link stringlengths 75 84 | letter stringclasses 5
values | answer float64 0 2,935,363,332B | problem stringlengths 14 5.33k | solution listlengths 1 13 |
|---|---|---|---|---|
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_1 | C | 10 | What is the value in simplest form of the following expression? \[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}\]
$\textbf{(A) }5 \qquad \textbf{(B) }4 + \sqrt{7} + \sqrt{10} \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$ | [
"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 $n^2$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_1 | C | 127 | What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$
$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$ | [
"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 ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_4 | D | 11 | What is the value of $(3x - 2)(4x + 1) - (3x - 2)4x + 1$ when $x=4$
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 11 \qquad\mathrm{(E)}\ 12$ | [
"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}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_2 | D | 11 | What is the value of $(3x - 2)(4x + 1) - (3x - 2)4x + 1$ when $x=4$
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 11 \qquad\mathrm{(E)}\ 12$ | [
"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}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | D | 2,015 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$ | [
"$(625^{\\log_5 2015})^\\frac{1}{4} = ((5^4)^{\\log_5 2015})^\\frac{1}{4} = (5^{4 \\cdot \\log_5 2015})^\\frac{1}{4} = (5^{\\log_5 2015 \\cdot 4})^\\frac{1}{4} = ((5^{\\log_5 2015})^4)^\\frac{1}{4} = (2015^4)^\\frac{1}{4} = \\boxed{2015}$",
"$(625^{\\log_5 2015})^{\\frac{1}{4}} = (625^{\\frac{1}{4}})^{\\log_5 201... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | null | 2,015 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$ | [
"We can rewrite $\\log_5 2015$ as as $5^x = 2015$ . Thus, $625^{x \\cdot \\frac{1}{4}} = 5^x = \\boxed{2015}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_1 | D | 18 | What is the value of $(8 \times 4 + 2) - (8 + 4 \times 2)$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 24$ | [
"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{... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5 | E | 1,010 | What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$
$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$ | [
"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 in... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5 | null | 1,010 | What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$
$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$ | [
"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 $1010^2-1009^2-100... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_1 | E | 11,110 | What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$ | [
"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 \\[12... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_1 | E | 11,110 | What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$ | [
"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 \\[12... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_3 | E | 2,000 | What is the value of $4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)$
$\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000$ | [
"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... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_1 | B | 100 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | [
"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 t... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_1 | B | 100 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | [
"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 t... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_1 | C | 49 | What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$ | [
"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^... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_1 | C | 49 | What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$ | [
"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^... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_1 | D | 10 | What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \frac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$ | [
"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 $\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_1 | D | 10 | What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \tfrac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$ | [
"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... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_4 | D | 32 | What is the value of $x$ if $|x-1|=|x-2|$
$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2$ | [
"$|x-1|$ 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 $-1=-2$ , w... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1 | D | 8 | What is the value of \[(2^2-2)-(3^2-3)+(4^2-4)\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$ | [
"$(4-2)-(9-3)+(16-4)=2-6+12=8.$ 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... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_14 | null | 2 | What is the value of \[(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)\] where $\log$ denotes the base-ten logarithm?
$\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3$ | [
"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... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_8 | B | 9,900 | What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$ | [
"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 n... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_1 | D | 5 | What is the value of \[1-(-2)-3-(-4)-5-(-6)?\]
$\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21$ | [
"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... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12 | D | 3,159 | What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$ | [
"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 \\cdo... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_1 | B | 50 | What is the value of \[2^{1+2+3}-(2^1+2^2+2^3)?\] $\textbf{(A) }0 \qquad \textbf{(B) }50 \qquad \textbf{(C) }52 \qquad \textbf{(D) }54 \qquad \textbf{(E) }57$ | [
"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}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_5 | B | 2 | What is the value of \[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?\] $\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}$ | [
"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\\ri... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_14 | E | 21,000 | What is the value of \[\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?\]
$\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2200\qquad \textbf{(E) }21000$ | [
"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*} \\lo... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_7 | C | 6 | What is the value of \[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\] $\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$ | [
"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 \\l... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_5 | B | 1,120 | What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$
$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$ | [
"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... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_3 | C | 8 | What is the value of the expression $\sqrt{16\sqrt{8\sqrt{4}}}$
$\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16$ | [
"$\\sqrt{16\\sqrt{8\\sqrt{4}}}$ $\\sqrt{16\\sqrt{8\\cdot 2}}$ $\\sqrt{16\\sqrt{16}}$ $\\sqrt{16\\cdot 4}$ $\\sqrt{64}$ $\\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_2 | A | 1 | What is the value of the following expression?
\[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}\]
$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{9951}{9950} \qquad \textbf{(C) } \frac{4780}{4779} \qquad \textbf{(D) } \frac{108}{107} \qquad \textbf{(E) } \frac{81}{80}$ | [
"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."
] |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_2 | D | 7 | What is the value of the product
\[\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?\]
$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(... | [
"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 an... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_2 | C | 6.54 | What is the value of this expression in decimal form? \[\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}\] $\textbf{(A) } 6.4\qquad\textbf{(B) } 6.504\qquad\textbf{(C) } 6.54\qquad\textbf{(D) } 6.9\qquad\textbf{(E) } 6.94$ | [
"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"
] |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_13 | C | 4 | What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2$ $AC = 3$ $AD = 4$ $BC = \sqrt{13}$ $BD = 2\sqrt{5}$ , and $CD = 5$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6$ | [
"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 $\\tri... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_6 | B | 27 | What non-zero real value for $x$ satisfies $(7x)^{14}=(14x)^7$
$\textbf{(A) } \frac17\qquad \textbf{(B) } \frac27\qquad \textbf{(C) } 1\qquad \textbf{(D) } 7\qquad \textbf{(E) } 14$ | [
"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, $x=\\boxed{27}$"
] |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_24 | C | 120 | What number is directly above $142$ in this array of numbers?
\[\begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array}\]
$\text{(A)}\ 99 \qquad \text{(B)}\ 119 \qquad \text{(C)}\ 120 \qquad \text{(D)}\ 121 \qquad \text{(E)}\ 122$ | [
"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 $12^2=144$ number... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_12 | B | 5 | What number should be removed from the list \[1,2,3,4,5,6,7,8,9,10,11\] so that the average of the remaining numbers is $6.1$
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | [
"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 $6.1$ . Se... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_4 | D | 2,400 | When $0.000315$ is multiplied by $7,928,564$ the product is closest to which of the following?
$\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000$ | [
"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 $\\boxed{2400}.$"
] |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_20 | D | 826 | When $10^{93}-93$ is expressed as a single whole number, the sum of the digits is
$\text{(A)}\ 10 \qquad \text{(B)}\ 93 \qquad \text{(C)}\ 819 \qquad \text{(D)}\ 826 \qquad \text{(E)}\ 833$ | [
"\\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 $9(91)+7=819+7=\\boxed{826}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_25 | A | 4 | When $15$ is appended to a list of integers, the mean is increased by $2$ . When $1$ is appended to the enlarged list, the mean of the enlarged list is decreased by $1$ . How many integers were in the original list?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm... | [
"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 th... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_24 | B | 1 | When $1999^{2000}$ is divided by $5$ , the remainder is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | [
"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 o... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11 | B | 84 | When $7$ fair standard $6$ -sided dice are thrown, the probability that the sum of the numbers on the top faces is $10$ can be written as \[\frac{n}{6^{7}},\] where $n$ is a positive integer. What is $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$ | [
"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... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11 | E | 84 | When $7$ fair standard $6$ -sided dice are thrown, the probability that the sum of the numbers on the top faces is $10$ can be written as \[\frac{n}{6^{7}},\] where $n$ is a positive integer. What is $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$ | [
"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 ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11 | null | 84 | When $7$ fair standard $6$ -sided dice are thrown, the probability that the sum of the numbers on the top faces is $10$ can be written as \[\frac{n}{6^{7}},\] where $n$ is a positive integer. What is $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$ | [
"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{... |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_12 | C | 10 | When $\left ( 1 - \frac{1}{a} \right ) ^6$ is expanded the sum of the last three coefficients is:
$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ -10\qquad\textbf{(E)}\ -11$ | [
"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 $\\binom{6}{2}-\\binom{6}{1}+\\binom{6}{0}=15-6+1=\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_23 | A | 11 | When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$
$\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9$ | [
"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 $1$... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_6 | C | 6 | When $x^5, x+\frac{1}{x}$ and $1+\frac{2}{x} + \frac{3}{x^2}$ are multiplied, the product is a polynomial of degree.
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$ | [
"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... |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_9 | B | 5 | When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:
$\textbf{(A)}\ \text{more than 5}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2$ | [
"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 $\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_20 | D | 2 | When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$ | [
"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 $x-1$ , mean... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_4 | B | 10 | When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes. As an old man, he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?
$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textb... | [
"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... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_4 | A | 1 | When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be leftover?
$\textbf{(A)}\ 1... | [
"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 ... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_10 | D | 15 | When Walter drove up to the gasoline pump, he noticed that his gasoline tank was 1/8 full. He purchased 7.5 gallons of gasoline for 10 dollars. With this additional gasoline, his gasoline tank was then 5/8 full. The number of gallons of gasoline his tank holds when it is full is $\text{(A)}\ 8.75 \qquad \text{(B)}\ ... | [
"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 $7.5 \\times 2 = 15$ ga... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_4 | E | 9 | When Yunji added all the integers from $1$ to $9$ , she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$ | [
"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.\... |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_5 | null | 283 | When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac ij$ , in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$ | [
"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(... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22 | E | 83 | When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.2... | [
"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... |
https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_2 | null | 26 | When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{ cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle? | [
"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... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_8 | E | 75 | When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\un... | [
"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} \\ \\und... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_5 | E | 75 | When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\un... | [
"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} \\ \\und... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_4 | D | 149 | When counting from $3$ to $201$ $53$ is the $51^{st}$ number counted. When counting backwards from $201$ to $3$ $53$ is the $n^{th}$ number counted. What is $n$
$\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150$ | [
"Note that $n$ is equal to the number of integers between $53$ and $201$ , inclusive. Thus, $n=201-53+1=\\boxed{149}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_3 | D | 149 | When counting from $3$ to $201$ $53$ is the $51^{st}$ number counted. When counting backwards from $201$ to $3$ $53$ is the $n^{th}$ number counted. What is $n$
$\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150$ | [
"Note that $n$ is equal to the number of integers between $53$ and $201$ , inclusive. Thus, $n=201-53+1=\\boxed{149}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_2 | null | 62 | When each of $702$ $787$ , and $855$ is divided by the positive integer $m$ , the remainder is always the positive integer $r$ . When each of $412$ $722$ , and $815$ is divided by the positive integer $n$ , the remainder is always the positive integer $s \neq r$ . Find $m+n+r+s$ | [
"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 o... |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_9 | C | 420 | When finding the sum $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}$ , the least common denominator used is
$\text{(A)}\ 120 \qquad \text{(B)}\ 210 \qquad \text{(C)}\ 420 \qquad \text{(D)}\ 840 \qquad \text{(E)}\ 5040$ | [
"We want the least common multiple of $2,3,4,5,6,7$ , which is $420$ , or choice $\\boxed{420}$"
] |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_14 | D | 24 | When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
$\text{(A)}\ 8 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 48$ | [
"Four gallons is $\\frac{1}{2}-\\frac{1}{3}= \\frac{1}{6}$ . Thus, capacity of the tank in gallons is $6 \\times 4= \\boxed{24}$"
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_33 | C | 25 | When one ounce of water is added to a mixture of acid and water, the new mixture is $20\%$ acid. When one ounce of acid is added to the new mixture, the result is $33\frac13\%$ acid. The percentage of acid in the original mixture is
$\textbf{(A)}\ 22\% \qquad \textbf{(B)}\ 24\% \qquad \textbf{(C)}\ 25\% \qquad \textbf{... | [
"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. \... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_5 | null | 29 | When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on op... | [
"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 pro... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_7 | B | 10 | When the World Wide Web first became popular in the $1990$ s, download speeds reached a maximum of about $56$ kilobits per second. Approximately how many minutes would the download of a $4.2$ -megabyte song have taken at that speed? (Note that there are $8000$ kilobits in a megabyte.)
$\textbf{(A) } 0.6 \qquad \textbf{... | [
"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... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_7 | C | 0.02 | When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?
$\textbf{(A) }0.0002\qquad\textbf{(B) }0.002\qquad\textbf{(C) }0.02\qquad\textbf{(D) }0.2\qquad\textbf{(E) }2$ | [
"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 i... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_2 | C | 19 | When the fraction $\dfrac{49}{84}$ is expressed in simplest form, then the sum of the numerator and the denominator will be
$\text{(A)}\ 11 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 133$ | [
"\\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 $7+12=19\\rightarrow \\boxed{19}$"
] |
https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_31 | C | 3 | When the number $2^{1000}$ is divided by $13$ , the remainder in the division is
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }7\qquad \textbf{(E) }11$ | [
"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 $\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_5 | A | 15 | When the repeating decimal $0.363636\ldots$ is written in simplest fractional form, the sum of the numerator and denominator is:
$\textbf{(A)}\ 15 \qquad \textbf{(B) }\ 45 \qquad \textbf{(C) }\ 114 \qquad \textbf{(D) }\ 135 \qquad \textbf{(E) }\ 150$ | [
"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 $\\boxed{15}$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_12 | C | 6 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\q... | [
"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 ... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_6 | C | 6 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\q... | [
"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 ... |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_7 | C | 30 | When three different numbers from the set $\{ -3, -2, -1, 4, 5 \}$ are multiplied, the largest possible product is
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$ | [
"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 $-2$ . T... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_3 | E | 4 | When three positive integers $a$ $b$ , and $c$ are multiplied together, their product is $100$ . Suppose $a < b < c$ . In how many ways can the numbers be chosen?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$ | [
"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 $a=1$ , t... |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_9 | D | 4 | When you simplify $\left[ \sqrt [3]{\sqrt [6]{a^9}} \right]^4\left[ \sqrt [6]{\sqrt [3]{a^9}} \right]^4$ , the result is:
$\textbf{(A)}\ a^{16} \qquad\textbf{(B)}\ a^{12} \qquad\textbf{(C)}\ a^8 \qquad\textbf{(D)}\ a^4 \qquad\textbf{(E)}\ a^2$ | [
"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}.$"
] |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_2 | A | 1 | Which digit of $.12345$ , when changed to $9$ , gives the largest number?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$ | [
"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 $9 \\rightarrow \\boxed{1}$"
] |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_6 | C | 7 | Which number in the array below is both the largest in its column and the smallest in its row? (Columns go up and down, rows go right and left.) \[\begin{tabular}[t]{ccccc} 10 & 6 & 4 & 3 & 2 \\ 11 & 7 & 14 & 10 & 8 \\ 8 & 3 & 4 & 5 & 9 \\ 13 & 4 & 15 & 12 & 1 \\ 8 & 2 & 5 & 9 & 3 \end{tabular}\]
$\text{(A)}\ 1 \qquad... | [
"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 $\\rightarrow \\boxed{7}$"
] |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_4 | E | 8 | Which of the following could not be the units digit [ones digit] of the square of a whole number?
$\text{(A)}\ 1 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | [
"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. Testin... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_9 | E | 0 | Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which \[\sin(x+y)\leq \sin(x)+\sin(y)\] for every $x$ between $0$ and $\pi$ , inclusive?
$\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \... | [
"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, \\[\\co... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_9 | null | 0 | Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which \[\sin(x+y)\leq \sin(x)+\sin(y)\] for every $x$ between $0$ and $\pi$ , inclusive?
$\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \... | [
"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 han... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21 | E | 12 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 ... | [
"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 ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_16 | E | 12 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 ... | [
"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 ... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | D | 100 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | [
"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 divisi... |
https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_19 | B | 0.13 | Which of the following is closest to $\sqrt{65}-\sqrt{63}$
$\textbf{(A)}\ .12 \qquad \textbf{(B)}\ .13 \qquad \textbf{(C)}\ .14 \qquad \textbf{(D)}\ .15 \qquad \textbf{(E)}\ .16$ | [
"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}... |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_5 | B | 0.11 | Which of the following is closest to the product $(.48017)(.48017)(.48017)$
$\text{(A)}\ 0.011 \qquad \text{(B)}\ 0.110 \qquad \text{(C)}\ 1.10 \qquad \text{(D)}\ 11.0 \qquad \text{(E)}\ 110$ | [
"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 $\\dfrac{1}{2}$... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_2 | C | 12 | Which of the following is equal to $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}$
$\text{(A) } \frac 14 \qquad \text{(B) } \frac 13 \qquad \text{(C) } \frac 12 \qquad \text{(D) } \frac 23 \qquad \text{(E) } \frac 34$ | [
"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}... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_2 | null | 12 | Which of the following is equal to $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}$
$\text{(A) } \frac 14 \qquad \text{(B) } \frac 13 \qquad \text{(C) } \frac 12 \qquad \text{(D) } \frac 23 \qquad \text{(E) } \frac 34$ | [
"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 $\\dfrac{4-3}{6-4} = \\boxed{12}$"
] |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_3 | A | 0.99 | Which of the following numbers is the largest?
$\text{(A)}\ .99 \qquad \text{(B)}\ .9099 \qquad \text{(C)}\ .9 \qquad \text{(D)}\ .909 \qquad \text{(E)}\ .9009$ | [
"We have $.99>.9099>.909>.9009>.9$ , so choice $\\boxed{.99}$ is the largest."
] |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_3 | B | 0.979 | Which of the following numbers is the largest?
$\text{(A)}\ 0.97 \qquad \text{(B)}\ 0.979 \qquad \text{(C)}\ 0.9709 \qquad \text{(D)}\ 0.907 \qquad \text{(E)}\ 0.9089$ | [
"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 ... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_2 | A | 2 | Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | [
"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 reci... |
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_19 | D | 1 | Which positive numbers $x$ satisfy the equation $(\log_3x)(\log_x5)=\log_35$
$\textbf{(A)}\ 3 \text{ and } 5 \text{ only} \qquad \textbf{(B)}\ 3, 5, \text{ and } 15 \text{ only} \qquad \\ \textbf{(C)}\ \text{only numbers of the form } 5^n \cdot 3^m, \text{ where } n \text{ and } m \text{ are positive integers} \qquad ... | [
"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... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_1 | D | 10 | While eating out, Mike and Joe each tipped their server $2$ dollars. Mike tipped $10\%$ of his bill and Joe tipped $20\%$ of his bill. What was the difference, in dollars between their bills?
$\textbf{(A) } 2\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 10\qquad \textbf{(E) } 20$ | [
"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 $m-j=20-10=10 \\Rightarrow \\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_2 | C | 50 | While exploring a cave, Carl comes across a collection of $5$ -pound rocks worth $$14$ each, $4$ -pound rocks worth $$11$ each, and $1$ -pound rocks worth $$2$ each. There are at least $20$ of each size. He can carry at most $18$ pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?
... | [
"The value of $5$ -pound rocks is $$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 va... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_10 | null | 900 | While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correct angular mode. He was lucky to get the right answer. The two least positive real values of $x$ for which the sine of $x$ degrees is the same as the sine of $x$ radians are $\frac{m\pi}{n-\pi}... | [
"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 $x > \\frac{\\pi x}{1... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_9 | null | 90 | While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next... | [
"We proceed with casework based on the person who sits first after the break.\n$\\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... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_3 | B | 23 | Wind chill is a measure of how cold people feel when exposed to wind outside. A good estimate for wind chill can be found using this calculation \[(\text{wind chill}) = (\text{air temperature}) - 0.7 \times (\text{wind speed}),\] where temperature is measured in degrees Fahrenheit $(^{\circ}\text{F})$ and the wind spee... | [
"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 t... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13 | null | 91 | With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$ | [
"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^... |
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_19 | B | 13 | With the use of three different weights, namely $1$ lb., $3$ lb., and $9$ lb., how many objects of different weights can be weighed, if the objects is to be weighed and the given weights may be placed in either pan of the scale? $\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 11\qquad\textbf{(D)}\ 9\qquad\t... | [
"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."
] |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_13 | B | 34 | Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to hi... | [
"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 hi... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_9 | B | 34 | Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to hi... | [
"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 hi... |
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