\\sqrt{a^2 - x^2} = x^2 - a$ . Now, we take square of both sides, and rearrange to obtain $x^4 - (2a - 1)x^2 = 0$ . Now, we may take the second derivative of the equation to obtain $6x^2 - (2a - 1) = 0$ . Now, we must take discriminant. Since we need the roots of that equation to be real and not repetitive (otherwise they would not intersect each other at three points), the discriminant must be greater than zero. Thus,\n\\[b^2 - 4ac > 0 \\rightarrow 0 - 4(6)(-(2a - 1)) > 0 \\rightarrow a > \\frac{1}{2}\\] The answer is $\\boxed{12}$ and we are done.", "Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$ , and they have symmetry across the $y$ -axis, thus, for them to intersect at exactly $3$ points, it suffices to find the $y$ solution.\nFirst, rewrite the second equation to $y=x^2-a\\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$ . This value must not be less than $0$ (It is the square root part of the quadratic formula). To find when it is $0$ , we find the roots: \\[4a^2-4a+1=0 \\implies a=\\frac{4\\pm\\sqrt{16-16}}{8}=\\frac{1}{2}\\] Since $\\lim_{a\\to \\infty}(4a^2-4a+1)=\\infty$ , our range is $\\boxed{12}$", "We can see that if $a = 1$ , we know that the points where the two curves intersect are $(0, -1), (1, 0)$ and $(-1, 0)$ .Because there are only $3$ intersections and $a > 1/2$ , as well as $a > 1/4$ we know that either $\\textbf{(D)}$ or $\\textbf{(E)}$ is the correct answer. Then we can test a number from $(1/2, 1/4)$ to eliminate the remaining answer. So $\\boxed{12}$ is the correct answer.", "Simply plug in $a = \\frac{1}{2}, \\frac{1}{4}, 1$ and solve the systems. (This shouldn't take too long.) And then realize that only $a=1$ yields three real solutions for $x$ , so we are done and the answer is $\\boxed{12}$", "An ideal solution come to mind is where they intersect at the $x$ -axis at the same time, which is $(a, 0)$ and $(-a, 0).$ Take the root of our $y=x^2-a$ we get $x=\\sqrt{a}$ , set them equal we get $a=\\sqrt{a}.$ The only answer is $1$ so it only left us with the answer choice $\\boxed{12}$" ]

[ "Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get \\[x^2+(x^2-a)^2=a^2 \\implies x^2+x^4-2ax^2=0 \\implies x^2(x^2-(2a-1))=0\\] Since this is a quartic, there are $4$ total roots (counting multiplicity). We see that $x=0$ always has at least one intersection at $(0,-a)$ (and is in fact a double root).\nThe other two intersection points have $x$ coordinates $\\pm\\sqrt{2a-1}$ . We must have $2a-1> 0;$ otherwise we are in the case where the parabola lies entirely above the circle (tangent at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see that $\\boxed{12}$", "Substituting $y = x^2 - a$ gives $x^2 + (x^2 - a)^2 = a^2$ , which simplifies to $x^2 + x^4 - 2x^2a + a^2 = a^2$ . This further simplifies to $x^2(1 + x^2 - 2a) = 0$ . Thus, either $x^2 = 0$ , or $x^2 - 2a + 1 = 0$\nSince we care about $a$ , we consider the second case. We solve in terms of $a$ , giving $a = \\frac{x^2}{2} + \\frac{1}{2}$ . We see that in order to find the range in which $a$ lies, we must find the vertex of this equation, which turns out to be $\\left(0, \\frac{1}{2}\\right)$ . Hence, we know that the minimum is $\\frac{1}{2}$ , which further implies that $\\boxed{12}$", "\nLooking at a graph, it is obvious that the two curves intersect at $(0, -a)$ . We also see that if the parabola goes \"in\" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.\nWe have $x^2 - a = -\\sqrt{a^2 - x^2}$ . Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$ . Since $x = 0$ is already accounted for, we only need to find one solution for $x^2 = 2a - 1$ , where the right hand side portion is obviously increasing. Since $a = \\frac{1}{2}$ begets $x = 0$ (an overcount), we have $\\boxed{12}$ as the right answer.", "This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is $2$ , the radius of the circle that matches it has a radius of $\\frac{1}{2}$ . This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only $1$ point. When a larger circle is used, it is tangent to $3$ points because the points on either side are now separated from the vertex. Therefore, $\\boxed{12}$ is correct.", "Now, let's graph these two equations. We want the blue parabola to be inside this red circle. Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$ . Expanding, we get $x^4-2ax^2+x^2=0$ . Factoring out the $x^2$ , we get $x^2(x^2-2a+1)=0$ . Then we find that $x=0$ or $x=\\pm\\sqrt{2a-1}$ . Therefore, $2a-1>0$ , which means $\\boxed{12}$", "In order to solve for the values of $a$ , we need to just count multiplicities of the roots when the equations are set equal to each other: in other words, take the derivative. We know that $\\sqrt{a^2 - x^2} = x^2 - a$ . Now, we take square of both sides, and rearrange to obtain $x^4 - (2a - 1)x^2 = 0$ . Now, we may take the second derivative of the equation to obtain $6x^2 - (2a - 1) = 0$ . Now, we must take discriminant. Since we need the roots of that equation to be real and not repetitive (otherwise they would not intersect each other at three points), the discriminant must be greater than zero. Thus,\n\\[b^2 - 4ac > 0 \\rightarrow 0 - 4(6)(-(2a - 1)) > 0 \\rightarrow a > \\frac{1}{2}\\] The answer is $\\boxed{12}$ and we are done.", "Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$ , and they have symmetry across the $y$ -axis, thus, for them to intersect at exactly $3$ points, it suffices to find the $y$ solution.\nFirst, rewrite the second equation to $y=x^2-a\\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$ . This value must not be less than $0$ (It is the square root part of the quadratic formula). To find when it is $0$ , we find the roots: \\[4a^2-4a+1=0 \\implies a=\\frac{4\\pm\\sqrt{16-16}}{8}=\\frac{1}{2}\\] Since $\\lim_{a\\to \\infty}(4a^2-4a+1)=\\infty$ , our range is $\\boxed{12}$", "We can see that if $a = 1$ , we know that the points where the two curves intersect are $(0, -1), (1, 0)$ and $(-1, 0)$ .Because there are only $3$ intersections and $a > 1/2$ , as well as $a > 1/4$ we know that either $\\textbf{(D)}$ or $\\textbf{(E)}$ is the correct answer. Then we can test a number from $(1/2, 1/4)$ to eliminate the remaining answer. So $\\boxed{12}$ is the correct answer.", "Simply plug in $a = \\frac{1}{2}, \\frac{1}{4}, 1$ and solve the systems. (This shouldn't take too long.) And then realize that only $a=1$ yields three real solutions for $x$ , so we are done and the answer is $\\boxed{12}$", "An ideal solution come to mind is where they intersect at the $x$ -axis at the same time, which is $(a, 0)$ and $(-a, 0).$ Take the root of our $y=x^2-a$ we get $x=\\sqrt{a}$ , set them equal we get $a=\\sqrt{a}.$ The only answer is $1$ so it only left us with the answer choice $\\boxed{12}$" ]

[ "Let our $4$ numbers be $n, n+2, n+4, n+6$ , where $n$ is odd. Then, our sum is $4n+12$ . The only answer choice that cannot be written as $4n+12$ , where $n$ is odd, is $\\boxed{100}$", "If the four consecutive odd integers are $2n-3,~ 2n-1, ~2n+1$ and $2n+3$ ; then, the sum is $8n$ . All the integers are divisible by $8$ except $\\boxed{100}$", "If the four consecutive odd integers are $a,~ a+2, ~a+4$ and $a+6$ , the sum is $4a+12$ , and $4a+12$ divided by $4$ gives $a+3$ . This means that $a+3$ must be even. The only integer that does not give an even integer when divided by $4$ is $100$ , so the answer is $\\boxed{100}$", "From Solution 1, we have the sum of the $4$ numbers to be equal to $4n + 12$ . Taking mod 8 gives us $4n + 4 \\equiv b \\pmod8$ for some residue $b$ and for some odd integer $n$ . Since $n \\equiv 1 \\pmod{2}$ , we can express it as the equation $n = 2a + 1$ for some integer $a$ . Multiplying 4 to each side of the equation yields $4n = 8a + 4$ , and taking mod 8 gets us $4n \\equiv 4 \\pmod{8}$ , so $b = 0$ . All the answer choices except choice D is a multiple of 8, and since 100 satisfies all the conditions the answer is $\\boxed{100}$", "Since they want CONSECUTIVE odd numbers, it won't be hard to just list the sums out: \\[16=1+3+5+7\\] \\[40=7+9+11+13\\] \\[72=15+17+19+21\\] \\[200=47+49+51+53\\] All of the answer choices can be a sum of consecutive odd numbers except $100$ , so the answer is $\\boxed{100}$" ]

[ "We have $\\sqrt{65} > 8 > 7.5$ . Also $7.5^2 = (7 + 0.5)^2 = 7^2 + 2 \\cdot 7 \\cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 < 63$ , so $\\sqrt{63} > 7.5$ . Thus $\\sqrt{65} + \\sqrt{63} > 7.5 + 7.5 = 15$ . Now notice that $\\sqrt{65} - \\sqrt{63} = \\frac{(\\sqrt{65} - \\sqrt{63})(\\sqrt{65} + \\sqrt{63})}{\\sqrt{65} + \\sqrt{63}} = \\frac{2}{\\sqrt{65} + \\sqrt{63}}$ , so $\\sqrt{65} - \\sqrt{63} < \\frac{2}{15} = 0.1333333...$ , so the answer must be $A$ or $B$ . To determine which, we write $\\sqrt{65} - \\sqrt{63} > 0.125 \\iff 65 - 2\\sqrt{65 \\cdot 63} + 63 > 0.015625 \\iff 128 - 0.015625 > 2\\sqrt{4095} \\iff \\sqrt{4095} < 64 - 0.0078125 \\iff 4095 < 4096 - 128 \\cdot 0.0078125 + 0.0078125^2 = 4096 - 1 + 0.0078125^2$ which is true. Hence as the expression is greater than $0.125$ , and less than or equal to $0.13$ (since we showed it is certainly less than $0.1333333...$ ), it is closest to $0.13$ , which is answer $\\boxed{.13}$" ]

[ "Clearly, \\[.4<.48017<.5\\] Since the function $f(x)=x^3$ is strictly increasing, we can say that \\[.4^3<.48017^3<.5^3\\] from which it follows that $\\text{A}$ is much too small and $\\text{C}$ is much too large, so $\\boxed{0.110}$ is the answer.", "Since $0.48017$ is quite close to $0.5$ , or $\\dfrac{1}{2}$ , we can look for the answer choice that is just below $(\\dfrac{1}{2})^3=\\dfrac{1}{8}=0.125$ , which would be $\\boxed{0.110}$" ]

[ "We write both the numerator and denominator with a denominator of $12$ first, since the LCM of $2,3,$ and $4$ is $3\\cdot4=12$ . \nNext, we multiply both the numerator and denominator by $12$ $\\dfrac{\\frac{1}{3}-\\frac{1}{4}}{\\frac{1}{2}-\\frac{1}{3}}=\\dfrac{\\frac{4}{12}-\\frac{3}{12}}{\\frac{6}{12}-\\frac{4}{12}}=\\dfrac{\\frac{1}{12}}{\\frac{2}{12}}=\\boxed{12}$ .\n-sosiaops" ]

[ "Multiplying the numerator and the denominator by the same value does not change the value of the fraction.\nWe can multiply both by $12$ , getting $\\dfrac{4-3}{6-4} = \\boxed{12}$" ]

[ "We have $.99>.9099>.909>.9009>.9$ , so choice $\\boxed{.99}$ is the largest." ]

[ "Using the process of elimination:\nTenths digit: \nAll tenths digits are equal, at $9$\nHundreths digit: $A$ $B$ , and $C$ all have the same hundreths digit of $7$ , and it is greater than the hundredths of $D$ or $E$ (which is $0$ ). Eliminate both $D$ and $E$\nThousandths digit: $B$ has the largest thousandths digit of the remaining answers, and is the correct answer. $A$ has an \"invisible\" thousandths digit of $0$ , while $C$ also has a thousdandths digit of $0$\n$\\boxed{0.979}$" ]

[ "The number $0$ has no reciprocal, and $1$ and $-1$ are their own reciprocals. This leaves only $2$ and $-2$ . The reciprocal of $2$ is $1/2$ , but $2$ is not less than $1/2$ . The reciprocal of $-2$ is $-1/2$ , and $-2$ is less than $-1/2$ , so it is $\\boxed{2}$", "The statement \"a number is less than its reciprocal\" can be translated as $x < \\frac{1}{x}$\nMultiplication by $x$ can be done if you do it in three parts: $x>0$ $x=0$ , and $x<0$ . You have to be careful about the direction of the inequality, as you do not know the sign of $x$\nIf $x>0$ , the sign of the inequality remains the same. Thus, we have $x^2 < 1$ when $x > 0$ . This leads to $0 < x < 1$\nIf $x=0$ , the inequality $x < \\frac{1}{x}$ is undefined.\nIf $x<0$ , the sign of the inequality must be switched. Thus, we have $x^2 > 1$ when $x < 0$ . This leads to $x < -1$\nPutting the solutions together, we have $x<-1$ or $0 < x < 1$ , or in interval notation, $(-\\infty, -1) \\cup(0, 1)$ . The only answer in that range is $\\boxed{2}$", "Starting again with $x < \\frac{1}{x}$ , we avoid multiplication by $x$ . Instead, move everything to the left, and find a common denominator:\n$x < \\frac{1}{x}$\n$x - \\frac{1}{x} < 0$\n$\\frac{x^2 - 1}{x} < 0$\n$\\frac{(x+1)(x-1)}{x} < 0$\nDivide this expression at $x=-1$ $x=0$ , and $x=1$ , as those are the three points where the expression on the left will \"change sign\".\nIf $x<-1$ , all three of those terms will be negative, and the inequality is true. Therefore, $(-\\infty, -1)$ is part of our solution set.\nIf $-1 < x < 0$ , the $(x+1)$ term will become positive, but the other two terms remain negative. Thus, there are no solutions in this region.\nIf $0 < x < 1$ , then both $(x+1)$ and $x$ are positive, while $(x-1)$ remains negative. Thus, the entire region $(0, -1)$ is part of the solution set.\nIf $1 < x$ , then all three terms are positive, and there are no solutions.\nAt all three \"boundary points\", the function is either $0$ or undefined. Therefore, the entire solution set is $(-\\infty, -1) \\cup (0, 1)$ , and the only option in that region is $x=-2$ , leading to $\\boxed{2}$", "We can find out all of their reciprocals. Now we compare and see that the answer is $\\boxed{2}$" ]

[ "By the change-of-base formula, we can simplify the left side of the equation: $(\\log_3x)(\\log_x5) = (\\frac{\\log_x}{\\log_3})(\\frac{\\log_5}{\\log_x}) = \\frac{\\log_5}{\\log_3}$\nWe see that this in fact simplifies to $\\log_35$ , which will always equal the right side of the equation, since they are the same exact expressions.\nBut we have to be careful because $x \\neq 1$ . Plugging in $x=1$ , the left side would equal $(\\log_31)(\\log_x5) = 0 \\cdot \\log_x5 = 0$ , and $\\log_35$ definitely does not equal $0$\nBesides $1$ $x$ can take on any positive value, and the equation would work. Therefore, the answer is $\\boxed{1}$ jiang147369" ]

[ "Let $m$ be Mike's bill and $j$ be Joe's bill.\n$\\frac{10}{100}m=2$\n$m=20$\n$\\frac{20}{100}j=2$\n$j=10$\nSo the desired difference is $m-j=20-10=10 \\Rightarrow \\boxed{10}$" ]

[ "The value of $5$ -pound rocks is $$14\\div5=$2.80$ per pound, and the value of $4$ -pound rocks is $$11\\div4=$2.75$ per pound. Clearly, Carl should not carry more than three $1$ -pound rocks. Otherwise, he can replace some $1$ -pound rocks with some heavier rocks, preserving the weight but increasing the total value.\nWe perform casework on the number of $1$ -pound rocks Carl can carry: \\[\\begin{array}{c|c|c||c} & & & \\\\ [-2.5ex] \\boldsymbol{1}\\textbf{-Pound Rocks} & \\boldsymbol{4}\\textbf{-Pound Rocks} & \\boldsymbol{5}\\textbf{-Pound Rocks} & \\textbf{Total Value} \\\\ \\textbf{(}\\boldsymbol{$2}\\textbf{ Each)} & \\textbf{(}\\boldsymbol{$11}\\textbf{ Each)} & \\textbf{(}\\boldsymbol{$14}\\textbf{ Each)} & \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 0 & 2 & 2 & $50 \\\\ & & & \\\\ [-2.25ex] 1 & 3 & 1 & $49 \\\\ & & & \\\\ [-2.25ex] 2 & 4 & 0 & $48 \\\\ & & & \\\\ [-2.25ex] 3 & 0 & 3 & $48 \\end{array}\\] Clearly, the maximum value of the rocks Carl can carry is $\\boxed{50}$ dollars.", "Since each rock is worth $1$ dollar less than $3$ times its weight (in pounds), the answer is just $3\\cdot 18=54$ minus the minimum number of rocks we need to make $18$ pounds. Note that we need at least $4$ rocks (two $5$ -pound rocks and two $4$ -pound rocks) to make $18$ pounds, so the answer is $54-4=\\boxed{50}.$" ]

[ "Note that $x$ degrees is equal to $\\frac{\\pi x}{180}$ radians. Also, for $\\alpha \\in \\left[0 , \\frac{\\pi}{2} \\right]$ , the two least positive angles $\\theta > \\alpha$ such that $\\sin{\\theta} = \\sin{\\alpha}$ are $\\theta = \\pi-\\alpha$ , and $\\theta = 2\\pi + \\alpha$\nClearly $x > \\frac{\\pi x}{180}$ for positive real values of $x$\n$\\theta = \\pi-\\alpha$ yields: $x = \\pi - \\frac{\\pi x}{180} \\Rightarrow x = \\frac{180\\pi}{180+\\pi} \\Rightarrow (p,q) = (180,180)$\n$\\theta = 2\\pi + \\alpha$ yields: $x = 2\\pi + \\frac{\\pi x}{180} \\Rightarrow x = \\frac{360\\pi}{180-\\pi} \\Rightarrow (m,n) = (360,180)$\nSo, $m+n+p+q = \\boxed{900}$" ]

[ "We proceed with casework based on the person who sits first after the break.\n$\\textbf{Case 1:}$ A is first. Then the first three people in the row can be ACE, ACF, ADB, ADF, AEB, AEC, AFB, AFC, or AFD, which yield 2, 1, 2, 2, 1, 2, 0, 1, and 1 possible configurations, respectively, implying 2 + 1 + 2 + 2 + 1 + 2 + 0 + 1 + 1 = 12 possible configurations in this case.\n$\\textbf{Case 2:}$ B is first. Then the first three people in the row can be BDA, BDF, BEA, BEC, BFA, BFC, or BFD, which yield 2, 4, 2, 4, 0, 1, and 2 possible configurations, respectively, implying 2 + 4 + 2 + 4 + 0 + 1 + 2 = 15 possible configurations in this case.\n$\\textbf{Case 3:}$ C is first. Then the first three people in the row can be CAD, CAE, CAF, CEA, CEB, CFA, CFB, or CFD, which yield 1, 2, 1, 4, 4, 2, 2, and 2 possible configurations, respectively, implying 1 + 2 + 1 + 4 + 4 + 2 + 2 + 2 = 18 possible configurations in this case.\nFinally, the number of valid configurations for A and F, B and E, and C and D are equal by symmetry, so our final answer is 2(12 + 15 + 18), which computes to be $\\boxed{090}.$ ~peace09", "We determine the order of A, B, C, relative to each other. Then, we will insert D, E, F into the alignment and calculate the total number of possibilities. This solution can be visualised as standing in lines rather than sitting on chairs.\nThere are 6 possible alignments for A, B, and C, but we only evaluate 3 because the other 3 cases can be mirrored to overlap these 3 cases.\n$\\textbf{Case 1: A B C}$ In this case, there must be a person standing between A and B and also between B and C. Also, D cannot be adjacent to C. There are 9 possibilities.\n$\\textbf{Case 2: A C B}$ In this case, there must be a person standing between B and C. Also, D cannot be adjacent to C. There are 12 possibilities.\n$\\textbf{Case 3: C A B}$ In this case, there must be a person standing between A and B. Also, D cannot be adjacent to C. There are 24 possibilities.\nSo the total number of cases is 2(9+12+24)= $\\boxed{090}$", "Consider arrangements of numbers $1, 2, 3, ..., n$\nLet $f(n,k)$ be the number of arrangements in which $1$ and $2$ $2$ and $3$ $3$ and $4$ , ..., $k-1$ and $k$ aren't together. We need to find $f(6,6)$\nLet $d(n,k)$ be the number of arrangements in which $1$ and $2$ $2$ and $3$ $3$ and $4$ , ..., $k-2$ and $k-1$ aren't together, but $(k-1)$ and $k$ are together.\nThen,\n\\[f(n,k+1) = f(n,k) - d(n,k+1) ......(1)\\]\n\\[d(n,k+1) = 2f(n-1,k) + d(n-1,k) ......(2)\\]\nHence, \\[f(n,k+1)=f(n,k)-f(n-1,k)-f(n-1,k-1)\\]\nAnd because $f(n,0) = f(n,1) = n!$ , it's easy to get $f(6,6) = \\boxed{090}$" ]

[ "By substitution, we have \\begin{align*} (\\text{wind chill}) &= 36 - 0.7 \\times 18 \\\\ &= 36 - 12.6 \\\\ &= 23.4 \\\\ &\\approx \\boxed{23} ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM", "$0.7$ is very close to $\\frac{2}{3}$ - therefore, we can substitute $\\frac{2}{3}$ into the equation to get $36 - \\frac{2}{3} * 18$ , which is $36 - 12 = 24$ . As $\\frac{2}{3}$ is slightly less than $0.7$ , the correct answer is slightly less than $24$ . Therefore, the answer is $\\boxed{23}$" ]

[ "Let $x = \\cos 1^\\circ + i \\sin 1^\\circ$ . Then from the identity \\[\\sin 1 = \\frac{x - \\frac{1}{x}}{2i} = \\frac{x^2 - 1}{2 i x},\\] we deduce that (taking absolute values and noticing $|x| = 1$ \\[|2\\sin 1| = |x^2 - 1|.\\] But because $\\csc$ is the reciprocal of $\\sin$ and because $\\sin z = \\sin (180^\\circ - z)$ , if we let our product be $M$ then \\[\\frac{1}{M} = \\sin 1^\\circ \\sin 3^\\circ \\sin 5^\\circ \\dots \\sin 177^\\circ \\sin 179^\\circ\\] \\[= \\frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \\dots |x^{354} - 1| |x^{358} - 1|\\] because $\\sin$ is positive in the first and second quadrants. Now, notice that $x^2, x^6, x^{10}, \\dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z - x^2)(z - x^6)\\dots (z - x^{358}) = z^{90} + 1$ , and so \\[\\frac{1}{M} = \\dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \\dots |1 - x^{358}| = \\dfrac{1}{2^{90}} |1^{90} + 1| = \\dfrac{1}{2^{89}}.\\] It is easy to see that $M = 2^{89}$ and that our answer is $2 + 89 = \\boxed{91}$", "Let $p=\\sin1\\sin3\\sin5...\\sin89$\n\\[p=\\sqrt{\\sin1\\sin3\\sin5...\\sin177\\sin179}\\]\n\\[=\\sqrt{\\frac{\\sin1\\sin2\\sin3\\sin4...\\sin177\\sin178\\sin179}{\\sin2\\sin4\\sin6\\sin8...\\sin176\\sin178}}\\]\n\\[=\\sqrt{\\frac{\\sin1\\sin2\\sin3\\sin4...\\sin177\\sin178\\sin179}{(2\\sin1\\cos1)\\cdot(2\\sin2\\cos2)\\cdot(2\\sin3\\cos3)\\cdot....\\cdot(2\\sin89\\cos89)}}\\]\n\\[=\\sqrt{\\frac{1}{2^{89}}\\frac{\\sin90\\sin91\\sin92\\sin93...\\sin177\\sin178\\sin179}{\\cos1\\cos2\\cos3\\cos4...\\cos89}}\\]\n$=\\sqrt{\\frac{1}{2^{89}}}$ because of the identity $\\sin(90+x)=\\cos(x)$\nwe want $\\frac{1}{p^2}=2^{89}$\nThus the answer is $2+89=\\boxed{091}$", "Similar to Solution $2$ , so we use $\\sin{2\\theta}=2\\sin\\theta\\cos\\theta$ and we find that: \\begin{align*}\\sin(4)\\sin(8)\\sin(12)\\sin(16)\\cdots\\sin(84)\\sin(88)&=(2\\sin(2)\\cos(2))(2\\sin(4)\\cos(4))(2\\sin(6)\\cos(6))(2\\sin(8)\\cos(8))\\cdots(2\\sin(42)\\cos(42))(2\\sin(44)\\cos(44))\\\\ &=(2\\sin(2)\\sin(88))(2\\sin(4))\\sin(86))(2\\sin(6)\\sin(84))(2\\sin(8)\\sin(82))\\cdots(2\\sin(42)\\sin(48))(2\\sin(44)\\sin(46))\\\\ &=2^{22}(\\sin(2)\\sin(88)\\sin(4)\\sin(86)\\sin(6)\\sin(84)\\sin(8)\\sin(82)\\cdots\\sin(42)\\sin(48)\\sin(44)\\sin(46))\\\\ &=2^{22}(\\sin(2)\\sin(4)\\sin(6)\\sin(8)\\cdots\\sin(82)\\sin(84)\\sin(86)\\sin(88))\\end{align*} Now we can cancel the sines of the multiples of $4$ \\[1=2^{22}(\\sin(2)\\sin(6)\\sin(10)\\sin(14)\\cdots\\sin(82)\\sin(86))\\] So $\\sin(2)\\sin(6)\\sin(10)\\sin(14)\\cdots\\sin(82)\\sin(86)=2^{-22}$ and we can apply the double-angle formula again: \\begin{align*}2^{-22}&=(\\sin(2)\\sin(6)\\sin(10)\\sin(14)\\cdots\\sin(82)\\sin(86)\\\\ &=(2\\sin(1)\\cos(1))(2\\sin(3)\\cos(3))(2\\sin(5)\\cos(5))(2\\sin(7)\\cos(7))\\cdots(2\\sin(41)\\cos(41))(2\\sin(43)\\cos(43))\\\\ &=(2\\sin(1)\\sin(89))(2\\sin(3)\\sin(87))(2\\sin(5)\\sin(85))(2\\sin(7)\\sin(87))\\cdots(2\\sin(41)\\sin(49))(2\\sin(43)\\sin(47))\\\\ &=2^{22}(\\sin(1)\\sin(89)\\sin(3)\\sin(87)\\sin(5)\\sin(85)\\sin(7)\\sin(83)\\cdots\\sin(41)\\sin(49)\\sin(43)\\sin(47))\\\\ &=2^{22}(\\sin(1)\\sin(3)\\sin(5)\\sin(7)\\cdots\\sin(41)\\sin(43))(\\sin(47)\\sin(49)\\cdots\\sin(83)\\sin(85)\\sin(87)\\sin(89))\\end{align*} Of course, $\\sin(45)=2^{-\\frac{1}{2}}$ is missing, so we multiply it to both sides: \\[2^{-22}\\sin(45)=2^{22}(\\sin(1)\\sin(3)\\sin(5)\\sin(7)\\cdots\\sin(41)\\sin(43))(\\sin(45))(\\sin(47)\\sin(49)\\cdots\\sin(83)\\sin(85)\\sin(87)\\sin(89))\\] \\[\\left(2^{-22}\\right)\\left(2^{-\\frac{1}{2}}\\right)=2^{22}(\\sin(1)\\sin(3)\\sin(5)\\sin(7)\\cdots\\sin(83)\\sin(85)\\sin(87)\\sin(89))\\] \\[2^{-\\frac{45}{2}}=2^{22}(\\sin(1)\\sin(3)\\sin(5)\\sin(7)\\cdots\\sin(83)\\sin(85)\\sin(87)\\sin(89))\\] Now isolate the product of the sines: \\[\\sin(1)\\sin(3)\\sin(5)\\sin(7)\\cdots\\sin(83)\\sin(85)\\sin(87)\\sin(89)=2^{-\\frac{89}{2}}\\] And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: \\[\\csc^2(1)\\csc^2(3)\\csc^2(5)\\csc^2(7)\\cdots\\csc^2(83)\\csc^2(85)\\csc^2(87)\\csc^2(89)=\\left(\\frac{1}{2^{-\\frac{89}{2}}}\\right)^2=\\left(2^{\\frac{89}{2}}\\right)^2=2^{89}\\] The answer is therefore $m+n=(2)+(89)=\\boxed{091}$", "Let $p=\\prod_{k=1}^{45} \\csc^2(2k-1)^\\circ$\nThen, $\\sqrt{\\frac{1}{p}}=\\prod_{k=1}^{45} \\sin(2k-1)^\\circ$\nSince $\\sin\\theta=\\cos(90^{\\circ}-\\theta)$ , we can multiply both sides by $\\frac{\\sqrt{2}}{2}$ to get $\\sqrt{\\frac{1}{2p}}=\\prod_{k=1}^{23} \\sin(2k-1)^\\circ\\cos(2k-1)^\\circ$\nUsing the double-angle identity $\\sin2\\theta=2\\sin\\theta\\cos\\theta$ , we get $\\sqrt{\\frac{1}{2p}}=\\frac{1}{2^{23}}\\prod_{k=1}^{23} \\sin(4k-2)^\\circ$\nNote that the right-hand side is equal to $\\frac{1}{2^{23}}\\prod_{k=1}^{45} \\sin(2k)^\\circ\\div \\prod_{k=1}^{22} \\sin(4k)^\\circ$ , which is equal to $\\frac{1}{2^{23}}\\prod_{k=1}^{45} \\sin(2k)^\\circ\\div \\prod_{k=1}^{22} 2\\sin(2k)^\\circ\\cos(2k)^\\circ$ , again, from using our double-angle identity.\nPutting this back into our equation and simplifying gives us $\\sqrt{\\frac{1}{2p}}=\\frac{1}{2^{45}}\\prod_{k=23}^{45} \\sin(2k)^\\circ\\div \\prod_{k=1}^{22} \\cos(2k)^\\circ$\nUsing the fact that $\\sin\\theta=\\cos(90^{\\circ}-\\theta)$ again, our equation simplifies to $\\sqrt{\\frac{1}{2p}}=\\frac{\\sin90^\\circ}{2^{45}}$ , and since $\\sin90^\\circ=1$ , it follows that $2p = 2^{90}$ , which implies $p=2^{89}$ . Thus, $m+n=2+89=\\boxed{091}$", "Recall that $\\sin\\alpha\\cdot \\sin(60^{\\circ}-\\alpha)\\cdot \\sin(60^{\\circ}+\\alpha)=\\frac{1}{4}\\cdot \\sin3\\alpha$ Since it is in csc, we can write in sin and then take reciprocal.\nWe can group them by threes, $P=(\\sin1^{\\circ}\\cdot \\sin59^{\\circ}\\cdot \\sin61^{\\circ})\\cdots(\\sin29^{\\circ}\\cdot \\sin31^{\\circ}\\cdot \\sin89^{\\circ})$ . Thus \\begin{align*} P &=\\frac{1}{4^{15}}\\cdot \\sin3^{\\circ}\\cdot \\sin9^{\\circ}\\cdots\\sin87^{\\circ}\\\\ &=\\frac{1}{4^{20}}\\cdot \\sin9^{\\circ}\\cdot \\sin27^{\\circ}\\cdot \\sin45^{\\circ}\\cdot \\sin63^{\\circ}\\cdot \\sin81^{\\circ}\\\\ &=\\frac{1}{4^{20}}\\cdot \\frac{\\sqrt{2}}{2}\\cdot \\sin9^{\\circ}\\cdot \\cos9^{\\circ}\\cdot \\sin27^{\\circ}\\cdot \\cos27^{\\circ}\\\\ &=\\frac{1}{4^{21}}\\cdot \\frac{\\sqrt{2}}{2}\\cdot \\sin18^{\\circ}\\cdot \\cos36^{\\circ}=\\frac{\\sqrt{2}}{2^{45}} \\end{align*} So we take reciprocal, $\\frac 1P=2^{\\frac{89}{2}}$ , the desired answer is $\\frac{1}{P^2}=2^{89}$ leads to answer $\\boxed{091}$", "We have\n\\[\\prod_{k=1}^{45} \\csc^2(2k-1)^\\circ = \\left(\\frac{1}{\\sin1^\\circ \\cdot \\sin3^\\circ \\cdots \\sin89^\\circ}\\right)^2.\\]\nMultiplying by $\\frac{\\sin2^\\circ \\cdot \\sin4^\\circ \\cdots \\sin88^\\circ}{\\sin2^\\circ \\cdot \\sin4^\\circ \\cdots \\sin88^\\circ}$ gives\n\\[\\left(\\frac{\\sin2^\\circ \\cdot \\sin4^\\circ \\cdots \\sin88^\\circ}{\\sin1^\\circ \\sin2^\\circ \\cdot \\sin3^\\circ \\cdots \\sin88^\\circ \\cdot \\sin89^\\circ}\\right)^2\\]\n\\[= \\left(\\frac{\\sin2^\\circ \\cdot \\sin4^\\circ \\cdots \\sin88^\\circ}{\\sin1^\\circ \\sin2^\\circ \\cdot \\sin3^\\circ \\cdots \\sin 45^\\circ \\cdot \\cos 44^\\circ \\cdot \\cos 43^\\circ \\cdots \\cos1^\\circ}\\right)^2.\\]\nUsing $\\sin\\alpha \\cos\\alpha = \\frac{1}{2}\\sin{2\\alpha}$ gives\n\\[\\left(\\frac{\\sin2^\\circ \\cdot \\sin4^\\circ \\cdots \\sin88^\\circ}{\\frac{1}{2} \\sin2^\\circ \\cdot \\frac{1}{2} \\sin4^\\circ \\cdots \\frac{1}{2} \\sin88^\\circ \\cdot \\sin45^\\circ}\\right) ^2\\]\n\\[= \\left(\\frac{1}{(\\frac{1}{2})^{44} \\cdot \\frac{\\sqrt{2}}{2}}\\right)^2\\]\n\\[= 2^{89}.\\]\nThus, the answer is $2+89 = \\boxed{091}.$" ]

[ "The heaviest object that could be weighed with this set weighs $1 + 3 + 9 = 13$ lb., and we can weigh any positive integer weight at most that. This means that $13$ different objects could be weighed, so our answer is $\\boxed{13}$ and we are done." ]

[ "Let $x$ represent the distance from home to the stadium, and let $r$ represent the distance from Yan to home. Our goal is to find $\\frac{r}{x-r}$ . If Yan walks directly to the stadium, then assuming he walks at a rate of $1$ , it will take him $x-r$ units of time. Similarly, if he walks back home it will take him $r + \\frac{x}{7}$ units of time. Because the two times are equal, we can create the following equation: $x-r = r + \\frac{x}{7}$ . We get $x-2r=\\frac{x}{7}$ , so $\\frac{6}{7}x = 2r$ , and $\\frac{x}{r} = \\frac{7}{3}$ . This minus one is the reciprocal of what we want to find: $\\frac{7}{3}-1 = \\frac{4}{3}$ , so the answer is $\\boxed{34}$", " Let $H$ represent Yan's home, $S$ represent the stadium, and $Y$ represent Yan's current position. If Yan walks directly to the stadium, he will reach Point $P$ the same time he will reach $H$ if he is walking home. \nSince he bikes $7$ times as fast as he walks and the time is the same, the distance from his home to the stadium must be $7$ times the distance from $P$ to the stadium. If $PS=x$ , then $HS=7x$ and $HP=6x$ . Since Y is the midpoint of $\\overline{HP}$ $HY=YP=3x$ . Therefore, the ratio of Yan's distance from his home to his distance from the stadium is $\\frac{YH}{YS}=\\frac{3x}{4x}=\\boxed{34}$" ]

[ "Let $x$ represent the distance from home to the stadium, and let $r$ represent the distance from Yan to home. Our goal is to find $\\frac{r}{x-r}$ . If Yan walks directly to the stadium, then assuming he walks at a rate of $1$ , it will take him $x-r$ units of time. Similarly, if he walks back home it will take him $r + \\frac{x}{7}$ units of time. Because the two times are equal, we can create the following equation: $x-r = r + \\frac{x}{7}$ . We get $x-2r=\\frac{x}{7}$ , so $\\frac{6}{7}x = 2r$ , and $\\frac{x}{r} = \\frac{7}{3}$ . This minus one is the reciprocal of what we want to find: $\\frac{7}{3}-1 = \\frac{4}{3}$ , so the answer is $\\boxed{34}$", " Let $H$ represent Yan's home, $S$ represent the stadium, and $Y$ represent Yan's current position. If Yan walks directly to the stadium, he will reach Point $P$ the same time he will reach $H$ if he is walking home. \nSince he bikes $7$ times as fast as he walks and the time is the same, the distance from his home to the stadium must be $7$ times the distance from $P$ to the stadium. If $PS=x$ , then $HS=7x$ and $HP=6x$ . Since Y is the midpoint of $\\overline{HP}$ $HY=YP=3x$ . Therefore, the ratio of Yan's distance from his home to his distance from the stadium is $\\frac{YH}{YS}=\\frac{3x}{4x}=\\boxed{34}$" ]