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/1984_AIME_Problems/Problem_13 | null | 15 | Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$ | [
"We know that $\\tan(\\arctan(x)) = x$ so we can repeatedly apply the addition formula, $\\tan(x+y) = \\frac{\\tan(x)+\\tan(y)}{1-\\tan(x)\\tan(y)}$ . Let $a = \\cot^{-1}(3)$ $b=\\cot^{-1}(7)$ $c=\\cot^{-1}(13)$ , and $d=\\cot^{-1}(21)$ . We have\nso\nand\nso\nThus our answer is $10\\cdot\\frac{3}{2}=\\boxed{015}$"... |
https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_1 | null | 93 | Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ $a_2$ $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$ | [
"One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of $a_1$ , then use that to calculate $a_2$ and sum another arithmetic series to get our answer.\nA somewhat quicker method is to do the following: for each $n \\geq 1$ , we have $a_{2n - 1} = a_{2n}... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_9 | B | 3 | Find the value of $x$ that satisfies the equation $25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}.$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 9$ | [
"Manipulate the powers of $5$ in order to get a clean expression.\n\\[\\frac{5^{\\frac{48}{x}}}{5^{\\frac{26}{x}} \\cdot 25^{\\frac{17}{x}}} = \\frac{5^{\\frac{48}{x}}}{5^{\\frac{26}{x}} \\cdot 5^{\\frac{34}{x}}} = 5^{\\frac{48}{x}-(\\frac{26}{x}+\\frac{34}{x})} = 5^{-\\frac{12}{x}}\\] \\[25^{-2} = (5^2)^{-2} = 5^{... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_8 | C | 50 | Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$ | [
"We can group each subtracting pair together: \\[(100-98)+(96-94)+(92-90)+ \\ldots +(8-6)+(4-2).\\] After subtracting, we have: \\[2+2+2+\\ldots+2+2=2(1+1+1+\\ldots+1+1).\\] There are $50$ even numbers, therefore there are $\\dfrac{50}{2}=25$ even pairs. Therefore the sum is $2 \\cdot 25=\\boxed{50}$",
"Since our... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_13 | D | 32 | Find the value(s) of $x$ such that $8xy - 12y + 2x - 3 = 0$ is true for all values of $y$
$\textbf{(A) } \frac23 \qquad \textbf{(B) } \frac32 \text{ or } -\frac14 \qquad \textbf{(C) } -\frac23 \text{ or } -\frac14 \qquad \textbf{(D) } \frac32 \qquad \textbf{(E) } -\frac32 \text{ or } -\frac14$ | [
"We have $8xy - 12y + 2x - 3 = 4y(2x - 3) + (2x - 3) = (4y + 1)(2x - 3)$\nAs $(4y + 1)(2x - 3) = 0$ must be true for all $y$ , we must have $2x - 3 = 0$ , hence $\\boxed{32}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_USAJMO_Problems/Problem_1 | null | 1 | Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square. | [
"We will first take the expression modulo $3$ . We get $2^n+12^n+2011^n \\equiv -1^n+1^n \\pmod 3$\nLemma 1: All perfect squares are equal to $0$ or $1$ modulo $3$ .\nWe can prove this by testing the residues modulo $3$ . We have $0^2 \\equiv 0 \\pmod 3$ $1^2 \\equiv 1 \\pmod 3$ , and $2^2 \\equiv 1 \\pmod 3$ , so ... |
https://artofproblemsolving.com/wiki/index.php/2016_USAJMO_Problems/Problem_4 | null | 6,097,392 | Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,...,N\}$ , one can still find $2016$ distinct numbers among the remaining elements with sum $N$ | [
"Since any $2016$ elements are removed, suppose we remove the integers from $1$ to $2016$ . Then the smallest possible sum of $2016$ of the remaining elements is \\[2017+2018+\\cdots + 4032 = 1008 \\cdot 6049 = 6097392\\] so clearly $N\\ge 6097392$ . We will show that $N=6097392$ works.\n$\\vspace{0.2 in}$\n$\\{1,2... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_16 | D | 2.2 | Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpo... | [
"After the first swap, we do casework on the next swap.\nCase 1: Silva swaps the two balls that were just swapped\nThere is only one way for Silva to do this, and it leaves 5 balls occupying their original position.\nCase 2: Silva swaps one ball that has just been swapped with one that hasn't been swapped\nThere ar... |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_25 | A | 3 | Five cards are lying on a table as shown.
\[\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\ \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}\] | [
"Using the answer choices, we see that P and Q are logically equivalent (both are non-vowel letters) and so are $4$ and $6$ (both are even numbers). Thus, if one of these is correct, then the other option in its pair must also be correct. There can't be 2 answers. So, the only remaining answer choice is $\\boxed{3}... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_23 | B | 150 | Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$ | [
"Firstly, observe that it is not possible for a single student to receive $4$ or $5$ awards because this would mean that one of the other students receives no awards. Thus, each student must receive either $1$ $2$ , or $3$ awards. If a student receives $3$ awards, then the other two students must each receive $1$ a... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_13 | B | 2 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved tw... | [
"Assume that Edie and Dee were originally in seats 3 and 4. If this were so, there is no possible position for which Bea can move 2 seats to the right. The same applies for seats 2 and 3. This means that either Edie or Dee was originally in an edge seat. If Edie and Dee were in seats 1 and 2, then Bea must have bee... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_13 | null | 2 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved tw... | [
"Note that the person (out of A,B,C) that moves the most, moves the amount equal to the sum of what the other 2 move. They essentially make a cycle. D & E are seat fillers and can be ignored. A,B,C take up either seats 1,2,3 or 2,4,5. In each case you find A was originally in seat $\\boxed{2}$",
"Note that the ne... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_10 | B | 2 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved tw... | [
"Assume that Edie and Dee were originally in seats 3 and 4. If this were so, there is no possible position for which Bea can move 2 seats to the right. The same applies for seats 2 and 3. This means that either Edie or Dee was originally in an edge seat. If Edie and Dee were in seats 1 and 2, then Bea must have bee... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_10 | null | 2 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved tw... | [
"Note that the person (out of A,B,C) that moves the most, moves the amount equal to the sum of what the other 2 move. They essentially make a cycle. D & E are seat fillers and can be ignored. A,B,C take up either seats 1,2,3 or 2,4,5. In each case you find A was originally in seat $\\boxed{2}$",
"Note that the ne... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 | null | 191 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | [
"For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are $14!$ arrangements without restrictions.\nFirst, there are $\\binom{7}{5}$ ways to choose the man-woman diameters. Then, there are $10\\cdot8\\cdot6\\cdot4\\cdot2$ ways to pl... |
https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_24 | B | 8 | Five people are sitting at a round table. Let $f\geq 0$ be the number of people sitting next to at least 1 female and $m\geq0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is
$\mathrm{(A) \ 7 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 9 } \qquad \mathrm{(D... | [
"Suppose there are more men than women; then there are between zero and two women.\nIf there are no women, the pair is $(0,5)$ . If there is one woman, the pair is $(2,5)$\nIf there are two women, there are two arrangements: one in which they are together, and one in which they are apart, giving the pairs $(4,5)$ a... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_13 | B | 171 | Five test scores have a mean (average score) of $90$ , a median (middle score) of $91$ and a mode (most frequent score) of $94$ . The sum of the two lowest test scores is
$\text{(A)}\ 170 \qquad \text{(B)}\ 171 \qquad \text{(C)}\ 176 \qquad \text{(D)}\ 177 \qquad \text{(E)}\ \text{not determined by the information giv... | [
"Because there was an odd number of scores, $91$ must be the middle score. Since there are two scores above $91$ and $94$ appears the most frequent (so at least twice) and $94>91$ $94$ appears twice. Also, the sum of the five numbers is $90 \\times 5 =450$ . Thus, the sum of the lowest two scores is $450-91-94-94= ... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_11 | null | 544 | Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way). | [
"It is obvious that any configuration of one-way roads which contains a town whose roads all lead into it or lead out of it cannot satisfy the given. We claim that any configuration which does not have a town whose roads all lead into it or lead out of it does satisfy the given conditions. Now we show that a loop o... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_17 | C | 23 | Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(... | [
"For $c\\geq 1.5$ the shaded area is at most $1.5$ , which is too little. Hence $c<1.5$ , and therefore the point $(2,1)$ is indeed inside the shaded part, as shown in the picture.\nThen the area of the shaded part is one less than the area of the triangle with vertices $(c,0)$ $(3,0)$ , and $(3,3)$ . The area of t... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_14 | C | 23 | Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(... | [
"For $c\\geq 1.5$ the shaded area is at most $1.5$ , which is too little. Hence $c<1.5$ , and therefore the point $(2,1)$ is indeed inside the shaded part, as shown in the picture.\nThen the area of the shaded part is one less than the area of the triangle with vertices $(c,0)$ $(3,0)$ , and $(3,3)$ . The area of t... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_17 | E | 12 | Flora the frog starts at 0 on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance $m$ with probability $\frac{1}{2^m}$
What is the probability that Flora will eventually land at 10?
$\textbf{(A)}~\frac{5}{512}\qquad\textbf{(... | [
"Initially, the probability of landing at $10$ and landing past $10$ (summing the infinte series) are exactly the same. Landing before 10 repeats this initial condition, with a different irrelevant scaling factor. Therefore, the probability must be $\\boxed{12}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_1 | null | 336 | For $-1<r<1$ , let $S(r)$ denote the sum of the geometric series \[12+12r+12r^2+12r^3+\cdots .\] Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a)=2016$ . Find $S(a)+S(-a)$ | [
"The sum of an infinite geometric series is $\\frac{a}{1-r}\\rightarrow \\frac{12}{1\\mp a}$ . The product $S(a)S(-a)=\\frac{144}{1-a^2}=2016$ $\\frac{12}{1-a}+\\frac{12}{1+a}=\\frac{24}{1-a^2}$ , so the answer is $\\frac{2016}{6}=\\boxed{336}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_15 | null | 863 | For $1 \leq i \leq 215$ let $a_i = \dfrac{1}{2^{i}}$ and $a_{216} = \dfrac{1}{2^{215}}$ . Let $x_1, x_2, ..., x_{216}$ be positive real numbers such that $\sum_{i=1}^{216} x_i=1$ and $\sum_{1 \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}$ . The maximum possible value of... | [
"Note that \\begin{align*}\\sum_{1 \\leq i < j \\leq 216} x_ix_j &= \\frac12\\left(\\left(\\sum_{i=1}^{216} x_i\\right)^2-\\sum_{i=1}^{216} x_i^2\\right)\\\\&=\\frac12\\left(1-\\sum x_i^2\\right).\\end{align*} Substituting this into the second equation and collecting $x_i^2$ terms, we find \\[\\sum_{i=1}^{216}\\fra... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_14 | null | 36 | For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \... | [
"Noticing the $\\sin$ and $\\cos$ in both $P$ and $Q,$ we think of the angle addition identities:\n\\[\\sin(a + b) = \\sin a \\cos b + \\cos a \\sin b, \\cos(a + b) = \\cos a \\cos b - \\sin a \\sin b\\]\nWith this in mind, we multiply $P$ by $\\sin \\theta$ and $Q$ by $\\cos \\theta$ to try and use some angle addi... |
https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13 | null | 448 | For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | [
"Let $S$ be a non- empty subset of $\\{1,2,3,4,5,6\\}$\nThen the alternating sum of $S$ , plus the alternating sum of $S \\cup \\{7\\}$ , is $7$ . This is because, since $7$ is the largest element, when we take an alternating sum, each number in $S$ ends up with the opposite sign of each corresponding element of $S... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_25 | B | 7 | For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 1... | [
"Notice that $2$ is a prime factor of an integer $n$ if and only if $n$ is even. Therefore, given any sufficiently high positive integral value of $k$ , dividing $I_k$ by $2^6$ yields a terminal digit of zero, and dividing by 2 again leaves us with $2^7 \\cdot a = I_k$ where $a$ is an odd integer.\nObserve then tha... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_25 | null | 7 | For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 1... | [
"The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\\cdot 2^{k+2} + 2^6$\nFor $k\\in\\{1,2,3\\}$ we have $I_k = 2^{k+2} \\left( 5^{k+2} + 2^{4-k} \\right)$ . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\\leq 5$\nFor $k>4$ we have $I_k=2^6 \... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_18 | B | 7 | For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 1... | [
"Notice that $2$ is a prime factor of an integer $n$ if and only if $n$ is even. Therefore, given any sufficiently high positive integral value of $k$ , dividing $I_k$ by $2^6$ yields a terminal digit of zero, and dividing by 2 again leaves us with $2^7 \\cdot a = I_k$ where $a$ is an odd integer.\nObserve then tha... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_18 | null | 7 | For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 1... | [
"The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\\cdot 2^{k+2} + 2^6$\nFor $k\\in\\{1,2,3\\}$ we have $I_k = 2^{k+2} \\left( 5^{k+2} + 2^{4-k} \\right)$ . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\\leq 5$\nFor $k>4$ we have $I_k=2^6 \... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_12 | null | 47 | For $n \ge 1$ call a finite sequence $(a_1, a_2 \ldots a_n)$ of positive integers progressive if $a_i < a_{i+1}$ and $a_i$ divides $a_{i+1}$ for all $1 \le i \le n-1$ . Find the number of progressive sequences such that the sum of the terms in the sequence is equal to $360$ | [
"If the first term is $x$ , then dividing through by $x$ , we see that we can find the number of progressive sequences whose sum is $\\frac{360}{x} - 1$ , and whose first term is not 1. If $a(k)$ denotes the number of progressive sequences whose sum is $k$ and whose first term is not 1, then we can express the answ... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_25 | C | 2 | For $n$ a positive integer, let $R(n)$ be the sum of the remainders when $n$ is divided by $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , and $10$ . For example, $R(15) = 1+0+3+0+3+1+7+6+5=26$ . How many two-digit positive integers $n$ satisfy $R(n) = R(n+1)\,?$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(... | [
"Note that we can add $9$ to $R(n)$ to get $R(n+1)$ , but must subtract $k$ for all $k|n+1$ . Hence, we see that there are four ways to do that because $9=7+2=6+3=5+4=4+3+2$ . Note that only $7+2$ is a plausible option, since $4+3+2$ indicates $n+1$ is divisible by $6$ $5+4$ indicates that $n+1$ is divisible by $2$... |
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_29 | B | 80,200 | For $p=1, 2, \cdots, 10$ let $S_p$ be the sum of the first $40$ terms of the arithmetic progression whose first term is $p$ and whose common difference is $2p-1$ ; then $S_1+S_2+\cdots+S_{10}$ is
$\mathrm{(A)\ } 80000 \qquad \mathrm{(B) \ }80200 \qquad \mathrm{(C) \ } 80400 \qquad \mathrm{(D) \ } 80600 \qquad \mathrm... | [
"The $40\\text{th}$ term of an arithmetic progression with a first term $p$ and a common difference $2p-1$ is $p+39(2p-1)=79p-39$ . Therefore, the sum of the first $40$ terms of such a progression is $\\frac{40}{2}(79p-39+p)=1600p-780$\nWe now want to evaluate $\\sum_{p=1}^{10}(1600p-780)$ \\[\\sum_{p=1}^{10}(1600p... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_14 | null | 905 | For $t = 1, 2, 3, 4$ , define $S_t = \sum_{i = 1}^{350}a_i^t$ , where $a_i \in \{1,2,3,4\}$ . If $S_1 = 513$ and $S_4 = 4745$ , find the minimum possible value for $S_2$ | [
"Because the order of the $a$ 's doesn't matter, we simply need to find the number of $1$ $2$ $3$ s and $4$ s that minimize $S_2$ . So let $w, x, y,$ and $z$ represent the number of $1$ s, $2$ s, $3$ s, and $4$ s respectively. Then we can write three equations based on these variables. Since there are a total of $3... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_14 | E | 10 | For a certain complex number $c$ , the polynomial \[P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)\] has exactly 4 distinct roots. What is $|c|$
$\textbf{(A) } 2 \qquad \textbf{(B) } \sqrt{6} \qquad \textbf{(C) } 2\sqrt{2} \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \sqrt{10}$ | [
"The polynomial can be factored further broken down into\n$P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$\nby using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of two... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_22 | B | 201,400 | For a certain positive integer $n$ less than $1000$ , the decimal equivalent of $\frac{1}{n}$ is $0.\overline{abcdef}$ , a repeating decimal of period of $6$ , and the decimal equivalent of $\frac{1}{n+6}$ is $0.\overline{wxyz}$ , a repeating decimal of period $4$ . In which interval does $n$ lie?
$\textbf{(A)}\ [1,200... | [
"Solution by e_power_pi_times_i\nIf $\\frac{1}{n} = 0.\\overline{abcdef}$ $n$ must be a factor of $999999$ . Also, by the same procedure, $n+6$ must be a factor of $9999$ . Checking through all the factors of $999999$ and $9999$ that are less than $1000$ , we see that $n = 297$ is a solution, so the answer is $\\bo... |
https://artofproblemsolving.com/wiki/index.php/1992_AHSME_Problems/Problem_21 | A | 991 | For a finite sequence $A=(a_1,a_2,...,a_n)$ of numbers, the Cesáro sum of A is defined to be $\frac{S_1+\cdots+S_n}{n}$ , where $S_k=a_1+\cdots+a_k$ and $1\leq k\leq n$ . If the Cesáro sum of
the 99-term sequence $(a_1,...,a_{99})$ is 1000, what is the Cesáro sum of the 100-term sequence $(1,a_1,...,a_{99})$
$\text{(A... | [
"Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is \\[S_1 + S_2 + S_3 + ... + S_n\\] \\[= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \\ldots + (a_1 + a_2 + \\ldots + a_n)\\] \\[= n \\cdot a_1 + (n - 1) \\cdot a_2 + \\ldots + a_n.\\] If we take this to be ... |
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_26 | C | 20.5 | For a given arithmetic series the sum of the first $50$ terms is $200$ , and the sum of the next $50$ terms is $2700$ .
The first term in the series is:
$\textbf{(A)}\ -1221 \qquad \textbf{(B)}\ -21.5 \qquad \textbf{(C)}\ -20.5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 3.5$ | [
"Let the first term of the arithmetic sequence be $a$ and the common difference be $d$\nThe $50^{\\text{th}}$ term of the sequence is $a+49d$ , so the sum of the first $50$ terms is $\\frac{50(a + a + 49d)}{2}$\nThe $51^{\\text{th}}$ term of the sequence is $a+50d$ and the $100^{\\text{th}}$ term of the sequence is... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_3 | null | 69 | For a positive integer $n$ , let $d_n$ be the units digit of $1 + 2 + \dots + n$ . Find the remainder when \[\sum_{n=1}^{2017} d_n\] is divided by $1000$ | [
"We see that $d_n$ appears in cycles of $20$ and the cycles are \\[1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\\] adding a total of $70$ each cycle.\nSince $\\left\\lfloor\\frac{2017}{20}\\right\\rfloor=100$ , we know that by $2017$ , there have been $100$ cycles and $7000$ has been added. This can be discarded as we'... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12 | A | 10 | For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$ | [
"We have $5! = 2 \\cdot 3 \\cdot 4 \\cdot 5$ , and $2 \\cdot 5 \\cdot 9! = 10 \\cdot 9! = 10!$ . Therefore, the equation becomes $3 \\cdot 4 \\cdot 10! = 12 \\cdot N!$ , and so $12 \\cdot 10! = 12 \\cdot N!$ . Cancelling the $12$ s, it is clear that $N=\\boxed{10}$",
"Since $5! = 120$ , we obtain $120\\cdot 9!=12... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_25 | D | 18 | For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is... | [
"By geometric series, we have \\begin{alignat*}{8} A_n&=a\\bigl(\\phantom{ }\\underbrace{111\\cdots1}_{n\\text{ digits}}\\phantom{ }\\bigr)&&=a\\left(1+10+10^2+\\cdots+10^{n-1}\\right)&&=a\\cdot\\frac{10^n-1}{9}, \\\\ B_n&=b\\bigl(\\phantom{ }\\underbrace{111\\cdots1}_{n\\text{ digits}}\\phantom{ }\\bigr)&&=b\\left... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_25 | D | 18 | For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is... | [
"By geometric series, we have \\begin{alignat*}{8} A_n&=a\\bigl(\\phantom{ }\\underbrace{111\\cdots1}_{n\\text{ digits}}\\phantom{ }\\bigr)&&=a\\left(1+10+10^2+\\cdots+10^{n-1}\\right)&&=a\\cdot\\frac{10^n-1}{9}, \\\\ B_n&=b\\bigl(\\phantom{ }\\underbrace{111\\cdots1}_{n\\text{ digits}}\\phantom{ }\\bigr)&&=b\\left... |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_8 | null | 132 | For a real number $a$ , let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$ . Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$ . The region $\mathcal{R}$ is completely contained in a disk of ra... | [
"The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$ , namely $(\\pm5,0), (0,\\pm5), (\\pm3,\\pm4), (\\pm4,\\pm3).$ Since the points themselves are symmetric about $(0,0)$ , the boxes are symmetric about $\\left(\\frac12,\\frac12\\right)$ . The distance from ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_4 | C | 10 | For a real number $x$ , define $\heartsuit(x)$ to be the average of $x$ and $x^2$ . What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 20$ | [
"The average of two numbers, $a$ and $b$ , is defined as $\\frac{a+b}{2}$ . Thus the average of $x$ and $x^2$ would be $\\frac{x(x+1)}{2}$ . With that said, we need to find the sum when we plug, $1$ $2$ and $3$ into that equation. So:\n\\[\\frac{1(1+1)}{2} + \\frac{2(2+1)}{2} + \\frac{3(3+1)}{2} = \\frac{2}{2} + \... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_14 | null | 10 | For a real number $x$ let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$ , and define $\{x\} = x - \lfloor x \rfloor$ to be the fractional part of $x$ . For example, $\{3\} = 0$ and $\{4.56\} = 0.56$ . Define $f(x)=x\{x\}$ , and let $N$ be the number of real-valued solutions to the equation $f(f(f... | [
"To solve $f(f(f(x)))=17$ , we need to solve $f(x) = y$ where $f(f(y))=17$ , and to solve that we need to solve $f(y) = z$ where $f(z) = 17$\nIt is clear to see for some integer $a \\geq 17$ there is exactly one value of $z$ in the interval $[a, a+1)$ where $f(z) = 17$ . To understand this, imagine the graph of $f(... |
https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_9 | C | 4 | For a sale, a store owner reduces the price of a $$10$ scarf by $20\%$ . Later the price is lowered again, this time by one-half the reduced price. The price is now
$\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qq... | [
"$100\\%-20\\%=80\\%$\n$10\\times80\\%=10\\times0.8$\n$10\\times0.8=8$\n$\\frac{8}{2}=4=\\boxed{4.00}$",
"The first discount has percentage 20, which is then discounted again for half of the already discounted price.\n$100-20=80$\n$\\frac{80}{2}=40$\n$40\\%\\times10=10\\times0.4=4=\\boxed{4.00}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_7 | null | 48 | For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hi... | [
"Let $x$ be the number of acorns that both animals had.\nSo by the info in the problem:\n$\\frac{x}{3}=\\left( \\frac{x}{4} \\right)+4$\nSubtracting $\\frac{x}{4}$ from both sides leaves\n$\\frac{x}{12}=4$\n$\\boxed{48}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_3 | D | 48 | For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide... | [
"If $x$ is the number of holes that the chipmunk dug, then $3x=4(x-4)$ , so $3x=4x-16$ , and $x=16$ . The number of acorns hidden by the chipmunk is equal to $3x = \\boxed{48}$",
"Trying answer choices, we see that $\\boxed{48}$ works. ~Extremelysupercooldude"
] |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_14 | D | 19 | For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$
$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$ | [
"It is possible to obtain $0$ $1$ $3$ $4$ $5$ , and $6$ points of intersection, as demonstrated in the following figures: \nIt is clear that the maximum number of possible intersections is ${4 \\choose 2} = 6$ , since each pair of lines can intersect at most once. We now prove that it is impossible to obtain two i... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_8 | D | 19 | For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$
$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$ | [
"It is possible to obtain $0$ $1$ $3$ $4$ $5$ , and $6$ points of intersection, as demonstrated in the following figures: \nIt is clear that the maximum number of possible intersections is ${4 \\choose 2} = 6$ , since each pair of lines can intersect at most once. We now prove that it is impossible to obtain two i... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_22 | B | 1 | For all integers $n$ greater than $1$ , define $a_n = \frac{1}{\log_n 2002}$ . Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$ . Then $b- c$ equals
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ \frac{1}{2002} \qquad\mathrm{(D)}\ \frac{1}{1001} \qquad\mathrm{(E)}\ \fr... | [
"Note that $\\frac{1}{\\log_a b}=\\log_b a$ . Thus $a_n=\\log_{2002} n$ . Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following:\n\\[\\log_{2002}\\left(\\frac{2*3*4*5}{10*11*12*13*14}=\\frac{1}{11*13*14}=\\frac{1}... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_22 | null | 1 | For all integers $n$ greater than $1$ , define $a_n = \frac{1}{\log_n 2002}$ . Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$ . Then $b- c$ equals
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ \frac{1}{2002} \qquad\mathrm{(D)}\ \frac{1}{1001} \qquad\mathrm{(E)}\ \fr... | [
"Note that $a_2 = \\frac{1}{\\log_2 2002}$ $1$ is also equal to $\\log_2 2$ . So $a_2 = \\frac{\\log_2 2}{\\log_2 2002}$ . By the change of bases formula, $a_2 = \\log_{2002} 2$ . Following the same reasoning, $a_3 = \\log_{2002} 3$ $a_4 = \\log_{2002} 4$ and so on.\n\\[b = \\log_{2002} 2 + \\log_{2002} 3 + .....+ ... |
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_2 | null | 1 | For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals
$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$ | [
"Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into \\[(x+1)(y-1) = -1\\] Plugging in $-1$ and $1$ as the $x$ and $y$ sides respectively, we get $x = -2$ and $y = 2$ . Plugging this in to $\\frac{1}{x}-\\frac{1}{y}$ gives us $\\boxed{1}$ as our ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_14 | D | 2 | For all positive integers $n$ , let $f(n)=\log_{2002} n^2$ . Let $N=f(11)+f(13)+f(14)$ . Which of the following relations is true?
$\text{(A) }N<1 \qquad \text{(B) }N=1 \qquad \text{(C) }1<N<2 \qquad \text{(D) }N=2 \qquad \text{(E) }N>2$ | [
"First, note that $2002 = 11 \\cdot 13 \\cdot 14$\nUsing the fact that for any base we have $\\log a + \\log b = \\log ab$ , we get that $N = \\log_{2002} (11^2 \\cdot 13^2 \\cdot 14^2) = \\log_{2002} 2002^2 = \\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_21 | A | 448 | For all positive integers $n$ less than $2002$ , let
\begin{eqnarray*} a_n =\left\{ \begin{array}{lr} 11, & \text{if\ }n\ \text{is\ divisible\ by\ }13\ \text{and\ }14;\\ 13, & \text{if\ }n\ \text{is\ divisible\ by\ }14\ \text{and\ }11;\\ 14, & \text{if\ }n\ \text{is\ divisible\ by\ }11\ \text{and\ }13;\\ 0, & \text{oth... | [
"Find the LCMs of the groups of the numbers.\nNotice that the groups are relatively prime.\nSo $a_n=$\n11 if $n$ is a multiple of 182.\n13 if $n$ is a multiple of 154.\n14 if $n$ is a multiple of 143.\nWhen do we see ambiguities (for example: $n$ is a multiple of 11, 13, and 14)? This is only done when $n$ is a mul... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_15 | null | 511 | For all positive integers $x$ , let \[f(x)=\begin{cases}1 & \text{if }x = 1\\ \frac x{10} & \text{if }x\text{ is divisible by 10}\\ x+1 & \text{otherwise}\end{cases}\] and define a sequence as follows: $x_1=x$ and $x_{n+1}=f(x_n)$ for all positive integers $n$ . Let $d(x)$ be the smallest $n$ such that $x_n=1$ . (For e... | [
"We backcount the number of ways. Namely, we start at $x_{20} = 1$ , which can only be reached if $x_{19} = 10$ , and then we perform $18$ operations that either consist of $A: (-1)$ or $B: (\\times 10)$ . We represent these operations in a string format, starting with the operation that sends $f(x_{18}) = x_{19}$ ... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | null | 454 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | [
"By PIE, $|A|+|B|-|A \\cap B| = |A \\cup B|$ . Substituting into the equation and factoring, we get that $(|A| - |A \\cap B|)(|B| - |A \\cap B|) = 0$ , so therefore $A \\subseteq B$ or $B \\subseteq A$ . WLOG $A\\subseteq B$ , then for each element there are $3$ possibilities, either it is in both $A$ and $B$ , it ... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_12 | null | 245 | For any finite set $X$ , let $| X |$ denote the number of elements in $X$ . Define \[S_n = \sum | A \cap B | ,\] where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$ .
For example, $S_2 = 4$ because the sum is taken over t... | [
"Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$ . The problem states that for $n=2, S_n$ is $4$ . Let's try it out for $n=3$\nLet's perform casework on the number of elements in $A, B$\n$\\textbf{Case 1:} |A| = |B| = 1$\nIn this case, the only possible equivale... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_15 | null | 149 | For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the sma... | [
"Note that for any $x_i$ , for any prime $p$ $p^2 \\nmid x_i$ . This provides motivation to translate $x_i$ into a binary sequence $y_i$\nLet the prime factorization of $x_i$ be written as $p_{a_1} \\cdot p_{a_2} \\cdot p_{a_3} \\cdots$ , where $p_i$ is the $i$ th prime number. Then, for every $p_{a_k}$ in the pri... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_19 | D | 26 | For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | [
"Factoring out $98!+99!+100!$ , we have $98! (1+99+99*100)$ , which is $98! (10000)$ . Next, $98!$ has $\\left\\lfloor\\frac{98}{5}\\right\\rfloor + \\left\\lfloor\\frac{98}{25}\\right\\rfloor = 19 + 3 = 22$ factors of $5$ . The $19$ is because of all the multiples of $5$ .The $3$ is because of all the multiples of... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_19 | null | 26 | For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | [
"Also, keep in mind that the number of $5$ ’s in $98! (10,000)$ is the same as the number of trailing zeros. The number of zeros is $98!$ , which means we need pairs of $5$ ’s and $2$ ’s; we know there will be many more $2$ ’s, so we seek to find the number of $5$ ’s in $98!$ , which the solution tells us. And, tha... |
https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_2 | null | 169 | For any positive integer $k$ , let $f_1(k)$ denote the square of the sum of the digits of $k$ . For $n \ge 2$ , let $f_n(k) = f_1(f_{n - 1}(k))$ . Find $f_{1988}(11)$ | [
"We see that $f_{1}(11)=4$\n$f_2(11) = f_1(4)=16$\n$f_3(11) = f_1(16)=49$\n$f_4(11) = f_1(49)=169$\n$f_5(11) = f_1(169)=256$\n$f_6(11) = f_1(256)=169$\nNote that this revolves between the two numbers. Since $1988$ is even, we thus have $f_{1988}(11) = f_{4}(11) = \\boxed{169}$"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_5 | null | 223 | For any positive integer $x_{}$ , let $S(x)$ be the sum of the digits of $x_{}$ , and let $T(x)$ be $|S(x+2)-S(x)|.$ For example, $T(199)=|S(201)-S(199)|=|3-19|=16.$ How many values of $T(x)$ do not exceed 1999? | [
"For most values of $x$ $T(x)$ will equal $2$ . For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take $T(a999)$ as an example, \\[|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|\\] And in general, the values of $T(x)$ will then be in the form of $|2 - 9n| = 9n - 2... |
https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_14 | A | 199 | For any real number a and positive integer k, define
${a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}$
What is
${-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}$
$\textbf{(A)}\ -199\qquad \textbf{(B)}\ -197\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 197\qquad \textbf{(E)}\ 199$ | [
"We expand both the numerator and the denominator.\n\\begin{align*} \\binom{-\\frac{1}{2}}{100}\\div\\binom{\\frac{1}{2}}{100} &= \\frac{ \\dfrac{ (-\\frac{1}{2}) (-\\frac{1}{2} - 1) (-\\frac{1}{2} - 2) \\cdots (-\\frac{1}{2} - (100 - 1)) }{\\cancel{(100)(99)\\cdots(1)}} }{ \\dfrac{ (\... |
https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | null | 819 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1... | [
"Note that the $\\Delta$ s are reminiscent of differentiation; from the condition $\\Delta(\\Delta{A}) = 1$ , we are led to consider the differential equation \\[\\frac{d^2 A}{dn^2} = 1\\] This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; \\[a_{n} = \\frac{1}{2}(n-19)(n-92)\\] as w... |
https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_30 | null | 97 | For any set $S$ , let $|S|$ denote the number of elements in $S$ , and let $n(S)$ be the number of subsets of $S$ , including the empty set and the set $S$ itself. If $A$ $B$ , and $C$ are sets for which $n(A)+n(B)+n(C)=n(A\cup B\cup C)$ and $|A|=|B|=100$ , then what is the minimum possible value of $|A\cap B\cap C|$
$... | [
"As $|A|=|B|=100$ $n(A)=n(B)=2^{100}$\nAs $n(A)+n(B)+n(C)=n(A \\cup B \\cup C)$ $2^{|A|}+2^{|B|}+2^{|C|}=2^{|A \\cup B \\cup C|}$ $2^{100}+2^{100}+2^{|C|}=2^{|A \\cup B \\cup C|}$\n$2^{101}+2^{|C|}=2^{|A \\cup B \\cup C|}$ as $|C|$ and $|A \\cup B \\cup C|$ are integers, $|C|=101$ and $|A \\cup B \\cup C| = 102$\nB... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_7 | null | 72 | For certain ordered pairs $(a,b)\,$ of real numbers , the system of equations
has at least one solution, and each solution is an ordered pair $(x,y)\,$ of integers. How many such ordered pairs $(a,b)\,$ are there? | [
"The equation $x^2+y^2=50$ is that of a circle of radius $\\sqrt{50}$ , centered at the origin. By testing integers until the left side becomes too big, we see that the lattice points on this circle are $(\\pm1,\\pm7)$ $(\\pm5,\\pm5)$ , and $(\\pm7,\\pm1)$ where the signs are all independent of each other, for a to... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_11 | null | 125 | For certain pairs $(m,n)$ of positive integers with $m\geq n$ there are exactly $50$ distinct positive integers $k$ such that $|\log m - \log k| < \log n$ . Find the sum of all possible values of the product $mn$ | [
"We have $\\log m - \\log k = \\log \\left( \\frac mk \\right)$ , hence we can rewrite the inequality as follows: \\[- \\log n < \\log \\left( \\frac mk \\right) < \\log n\\] We can now get rid of the logarithms, obtaining: \\[\\frac 1n < \\frac mk < n\\] And this can be rewritten in terms of $k$ as \\[\\frac mn < ... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_24 | C | 7,007 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | [
"$f(x)$ must have four roots, three of which are roots of $g(x)$ . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that\n\\[f(x)=g(x)(x+p)\\]\nwhere $-p\\in\\mathbb{R}$ is the fourth root of $f(x)$ . \n(Usin... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | C | 7,007 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | [
"Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$ . Let $r_4$ be the additional root of $f(x)$ . Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$ , we obtain the following:\n\\begin{align*} r_1+r_2+r_3&=-a \\\\ r_1+r_2+r_3+r_4&=-1 \\end{align*}\nThus $r_4=a-1$\nNow applying Vi... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | null | 7,007 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | [
"First off, let's get rid of the $x^4$ term by finding $h(x)=f(x)-xg(x)$ . This polynomial consists of the difference of two polynomials with $3$ common factors, so it must also have these factors. The polynomial is $h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c$ , and must be equal to $(1-a)g(x)$ . Equating the coefficients,... |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_5 | null | 51 | For certain real values of $a, b, c,$ and $d_{},$ the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots. The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$ Find $b.$ | [
"Since the coefficients of the polynomial are real, it follows that the non-real roots must come in complex conjugate pairs. Let the first two roots be $m,n$ . Since $m+n$ is not real, $m,n$ are not conjugates, so the other pair of roots must be the conjugates of $m,n$ . Let $m'$ be the conjugate of $m$ , and $n'$ ... |
https://artofproblemsolving.com/wiki/index.php/2012_USAJMO_Problems/Problem_5 | null | 502 | For distinct positive integers $a$ $b < 2012$ , define $f(a,b)$ to be the number of integers $k$ with $1 \le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012. Let $S$ be the minimum value of $f(a,b)$ , where $a$ and $b$ range over all pairs of distinct positive i... | [
"Let $ak \\equiv r_{a} \\pmod{2012}$ and $bk \\equiv r_{b} \\pmod{2012}$ . Notice that this means $a(2012 - k) \\equiv 2012 - r_{a} \\pmod{2012}$ and $b(2012 - k) \\equiv 2012 - r_{b} \\pmod{2012}$ . Thus, for every value of $k$ where $r_{a} > r_{b}$ , there is a value of $k$ where $r_{b} > r_{a}$ . Therefore, we m... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13 | null | 899 | For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square | [
"Given $g : x \\mapsto \\max_{j : 2^j | x} 2^j$ , consider $S_n = g(2) + \\cdots + g(2^n)$ . Define $S = \\{2, 4, \\ldots, 2^n\\}$ . There are $2^0$ elements of $S$ that are divisible by $2^n$ $2^1 - 2^0 = 2^0$ elements of $S$ that are divisible by $2^{n-1}$ but not by $2^n, \\ldots,$ and $2^{n-1}-2^{n-2} = 2^{n-2}... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_14 | null | 483 | For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with ... | [
"Let $n\\ge 2$ and define $a(n) = \\left\\lfloor \\sqrt n \\right\\rfloor$ . For $2\\le n \\le 1000$ , we have $1\\le a(n)\\le 31$\nFor $a^2 \\le x < (a+1)^2$ we have $y=ax$ . Thus $A(n+1)-A(n)=a(n+\\tfrac 12) = \\Delta_n$ (say), and $\\Delta_n$ is an integer if $a$ is even; otherwise $\\Delta_n$ is an integer plus... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_22 | B | 197 | For each integer $n\geq 2$ , let $S_n$ be the sum of all products $jk$ , where $j$ and $k$ are integers and $1\leq j<k\leq n$ . What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by $3$
$\textbf{(A)}\ 196\qquad\textbf{(B)}\ 197\qquad\textbf{(C)}\ 198\qquad\textbf{(D)}\ 199\qquad\textbf{(E)}\ 200... | [
"To get from $S_n$ to $S_{n+1}$ , we add $1(n+1)+2(n+1)+\\cdots +n(n+1)=(1+2+\\cdots +n)(n+1)=\\frac{n(n+1)^2}{2}$\nNow, we can look at the different values of $n$ mod $3$ . For $n\\equiv 0\\pmod{3}$ and $n\\equiv 2\\pmod{3}$ , then we have $\\frac{n(n+1)^2}{2}\\equiv 0\\pmod{3}$ . However, for $n\\equiv 1\\pmod{3}... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_25 | null | 962 | For each integer $n\geq 4$ , let $a_n$ denote the base- $n$ number $0.\overline{133}_n$ . The product $a_4a_5\cdots a_{99}$ can be expressed as $\frac {m}{n!}$ , where $m$ and $n$ are positive integers and $n$ is as small as possible. What is $m$
$\text {(A)} 98 \qquad \text {(B)} 101 \qquad \text {(C)} 132\qquad \text... | [
"Note that \\[0.\\overline{133}_n = \\frac{n^2+3n+3}{n^3-1},\\] by geometric series.\nThus, we're aiming to find the value of \\[\\prod_{k=4}^{99} \\frac{k^2+3k+3}{k^3 - 1}.\\] Expanding the product out, this is equivalent to \\[\\frac{4^2+3(4)+3}{4^3 - 1} \\cdot \\frac{5^2+3(5)+3}{5^3 - 1} \\cdot \\frac{6^2+3(6)+3... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_13 | null | 245 | For each integer $n\geq3$ , let $f(n)$ be the number of $3$ -element subsets of the vertices of the regular $n$ -gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$ | [
"Considering $n \\pmod{6}$ , we have the following formulas:\n$n\\equiv 0$ $\\frac{n(n-4)}{2} + \\frac{n}{3}$\n$n\\equiv 2, 4$ $\\frac{n(n-2)}{2}$\n$n\\equiv 3$ $\\frac{n(n-3)}{2} + \\frac{n}{3}$\n$n\\equiv 1, 5$ $\\frac{n(n-1)}{2}$\nTo derive these formulas, we note the following:\nAny isosceles triangle formed by... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_5 | null | 189 | For each ordered pair of real numbers $(x,y)$ satisfying \[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\] there is a real number $K$ such that \[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\] Find the product of all possible values of $K$ | [
"Using the logarithmic property $\\log_{a^n}b^n = \\log_{a}b$ , we note that \\[(2x+y)^2 = x^2+xy+7y^2\\] That gives \\[x^2+xy-2y^2=0\\] upon simplification and division by $3$ . Factoring $x^2+xy-2y^2=0$ gives \\[(x+2y)(x-y)=0\\] Then, \\[x=y \\text{ or }x=-2y\\] From the second equation, \\[9x^2+6xy+y^2=3x^2+4xy+... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_2 | C | 0 | For each pair of real numbers $a \neq b$ , define the operation $\star$ as
$(a \star b) = \frac{a+b}{a-b}$
What is the value of $((1 \star 2) \star 3)$
$\textbf{(A) } -\frac{2}{3}\qquad \textbf{(B) } -\frac{1}{5}\qquad \textbf{(C) } 0\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \textrm{This\, value\, is\, not\,... | [
"$((1 \\star 2) \\star 3) = \\left(\\left(\\frac{1+2}{1-2}\\right) \\star 3\\right) = (-3 \\star 3) = \\frac{-3+3}{-3-3} = 0 \\Longrightarrow \\boxed{0}$"
] |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_12 | null | 58 | For each permutation $a_1,a_2,a_3,\cdots,a_{10}$ of the integers $1,2,3,\cdots,10$ , form the sum
\[|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|.\]
The average value of all such sums can be written in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | [
"Because of symmetry, we may find all the possible values for $|a_n - a_{n - 1}|$ and multiply by the number of times this value appears. Each occurs $5 \\cdot 8!$ , because if you fix $a_n$ and $a_{n + 1}$ there are still $8!$ spots for the others and you can do this $5$ times because there are $5$ places $a_n$ an... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_2 | null | 12 | For each positive integer $k$ , let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$ . For example, $S_3$ is the sequence $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term $2005$ | [
"Suppose that the $n$ th term of the sequence $S_k$ is $2005$ . Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2\\cdot 3\\cdot 167$ . The ordered pairs $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$ $(2,1002)$ $(3,668)$ $(4,501)$ $(6,334)$ $(12,167)$ $(167,12)$ $(334,6)$ $(501,4)$ $(668,3)$ $... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_24 | B | 1 | For each positive integer $n > 1$ , let $P(n)$ denote the greatest prime factor of $n$ . For how many positive integers $n$ is it true that both $P(n) = \sqrt{n}$ and $P(n+48) = \sqrt{n+48}$
$\textbf{(A) } 0\qquad \textbf{(B) } 1\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | [
"If $P(n) = \\sqrt{n}$ , then $n = p_{1}^{2}$ , where $p_{1}$ is a prime number\nIf $P(n+48) = \\sqrt{n+48}$ , then $n + 48$ is a square, but we know that n is $p_{1}^{2}$\nThis means we just have to check for squares of primes, add $48$ and look whether the root is a prime number.\nWe can easily see that the diffe... |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_30 | null | 9 | For each positive integer $n$ , let
Let $k$ denote the smallest positive integer for which the rightmost nonzero digit of $a_k$ is odd. The rightmost nonzero digit of $a_k$ is
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ } 9$ | [
"We have $a_n = n(n+1)\\dots (n+9)$\nThe value $a_n$ can be written as $2^{x_n} 5^{y_n} r_n$ , where $r_n$ is not divisible by 2 and 5. The number of trailing zeroes is $z_n = \\min(x_n,y_n)$ . The last non-zero digit is the last digit of $2^{x_n-z_n} 5^{y_n-z_n} r_n$\nClearly, the last non-zero digit is even iff $... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_22 | null | 8 | For each positive integer $n$ , let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B$ , with no more than three $A$ s in a row and no more than three $B$ s in a row. What is the remainder when $S(2015)$ is divided by $12$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}... | [
"The huge $n$ value in place, as well as the \"no more than... in a row\" are key phrases that indicate recursion is the right way to go. \nLet's go with finding the case of $S(n)$ from previous cases.\nSo how can we make the words of $S(n)$ ? Do we choose 3-in-a-row of one letter, $A$ or $B$ , or do we want $2$ co... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_25 | D | 4 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | [
"It is well-known that $n \\equiv S(n)\\equiv S(S(n)) \\pmod{9}.$ Substituting, we have that \\[n+n+n \\equiv 2007 \\pmod{9} \\implies n \\equiv 0 \\pmod{3}.\\] Since $n \\leq 2007,$ we must have that $\\max S(n)=1+9+9+9=28.$ Now, we list out the possible vales for $S(n)$ in a table, noting that it is a multiple of... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_25 | null | 4 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | [
"Claim. The only positive integers $n$ that satisfy the condition are perfect multiples of $3$\nProof of claim:\nWe examine the positive integers mod $9$ . Here are the cases.\nCase 1. $n \\equiv 1 \\pmod 9$ . Now, we examine $S(n)$ modulo $9$ .\nCase 1.1. The tens digit of $n$ is different from the tens digit of t... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_22 | D | 4 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | [
"It is well-known that $n \\equiv S(n)\\equiv S(S(n)) \\pmod{9}.$ Substituting, we have that \\[n+n+n \\equiv 2007 \\pmod{9} \\implies n \\equiv 0 \\pmod{3}.\\] Since $n \\leq 2007,$ we must have that $\\max S(n)=1+9+9+9=28.$ Now, we list out the possible vales for $S(n)$ in a table, noting that it is a multiple of... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_22 | null | 4 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | [
"Claim. The only positive integers $n$ that satisfy the condition are perfect multiples of $3$\nProof of claim:\nWe examine the positive integers mod $9$ . Here are the cases.\nCase 1. $n \\equiv 1 \\pmod 9$ . Now, we examine $S(n)$ modulo $9$ .\nCase 1.1. The tens digit of $n$ is different from the tens digit of t... |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_14 | null | 109 | For each positive integer $n$ , let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$ . Find the largest value of $n$ for which $f(n) \le 300$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | [
"Observe that $f$ is strictly increasing in $n$ . We realize that we need $100$ terms to add up to around $300$ , so we need some sequence of $2$ s, $3$ s, and then $4$ s.\nIt follows that $n \\approx 100$ (alternatively, use binary search to get to this, with $n\\le 1000$ ). Manually checking shows that $f(109) = ... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_19 | null | 802 | For each positive integer $n$ , let $f(n) = n^4 - 360n^2 + 400$ . What is the sum of all values of $f(n)$ that are prime numbers?
$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$ | [
"To find the answer it was enough to play around with $f$ . One can easily find that $f(1)=41$ is a prime, then $f$ becomes negative for $n$ between $2$ and $18$ , and then $f(19)=761$ is again a prime number. And as $41 + 761 = 802$ is already the largest option, the answer must be $\\boxed{802}$",
"We will now ... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_23 | D | 10 | For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | [
"First, we can test values that would make $f(x)=12$ true. For this to happen $x$ must have $6$ divisors, which means its prime factorization is in the form $pq^2$ or $p^5$ , where $p$ and $q$ are prime numbers. Listing out values less than $50$ which have these prime factorizations, we find $12,18,20,28,44,45,50$ ... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_20 | D | 10 | For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | [
"First, we can test values that would make $f(x)=12$ true. For this to happen $x$ must have $6$ divisors, which means its prime factorization is in the form $pq^2$ or $p^5$ , where $p$ and $q$ are prime numbers. Listing out values less than $50$ which have these prime factorizations, we find $12,18,20,28,44,45,50$ ... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_12 | B | 4,015 | For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008$ th term of the sequence?
$\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064$ | [
"Letting the sum of the sequence equal $a_1+a_2+\\cdots+a_n$ yields the following two equations:\n$\\frac{a_1+a_2+\\cdots+a_{2008}}{2008}=2008$ and\n$\\frac{a_1+a_2+\\cdots+a_{2007}}{2007}=2007$\nTherefore:\n$a_1+a_2+\\cdots+a_{2008}=2008^2$ and $a_1+a_2+\\cdots+a_{2007}=2007^2$\nHence, by substitution, $a_{2008}=2... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_13 | B | 4,015 | For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008^{\text{th}}$ term of the sequence?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$ | [
"Since the mean of the first $n$ terms is $n$ , the sum of the first $n$ terms is $n^2$ . Thus, the sum of the first $2007$ terms is $2007^2$ and the sum of the first $2008$ terms is $2008^2$ . Hence, the $2008^{\\text{th}}$ term of the sequence is $2008^2-2007^2=(2008+2007)(2008-2007)=4015\\Rightarrow \\boxed{4015... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_15 | null | 363 | For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n \equiv 1 \pmod{2^n}.$ Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n = a_{n+1}.$ | [
"Denote $a_n = 23 b_n$ .\nThus, for each $n$ , we need to find smallest positive integer $k_n$ , such that \\[ 23 b_n = 2^n k_n + 1 . \\]\nThus, we need to find smallest $k_n$ , such that \\[ 2^n k_n \\equiv - 1 \\pmod{23} . \\]\nNow, we find the smallest $m$ , such that $2^m \\equiv 1 \\pmod{23}$ .\nBy Fermat's Th... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_11 | null | 955 | For each positive integer $p$ , let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \frac{1}{2}$ . For example, $b(6) = 2$ and $b(23) = 5$ . If $S = \sum_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000. | [
"$\\left(k- \\frac 12\\right)^2=k^2-k+\\frac 14$ and $\\left(k+ \\frac 12\\right)^2=k^2+k+ \\frac 14$ . Therefore $b(p)=k$ if and only if $p$ is in this range, or $k^2-k<p\\leq k^2+k$ . There are $2k$ numbers in this range, so the sum of $b(p)$ over this range is $(2k)k=2k^2$ $44<\\sqrt{2007}<45$ , so all numbers $... |
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_29 | null | 6 | For each positive number $x$ , let $f(x)=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2} {\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}$ .
The minimum value of $f(x)$ is
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }6$ | [
"Let $a = \\left( x + \\frac{1}{x} \\right)^3$ and $b = x^3 + \\frac{1}{x^3}$ . Then \\begin{align*} f(x) &= \\frac{\\left( x + \\frac{1}{x} \\right)^6 - (x^6 + \\frac{1}{x^6}) - 2}{\\left( x + \\frac{1}{x} \\right)^3 + (x^3 + \\frac{1}{x^3})} \\\\ &= \\frac{\\left( x + \\frac{1}{x} \\right)^6 - (x^6 + 2 + \\frac{1... |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_2 | null | 340 | For each real number $x$ , let $\lfloor x \rfloor$ denote the greatest integer that does not exceed x. For how many positive integers $n$ is it true that $n<1000$ and that $\lfloor \log_{2} n \rfloor$ is a positive even integer? | [
"For integers $k$ , we want $\\lfloor \\log_2 n\\rfloor = 2k$ , or $2k \\le \\log_2 n < 2k+1 \\Longrightarrow 2^{2k} \\le n < 2^{2k+1}$ . Thus, $n$ must satisfy these inequalities (since $n < 1000$ ):\nThere are $4$ for the first inequality, $16$ for the second, $64$ for the third, and $256$ for the fourth, so the ... |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_21 | A | 8 | For every $3^\circ$ rise in temperature, the volume of a certain gas expands by $4$ cubic centimeters. If the volume of the gas is $24$ cubic centimeters when the temperature is $32^\circ$ , what was the volume of the gas in cubic centimeters when the temperature was $20^\circ$
$\text{(A)}\ 8 \qquad \text{(B)}\ 12 \qq... | [
"We know that \\[T=32-3k\\Rightarrow V=24-4k.\\] Setting $k=4$ , we get the volume when the temperature is $20^\\circ$ , which is $24-4(4)=8\\rightarrow \\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_13 | null | 59 | For every $m \geq 2$ , let $Q(m)$ be the least positive integer with the following property: For every $n \geq Q(m)$ , there is always a perfect cube $k^3$ in the range $n < k^3 \leq m \cdot n$ . Find the remainder when \[\sum_{m = 2}^{2017} Q(m)\] is divided by 1000. | [
"We claim that $Q(m) = 1$ when $m \\ge 8$\nWhen $m \\ge 8$ , for every $n \\ge Q(m) = 1$ , we need to prove there exists an integer $k$ , such that $n < k^3 \\le m \\cdot n$\nThats because $\\sqrt[3]{m \\cdot n} - \\sqrt[3]{n} \\ge 2\\sqrt[3]{n} - \\sqrt[3]{n} = \\sqrt[3]{n} \\ge 1$ , so k exists between $\\sqrt[3]... |
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