link stringlengths 75 84 | letter stringclasses 5
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https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_25 | C | 2 | Let $f(x)= \sqrt{ax^2+bx}$ . For how many real values of $a$ is there at least one positive value of $b$ for which the domain of $f$ and the range of $f$ are the same set
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} }$ | [
"If $f(x)=y$ , then squaring both sides of the given equation and subtracting $ax^2$ and $bx$ yields $y^2-ax^2-bx=0$ . Completing the square, we get $(x+\\frac{b}{2a})^2-\\frac{y^2}{a}=\\frac{b^2}{4a^2}$ where $y\\geq 0$ . Divide out by $\\frac{b^2}{4a^2}$ to put the equation in the standard form of an ellipse or h... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_7 | null | 21 | Let $f(x)=(x^2+3x+2)^{\cos(\pi x)}$ . Find the sum of all positive integers $n$ for which \[\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.\] | [
"First, let's split it into two cases to get rid of the absolute value sign\n$\\left |\\sum_{k=1}^n\\log_{10}f(k)\\right|=1 \\iff \\sum_{k=1}^n\\log_{10}f(k)=1,-1$\nNow we simplify using product-sum logarithmic identites:\n$\\log_{10}{f(1)}+\\log_{10}{f(2)}+...+\\log_{10}{f(n)}=\\log_{10}{f(1)\\cdot f(2) \\cdot ...... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_20 | C | 3 | Let $f(x)=ax^2+bx+c$ , where $a$ $b$ , and $c$ are integers. Suppose that $f(1)=0$ $50<f(7)<60$ $70<f(8)<80$ $5000k<f(100)<5000(k+1)$ for some integer $k$ . What is $k$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | [
"From $f(1) = 0$ , we know that $a+b+c = 0$\nFrom the first inequality, we get $50 < 49a+7b+c < 60$ . Subtracting $a+b+c = 0$ from this gives us $50 < 48a+6b < 60$ , and thus $\\frac{25}{3} < 8a+b < 10$ . Since $8a+b$ must be an integer, it follows that $8a+b = 9$\nSimilarly, from the second inequality, we get $70 ... |
https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_16 | E | 17 | Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0, 1, 2, \dots , 25 \}$ .
The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:
$\textbf{(A)}\ 25\qquad \textbf{(B)}\ 22\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 17$ | [
"Note that for all polynomials $f(x)$ $f(x + 6) \\equiv f(x) \\pmod 6$\nProof:\nIf $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ , then $f(x+6) = a_n(x+6)^n + a_{n-1}(x+6)^{n-1} +...+ a_0$ . In the second equation, we can use the binomial expansion to expand every term, and then subtract off all terms that have a f... |
https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_19 | B | 2 | Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$ . The sum of the largest and smallest values of $f(x)$ is
$\textbf {(A)}\ 1 \qquad \textbf {(B)}\ 2 \qquad \textbf {(C)}\ 4 \qquad \textbf {(D)}\ 6 \qquad \textbf {(E)}\ \text{none of these}$ | [
"Note that at $x=2,3,4,$ one of the three absolute values is equal to $0.$\nWithout using absolute values, we rewrite $f(x)$ as a piecewise function: \\[f(x) = \\begin{cases} (x-2)+(4-x)-(6-2x) & \\mathrm{if} \\ 2\\leq x<3 \\\\ (x-2)+(4-x)-(2x-6) & \\mathrm{if} \\ 3\\leq x<4 \\\\ (x-2)+(x-4)-(2x-6) & \\mathrm{if} \... |
https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_2 | null | 15 | Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$ | [
"It is best to get rid of the absolute values first.\nUnder the given circumstances, we notice that $|x-p|=x-p$ $|x-15|=15-x$ , and $|x-p-15|=15+p-x$\nAdding these together, we find that the sum is equal to $30-x$ , which attains its minimum value (on the given interval $p \\leq x \\leq 15$ ) when $x=15$ , giving a... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_24 | C | 301 | Let $f_0(x)=x+|x-100|-|x+100|$ , and for $n\geq 1$ , let $f_n(x)=|f_{n-1}(x)|-1$ . For how many values of $x$ is $f_{100}(x)=0$
$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$ | [
"1. Draw the graph of $f_0(x)$ by dividing the domain into three parts. \n2. Apply the recursive rule a few times to find the pattern.\nNote: $f_n(x) = |f_{n-1}(x)| - 10$ is used to enlarge the difference, but the reasoning is the same. \n3. Extrapolate to $f_{100}$ . Notice that the summits start $100$ away from $... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_11 | null | 8 | Let $f_1(x) = \frac23 - \frac3{3x+1}$ , and for $n \ge 2$ , define $f_n(x) = f_1(f_{n-1}(x))$ . The value of $x$ that satisfies $f_{1001}(x) = x-3$ can be expressed in the form $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [
"After evaluating the first few values of $f_k (x)$ , we obtain $f_4(x) = f_1(x) = \\frac{2}{3} - \\frac{3}{3x+1} = \\frac{6x-7}{9x+3}$ . Since $1001 \\equiv 2 \\mod 3$ $f_{1001}(x) = f_2(x) = \\frac{3x+7}{6-9x}$ . We set this equal to $x-3$ , i.e.\n$\\frac{3x+7}{6-9x} = x-3 \\Rightarrow x = \\frac{5}{3}$ . The ans... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_10 | E | 8 | Let $f_n (x) = \text{sin}^n x + \text{cos}^n x.$ For how many $x$ in $[0,\pi]$ is it true that
\[6f_{4}(x)-4f_{6}(x)=2f_{2}(x)?\]
$\text{(A) }2 \qquad \text{(B) }4 \qquad \text{(C) }6 \qquad \text{(D) }8 \qquad \text{(E) more than }8$ | [
"Divide by 2 on both sides to get \\[3f_{4}(x)-2f_{6}(x)=f_{2}(x)\\] Substituting the definitions of $f_{2}(x)$ $f_{4}(x)$ , and $f_{6}(x)$ , we may rewrite the expression as \\[3(\\text{sin}^4{x} + \\text{cos}^4{x}) - 2(\\text{sin}^6{x} + \\text{cos}^6{x}) = 1\\] We now simplify each term separately using some alg... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_21 | A | 226 | Let $f_{1}(x)=\sqrt{1-x}$ , and for integers $n \geq 2$ , let $f_{n}(x)=f_{n-1}(\sqrt{n^2 - x})$ . If $N$ is the largest value of $n$ for which the domain of $f_{n}$ is nonempty, the domain of $f_{N}$ is $[c]$ . What is $N+c$
$\textbf{(A)}\ -226 \qquad \textbf{(B)}\ -144 \qquad \textbf{(C)}\ -20 \qquad \textbf{(D)}\ 20... | [
"The domain of $f_{1}(x)=\\sqrt{1-x}$ is defined when $x\\leq1$ \\[f_{2}(x)=f_{1}\\left(\\sqrt{4-x}\\right)=\\sqrt{1-\\sqrt{4-x}}\\]\nApplying the domain of $f_{1}(x)$ and the fact that square roots must be positive, we get $0\\leq\\sqrt{4-x}\\leq1$ . Simplifying, the domain of $f_{2}(x)$ becomes $3\\leq x\\leq4$\n... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_25 | null | 399 | Let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = z^2 + iz + 1$ . How many complex numbers $z$ are there such that $\text{Im}(z) > 0$ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $10$
$\textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \... | [
"Suppose $f(z)=z^2+iz+1=c=a+bi$ . We look for $z$ with $\\operatorname{Im}(z)>0$ such that $a,b$ are integers where $|a|, |b|\\leq 10$\nFirst, use the quadratic formula:\n$z = \\frac{1}{2} (-i \\pm \\sqrt{-1-4(1-c)}) = -\\frac{i}{2} \\pm \\sqrt{ -\\frac{5}{4} + c }$\nGenerally, consider the imaginary part of a radi... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_16 | A | 1 | Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1<a<b<c.$ What is $g(1)$ in terms of $a,b,$ and $c?$
$\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\fra... | [
"Note that $f(1/x)$ has the same roots as $g(x)$ , if it is multiplied by some monomial so that the leading term is $x^3$ they will be equal. We have \\[f(1/x) = \\frac{1}{x^3} + \\frac{a}{x^2}+\\frac{b}{x} + c\\] so we can see that \\[g(x) = \\frac{x^3}{c}f(1/x)\\] Therefore \\[g(1) = \\frac{1}{c}f(1) = \\boxed{1}... |
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_28 | null | 6 | Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$ | [
"Let $r(x)$ be the remainder when $g(x^{12})$ is divided by $g(x)$ . Then $r(x)$ is the unique polynomial such that \\[g(x^{12}) - r(x) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 - r(x)\\] is divisible by $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1$ , and $\\deg r(x) < 5$\nNote that $(x - 1)(x^5 + x^4 + x^3 + x^2 +... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_23 | D | 8 | Let $h_n$ and $k_n$ be the unique relatively prime positive integers such that \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{h_n}{k_n}.\] Let $L_n$ denote the least common multiple of the numbers $1, 2, 3, \ldots, n$ . For how many integers with $1\le{n}\le{22}$ is $k_n<L_n$
$\textbf{(A) }0 \qquad\text... | [
"We are given that \\[\\sum_{i=1}^{n}\\frac1i = \\frac{1}{L_n}\\sum_{i=1}^{n}\\frac{L_n}{i} = \\frac{h_n}{k_n}.\\] Since $k_n < L_n,$ we need $\\gcd\\left(\\sum_{i=1}^{n}\\frac{L_n}{i}, L_n\\right)>1.$\nFor all primes $p$ such that $p\\leq n,$ let $v_p(L_n)=e\\geq1$ be the largest power of $p$ that is a factor of $... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_25 | E | 8,064 | Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silv... | [
"Consider $f(2)$ . The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$ . Since $100\\le (x+1)^2-x^2=2x+1$ , this first happens at $x\\ge \\lfloor 99/2\\rfloor = 50$ . The per... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_24 | D | 6 | Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$ | [
"$k \\equiv 2008^2 + 2^{2008} \\equiv 8^2 + 2^4 \\equiv 4+6 \\equiv 0 \\pmod{10}$\nSo, $k^2 \\equiv 0 \\pmod{10}$ . Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \\equiv 2^4 \\equiv 6 \\pmod{10}$\nTherefore, $k^2+2^k \\equiv 0+6 \\equiv 6 \\pmo... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_15 | D | 6 | Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$ | [
"$k \\equiv 2008^2 + 2^{2008} \\equiv 8^2 + 2^4 \\equiv 4+6 \\equiv 0 \\pmod{10}$\nSo, $k^2 \\equiv 0 \\pmod{10}$ . Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \\equiv 2^4 \\equiv 6 \\pmod{10}$\nTherefore, $k^2+2^k \\equiv 0+6 \\equiv 6 \\pmo... |
https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_12 | null | 243 | Let $m \ge 3$ be an integer and let $S = \{3,4,5,\ldots,m\}$ . Find the smallest value of $m$ such that for every partition of $S$ into two subsets, at least one of the subsets contains integers $a$ $b$ , and $c$ (not necessarily distinct) such that $ab = c$
Note : a partition of $S$ is a pair of sets $A$ $B$ such that... | [
"We claim that $243$ is the minimal value of $m$ . Let the two partitioned sets be $A$ and $B$ ; we will try to partition $3, 9, 27, 81,$ and $243$ such that the $ab=c$ condition is not satisfied. Without loss of generality , we place $3$ in $A$ . Then $9$ must be placed in $B$ , so $81$ must be placed in $A$ , and... |
https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_25 | D | 8 | Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$ .
The largest integer which divides all possible numbers of the form $m^2-n^2$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$ | [
"First, factor the difference of squares \\[(m+n)(m-n)\\] Since $m$ and $n$ are odd numbers, let $m=2a+1$ and $n=2b+1$ , where $a$ and $b$ can be any integer. \\[(2a+2b+2)(2a-2b)\\] Factor the resulting expression. \\[4(a+b+1)(a-b)\\] If $a$ and $b$ are both even, then $a-b$ is even. If $a$ and $b$ are both odd, t... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_12 | null | 248 | Let $m$ and $n$ be odd integers greater than $1.$ An $m\times n$ rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers $1$ through $n$ , those in the second row are numbered left to right with the integers $n + 1$ through $2n$ , and so on. Square $200$ is in ... | [
"Let us take some cases. Since $m$ and $n$ are odds, and $200$ is in the top row and $2000$ in the bottom, $m$ has to be $3$ $5$ $7$ , or $9$ . Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of $< 1$ . Therefore, $m < 1800 \\mod n < 1800-m$\nIf $m=3$... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_10 | null | 407 | Let $m$ and $n$ be positive integers satisfying the conditions
$\quad\bullet\ \gcd(m+n,210)=1,$
$\quad\bullet\ m^m$ is a multiple of $n^n,$ and
$\quad\bullet\ m$ is not a multiple of $n.$
Find the least possible value of $m+n.$ | [
"Taking inspiration from $4^4 \\mid 10^{10}$ we are inspired to take $n$ to be $p^2$ , the lowest prime not dividing $210$ , or $11 \\implies n = 121$ . Now, there are $242$ factors of $11$ , so $11^{242} \\mid m^m$ , and then $m = 11k$ for $k \\geq 22$ . Now, $\\gcd(m+n, 210) = \\gcd(11+k,210) = 1$ . Noting $k = 2... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_11 | null | 889 | Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$ | [
"For $0 < k < m$ , we have\nThus the product $a_{k}a_{k+1}$ is a monovariant : it decreases by 3 each time $k$ increases by 1. For $k = 0$ we have $a_{k}a_{k+1} = 37\\cdot 72$ , so when $k = \\frac{37 \\cdot 72}{3} = 888$ $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m = \\boxed{889}$ , our a... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_14 | null | 263 | Let $m$ be the largest real solution to the equation
$\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4$
There are positive integers $a$ $b$ , and $c$ such that $m=a+\sqrt{b+\sqrt{c}}$ . Find $a+b+c$ | [
"The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $\\frac{3}{x-3}$ , then the fraction becomes of the form $\\frac{x}{x - 3}$ . A similar cancellation happens with the other four terms. If we assume $x = 0$ is not the highest solution ... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3 | null | 476 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ | [
"The three-digit integers divisible by $17$ , and their digit sum: \\[\\begin{array}{c|c} m & s(m)\\\\ \\hline 102 & 3 \\\\ 119 & 11\\\\ 136 & 10\\\\ 153 & 9\\\\ 170 & 8\\\\ 187 & 16\\\\ 204 & 6\\\\ 221 & 5\\\\ 238 & 13\\\\ 255 & 12\\\\ 272 & 11\\\\ 289 & 19\\\\ 306 & 9\\\\ 323 & 8\\\\ 340 & 7\\\\ 357 & 15\\\\ 374 ... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_6 | null | 750 | Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$ | [
"We can use complementary counting. We can choose a five-element subset in ${14\\choose 5}$ ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 $A$ s and 5 $B$ s, thereby showing that there are ${10\\choose 5... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_9 | null | 0 | Let $m$ be the number of solutions in positive integers to the equation $4x+3y+2z=2009$ , and let $n$ be the number of solutions in positive integers to the equation $4x+3y+2z=2000$ . Find the remainder when $m-n$ is divided by $1000$ | [
"We can avoid computing $m$ and $n$ , instead we will compute $m-n$ directly.\nNote that $4x+3y+2z=2009$ if and only if $4(x-1)+3(y-1)+2(z-1)=2000$ . Hence there is an almost 1-to-1 correspondence between the positive integer solutions of the two equations. The only exceptions are the solutions of the first equatio... |
https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_12 | null | 19 | Let $m$ be the smallest integer whose cube root is of the form $n+r$ , where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$ . Find $n$ | [
"In order to keep $m$ as small as possible, we need to make $n$ as small as possible.\n$m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3$ . Since $r < \\frac{1}{1000}$ and $m - n^3 = r(3n^2 + 3nr + r^2)$ is an integer, we must have that $3n^2 + 3nr + r^2 \\geq \\frac{1}{r} > 1000$ . This means that the smallest possible... |
https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_5 | null | 634 | Let $m/n$ , in lowest terms, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$ . Find $m + n$ | [
"$10^{99} = 2^{99}5^{99}$ , so it has $(99 + 1)(99 + 1) = 10000$ factors. Out of these, we only want those factors of $10^{99}$ which are divisible by $10^{88}$ ; it is easy to draw a bijection to the number of factors that $10^{11} = 2^{11}5^{11}$ has, which is $(11 + 1)(11 + 1) = 144$ . Our probability is $\\frac... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_22 | A | 12 | Let $m>1$ and $n>1$ be integers. Suppose that the product of the solutions for $x$ of the equation
\[8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0\]
is the smallest possible integer. What is $m+n$
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272$ | [
"Rearranging logs, the original equation becomes\n\\[\\frac{8}{\\log n \\log m}(\\log x)^2 - \\left(\\frac{7}{\\log n}+\\frac{6}{\\log m}\\right)\\log x - 2013 = 0\\]\nBy Vieta's Theorem, the sum of the possible values of $\\log x$ is $\\frac{\\frac{7}{\\log n}+\\frac{6}{\\log m}}{\\frac{8}{\\log n \\log m}} = \\fr... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_25 | E | 11 | Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial \[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $... | [
"For a fixed value of $m,$ there is a total of $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1, a_2, a_3, a_4).$\nLet $S=a_1+a_2+a_3+a_4.$ We claim that exactly $\\frac1m$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S:$\nSince $\\gcd(m,4)=1,$ we conclude that \\[\\{k+4(0),k+4(1),k+... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_25 | B | 8,181 | Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$ . For how many values of $n$ is $q+r$ divisible by $11$
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$ | [
"$11|(q+r)$ implies that $11|(99q+q+r)$ and therefore $11|(100q+r)$ , so $11|n$ . Then, $n$ can range from $10010$ to $99990$ for a total of $\\boxed{8181}$ numbers."
] |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_18 | B | 8,181 | Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$ . For how many values of $n$ is $q+r$ divisible by $11$
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$ | [
"When a $5$ -digit number is divided by $100$ , the first $3$ digits become the quotient, $q$ , and the last $2$ digits become the remainder, $r$\nTherefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive.\nFor each of the $9\\cdot10\\cdot10=900$ possible... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_13 | B | 11 | Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\te... | [
"We can start by setting up an equation to convert $\\underline{32d}$ base $n$ to base 10. To convert this to base 10, it would be $3{n}^2+2n+d.$ Because it is equal to 263, we can set this equation to 263. Finally, subtract $d$ from both sides to get $3{n}^2+2n = 263-d$\nWe can also set up equations to convert $\\... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | B | 8 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textb... | [
"A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have:\n\\[\\begin{array}{c|c|c|c|c|c|c} \\mathrm{Factorial}&0!-4!&5!-9!&10!-14!&15!-19!&20!-24!&25!-2... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_7 | E | 84 | Let $n$ be a positive integer such that $\frac 12 + \frac 13 + \frac 17 + \frac 1n$ is an integer. Which of the following statements is not true:
$\mathrm{(A)}\ 2\ \text{divides\ }n \qquad\mathrm{(B)}\ 3\ \text{divides\ }n \qquad\mathrm{(C)}$ $\ 6\ \text{divides\ }n \qquad\mathrm{(D)}\ 7\ \text{divides\ }n \qquad\math... | [
"Since $\\frac 12 + \\frac 13 + \\frac 17 = \\frac {41}{42}$ $0 < \\lim_{n \\rightarrow \\infty} \\left(\\frac{41}{42} + \\frac{1}{n}\\right) < \\frac {41}{42} + \\frac 1n < \\frac{41}{42} + \\frac 11 < 2$\nFrom which it follows that $\\frac{41}{42} + \\frac 1n = 1$ and $n = 42$ . The only answer choice that is no... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_4 | E | 84 | Let $n$ be a positive integer such that $\frac 12 + \frac 13 + \frac 17 + \frac 1n$ is an integer. Which of the following statements is not true:
$\mathrm{(A)}\ 2\ \text{divides\ }n \qquad\mathrm{(B)}\ 3\ \text{divides\ }n \qquad\mathrm{(C)}$ $\ 6\ \text{divides\ }n \qquad\mathrm{(D)}\ 7\ \text{divides\ }n \qquad\math... | [
"Since $\\frac 12 + \\frac 13 + \\frac 17 = \\frac {41}{42}$ $0 < \\lim_{n \\rightarrow \\infty} \\left(\\frac{41}{42} + \\frac{1}{n}\\right) < \\frac {41}{42} + \\frac 1n < \\frac{41}{42} + \\frac 11 < 2$\nFrom which it follows that $\\frac{41}{42} + \\frac 1n = 1$ and $n = 42$ . The only answer choice that is no... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_14 | A | 12 | Let $n$ be the largest integer that is the product of exactly 3 distinct prime numbers $d$ $e$ , and $10d+e$ , where $d$ and $e$ are single digits. What is the sum of the digits of $n$
$\mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24$ | [
"Since we want $n$ to be as large as possible, we would like $d$ in $10d+e$ to be as large as possible. So, $d=7,$ the greatest single-digit prime. Then, $e$ cannot be $5$ because $10(7)+5 = 75,$ which is not prime. So $e = 3$ . Therefore, $d \\cdot e \\cdot (10d+e) = 7 \\cdot 3 \\cdot 73 = 1533$ .\nSo, the sum of ... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_12 | null | 270 | Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$ | [
"As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$ Lifting the Exponent shows that \\[3=v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1\\] so thus, $3^2$ divides $n$ . It also shows that \\[7=v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2\\] so thus, $7^5$ divides $n$\nNow, setting $n = 4c$ (ne... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_24 | C | 18 | Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$ | [
"We know that $\\gcd(n+57,63)=21$ and $\\gcd(n-57, 120)= 60$ by the Euclidean Algorithm. Hence, let $n+57=21\\alpha$ and $n-57=60 \\gamma$ , where $\\gcd(\\alpha,3)=1$ and $\\gcd(\\gamma,2)=1$ . Subtracting the two equations, $38=7\\alpha-20\\gamma$ . Letting $\\gamma = 2s+1$ , we get $58=7\\alpha-40s$ . Taking mod... |
https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_40 | D | 1 | Let $n$ be the number of integer values of $x$ such that $P = x^4 + 6x^3 + 11x^2 + 3x + 31$ is the square of an integer. Then $n$ is:
$\textbf{(A)}\ 4 \qquad \textbf{(B) }\ 3 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 1 \qquad \textbf{(E) }\ 0$ | [
"First, we wish to factor $P$ into a more manageable form. \nFrom the beginning of $P$ , we notice $x^4+6x^3$ , which gives us the idea to use $(x^2+3x)^2=x^4+6x^3+9x^2$\nThis gives us \\[P=(x^2+3x)^2+2x^2+3x+31\\] This is not useful, but it gives us a place to start from.\nWe can then try $(x^2+3x+1)=x^4+6x^3+11x^... |
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7 | null | 196 | Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$ | [
"We want $x_1 +x_2+x_3+x_4 =98$ . This seems like it can be solved with stars and bars, however note that the quadruples all need to be odd. This motivates us to set $x_i= 2y_i +1$ , as for all integers $y_i$ $2y_i + 1$ will be odd. Substituting we get \\[2y_1+2y_2+2y_3+2y_4 +4 = 98 \\implies y_1+y_2+y_3+y_4 =47\\]... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_9 | E | 7 | Let $n$ be the smallest positive integer such that $n$ is divisible by $20$ $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$ | [
"We know that $n^2 = k^3$ and $n^3 = m^2$ . Cubing and squaring the equalities respectively gives $n^6 = k^9 = m^4$ . Let $a = n^6$ . Now we know $a$ must be a perfect $36$ -th power because $lcm(9,4) = 36$ , which means that $n$ must be a perfect $6$ -th power. The smallest number whose sixth power is a multiple o... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_24 | C | 4,944 | Let $n$ denote the smallest positive integer that is divisible by both $4$ and $9,$ and whose base- $10$ representation consists of only $4$ 's and $9$ 's, with at least one of each. What are the last four digits of $n?$
$\textbf{(A) } 4444 \qquad\textbf{(B) } 4494 \qquad\textbf{(C) } 4944 \qquad\textbf{(D) } 9444 \qqu... | [
"For a number to be divisible by $4,$ the last two digits have to be divisible by $4.$ That means the last two digits of this integer must be $4.$\nFor a number to be divisible by $9,$ the sum of all the digits must be divisible by $9.$ The only way to make this happen is with nine $4$ 's. However, we also need one... |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_6 | null | 589 | Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$ | [
"We know that $n^2 = 2^{62}3^{38}$ must have $(62+1)\\times (38+1)$ factors by its prime factorization . If we group all of these factors (excluding $n$ ) into pairs that multiply to $n^2$ , then one factor per pair is less than $n$ , and so there are $\\frac{63\\times 39-1}{2} = 1228$ factors of $n^2$ that are les... |
https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_5 | null | 432 | Let $n^{}_{}$ be the smallest positive integer that is a multiple of $75_{}^{}$ and has exactly $75_{}^{}$ positive integral divisors, including $1_{}^{}$ and itself. Find $\frac{n}{75}$ | [
"The prime factorization of $75 = 3^15^2 = (2+1)(4+1)(4+1)$ . For $n$ to have exactly $75$ integral divisors, we need to have $n = p_1^{e_1-1}p_2^{e_2-1}\\cdots$ such that $e_1e_2 \\cdots = 75$ . Since $75|n$ , two of the prime factors must be $3$ and $5$ . To minimize $n$ , we can introduce a third prime factor, $... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_24 | B | 244 | Let $p$ $q$ , and $r$ be the distinct roots of the polynomial $x^3 - 22x^2 + 80x - 67$ . It is given that there exist real numbers $A$ $B$ , and $C$ such that \[\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \frac{C}{s-r}\] for all $s\not\in\{p,q,r\}$ . What is $\tfrac1A+\tfrac1B+\tfrac1C$
$\text... | [
"Multiplying both sides by $(s-p)(s-q)(s-r)$ yields \\[1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)\\] As this is a polynomial identity, and it is true for infinitely many $s$ , it must be true for all $s$ (since a polynomial with infinitely many roots must in fact be the constant polynomial $0$ ). This means we can... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_17 | A | 7 | Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$ | [
"More generally, let $a,b,c,d,p,$ and $q$ be positive integers such that $bc-ad=1$ and \\[\\frac ab < \\frac pq < \\frac cd.\\] From $\\frac ab < \\frac pq,$ we have $bp-aq>0,$ or \\[bp-aq\\geq1. \\hspace{15mm} (1)\\] From $\\frac pq < \\frac cd,$ we have $cq-dp>0,$ or \\[cq-dp\\geq1. \\hspace{15mm} (2)\\] Since $b... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_13 | null | 110 | Let $p$ be the least prime number for which there exists a positive integer $n$ such that $n^{4}+1$ is divisible by $p^{2}$ . Find the least positive integer $m$ such that $m^{4}+1$ is divisible by $p^{2}$ | [
"If \\(p=2\\), then \\(4\\mid n^4+1\\) for some integer \\(n\\). But \\(\\left(n^2\\right)^2\\equiv0\\) or \\(1\\pmod4\\), so it is impossible. Thus \\(p\\) is an odd prime.\nFor integer \\(n\\) such that \\(p^2\\mid n^4+1\\), we have \\(p\\mid n^4+1\\), hence \\(p\\nmid n^4-1\\), but \\(p\\mid n^8-1\\). By Fermat'... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_21 | C | 8 | Let $p(x) = x^3 + ax^2 + bx + c$ , where $a$ $b$ , and $c$ are complex numbers. Suppose that
What is the number of nonreal zeros of $x^{12} + ax^8 + bx^4 + c$
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12$ | [
"Consider the graph of $x^4$ . It is similar to a parabola, but with a wider \"base\". First examine $x^4-2009$ and $x^4-9002$ . Clearly they are just being translated down some large amount. This will result in the $x$ -axis being crossed twice, indicating $2$ real zeroes. From the Fundamental Theorem of Algebra w... |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_15 | null | 37 | Let $p_{}$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of $5$ heads before one encounters a run of $2$ tails. Given that $p_{}$ can be written in the form $m/n$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n$ | [
"Think of the problem as a sequence of 's and 's. No two 's can occur in a row, so the sequence is blocks of $1$ to $4$ 's separated by 's and ending in $5$ 's. Since the first letter could be or the sequence could start with a block of 's, the total probability is that $3/2$ of it has to start with an\nThe answer ... |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_14 | B | 2.4 | Let $s$ be the limiting sum of the geometric series $4- \frac83 + \frac{16}{9} - \dots$ , as the number of terms increases without bound. Then $s$ equals:
$\textbf{(A)}\ \text{a number between 0 and 1}\qquad\textbf{(B)}\ 2.4\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.6\qquad\textbf{(E)}\ 12$ | [
"The infinite sum of a geometric series with first term $a$ and common ratio $r$ $-1<r<1$ ) is $\\frac{a}{1-r}$ .\nNow, in this geometric series, $a=4$ , and $r=-\\frac23$ . Plugging these into the formula, we get $\\frac4{1-(-\\frac23)}$ , which simplifies to $\\frac{12}5$ , or $\\boxed{2.4}$"
] |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_17 | D | 10 | Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \te... | [
"Let $p, q$ , and $r$ be the roots of the polynomial. Then,\n$p^3 - 5p^2 + 8p - 13 = 0$\n$q^3 - 5q^2 + 8q - 13 = 0$\n$r^3 - 5r^2 + 8r - 13 = 0$\nAdding these three equations, we get\n$(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0$\n$s_3 - 5s_2 + 8s_1 = 39$\n$39$ can be written as $13s_0$ , giving\n... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_2 | null | 21 | Let $u$ and $v$ be integers satisfying $0 < v < u$ . Let $A = (u,v)$ , let $B$ be the reflection of $A$ across the line $y = x$ , let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ i... | [
"Since $A = (u,v)$ , we can find the coordinates of the other points: $B = (v,u)$ $C = (-v,u)$ $D = (-v,-u)$ $E = (v,-u)$ . If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and $ABE$ is a triangle. The area of $BCDE$ is $(2u)(2v) = 4... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_14 | null | 582 | Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers . Find $m+n$ | [
"Define $\\theta = 2\\pi/1997$ . By De Moivre's Theorem the roots are given by\nNow, let $v$ be the root corresponding to $m\\theta=2m\\pi/1997$ , and let $w$ be the root corresponding to $n\\theta=2n\\pi/ 1997$ . Then \\begin{align*} |v+w|^2 &= \\left(\\cos(m\\theta) + \\cos(n\\theta)\\right)^2 + \\left(\\sin(m\\t... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_4 | null | 834 | Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$ | [
"We rewrite $w$ and $z$ in polar form: \\begin{align*} w &= e^{i\\cdot\\frac{\\pi}{6}}, \\\\ z &= e^{i\\cdot\\frac{2\\pi}{3}}. \\end{align*} The equation $i \\cdot w^r = z^s$ becomes \\begin{align*} e^{i\\cdot\\frac{\\pi}{2}} \\cdot \\left(e^{i\\cdot\\frac{\\pi}{6}}\\right)^r &= \\left(e^{i\\cdot\\frac{2\\pi}{3}}\\... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_17 | A | 21 | Let $w$ $x$ $y$ , and $z$ be whole numbers. If $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$ , then what does $2w + 3x + 5y + 7z$ equal?
$\textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56$ | [
"The prime factorization of $588$ is $2^2\\cdot3\\cdot7^2.$ We can see $w=2, x=1,$ and $z=2.$ Because $5^0=1, y=0.$\n\\[2w+3x+5y+7z=4+3+0+14=\\boxed{21}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_7 | null | 100 | Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denote... | [
"Let $w = \\operatorname{cis}{(\\alpha)}$ and $z = 10\\operatorname{cis}{(\\beta)}$ . Then, $\\dfrac{w - z}{z} = \\dfrac{\\operatorname{cis}{(\\alpha)} - 10\\operatorname{cis}{(\\beta)}}{10\\operatorname{cis}{\\beta}}$\nMultiplying both the numerator and denominator of this fraction by $\\operatorname{cis}{(-\\beta... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15 | null | 169 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and ... | [
"Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$\nLet $w_3$ have center $(x,y)$ and radius $r$ . Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$ , and if they are internally tangent, it is $|r_1 - ... |
https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_11 | null | 163 | Let $w_1, w_2, \dots, w_n$ be complex numbers . A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[\sum_{k = 1}^n (z_k - w_k) = 0.\] For the numbers $w_1 = 32 + 170i$ $w_2 = - 7 + 64i$ $w_3 = - 9 + 200i... | [
"We know that\n$\\sum_{k=1}^5 w_k = 3 + 504i$\nAnd because the sum of the 5 $z$ 's must cancel this out,\n$\\sum_{k=1}^5 z_k = 3 + 504i$\nWe write the numbers in the form $a + bi$ and we know that\n$\\sum_{k=1}^5 a_k = 3$ and $\\sum_{k=1}^5 b_k = 504$\nThe line is of equation $y=mx+3$ . Substituting in the polar co... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_16 | D | 7 | Let $x = .123456789101112....998999$ , where the digits are obtained by writing the integers $1$ through $999$ in order.
The $1983$ rd digit to the right of the decimal point is
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$ | [
"We consider the first $1983$ digits, letting the $1983$ rd digit be $z$ . We can break the string of digits into three segments: let $A$ denote $123456789$ (the $1$ -digit numbers), let $B$ denote $1011...9899$ (the $2$ -digit numbers), and let $C$ denote $100101...z$ (the $3$ -digit numbers). Clearly there are $9... |
https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_1 | null | 60 | Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$ | [
"The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.\n$x^{24}=w$ $y^{40}=w$ , and $(xyz)^{12}=w$ . If we now convert everything to a power of $120$ , it will be easy to isolate $z$ and $w$\n$x^{120}=w^5$ $y^{120}=w^3$ , and $(xyz)^{120}=w^{10}$\nWith... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_18 | B | 31 | Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$ | [
"Substitute $a^cb^d$ into $x$ . We then have $7(a^{5c}b^{5d}) = 11y^{13}$ . Divide both sides by $7$ , and it follows that:\n\\[(a^{5c}b^{5d}) = \\frac{11y^{13}}{7}.\\]\nNote that because $11$ and $7$ are prime, the minimum value of $x$ must involve factors of $7$ and $11$ only. Thus, we try to look for the lowest ... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_18 | D | 31 | Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$ | [
"According to the problem, we have that $x^5$ and $y^{13}$ must be a multiple of both $7$ and $11$ . Thus, in their prime factorisation, there must be $7$ and $11$ . Thus, we have $a=7$ and $b=11$ . Now, let $x=7^c11^d\\implies x^5=7^{5c}11^{5d}\\implies7x^5=7^{5c+1}11^{5d}$ .\nWe can now divide both sides by 11 in... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14 | null | 89 | Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ | [
"The expression we want to find is $2(x^3+y^3) + x^3y^3$\nFactor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x^3+y^3)=945$ , respectively. Dividing the latter by the former equation yields $\\frac{x^2-xy+y^2}{xy} = \\frac{945}{810}$ . Adding 3 to both sides and simplifying yields $\\frac{(x+y)^2}{xy} = \... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_9 | null | 107 | Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$ . The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | [
"Let the equation $\\frac{\\sin x}{\\sin y} = 3$ be equation 1, and let the equation $\\frac{\\cos x}{\\cos y} = \\frac12$ be equation 2.\nHungry for the widely-used identity $\\sin^2\\theta + \\cos^2\\theta = 1$ , we cross multiply equation 1 by $\\sin y$ and multiply equation 2 by $\\cos y$\nEquation 1 then becom... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_24 | E | 154 | Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$ | [
"Let $x = 10a+b, y = 10b+a$ . The given conditions imply $x>y$ , which implies $a>b$ , and they also imply that both $a$ and $b$ are nonzero.\nThen, $x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$\nSince this must be a perfect square, all the exponents in its prime factorization must be even. $... |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_15 | E | 0.5 | Let $x$ be a real number such that $\sec x - \tan x = 2$ . Then $\sec x + \tan x =$
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2 \qquad \textbf{(C)}\ 0.3 \qquad \textbf{(D)}\ 0.4 \qquad \textbf{(E)}\ 0.5$ | [
"$(\\sec x - \\tan x)(\\sec x + \\tan x) = \\sec^{2} x - \\tan^{2} x = 1$ , so $\\sec x + \\tan x = \\boxed{0.5}$",
"Note that $\\sec x - \\tan x = (1-\\sin x)/\\cos x$ and $\\sec x + \\tan x = (1+\\sin x)/\\cos x$ . Let $(1+\\sin x)/\\cos x = y$ . Multiplying, we get $(1-\\sin^{2}x)/\\cos^{2}x = 1$ .Then, $2y = ... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_8 | null | 67 | Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [
"We can substitute $y = \\sin^2{x}$ . Since we know that $\\cos^2{x}=1-\\sin^2{x}$ , we can do some simplification.\nThis yields $y^5+(1-y)^5=\\frac{11}{36}$ . From this, we can substitute again to get some cancellation through binomials. If we let $z=\\frac{1}{2}-y$ , we can simplify the equation to: \\[\\left(\\f... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_20 | null | 16 | Let $x$ be chosen at random from the interval $(0,1)$ . What is the probability that $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ ?
Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$
$\mathrm{(A)}\ \frac 18 \qquad \mathrm{(B)}\ \frac 3{20} \qquad \mathrm{(C)}\ \frac 16 ... | [
"Let $k$ be an arbitrary integer. For which $x$ do we have $\\lfloor\\log_{10}4x\\rfloor = \\lfloor\\log_{10}x\\rfloor = k$\nThe equation $\\lfloor\\log_{10}x\\rfloor = k$ can be rewritten as $10^k \\leq x < 10^{k+1}$ . The second one gives us $10^k \\leq 4x < 10^{k+1}$ . Combining these, we get that both hold at t... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_19 | B | 13 | Let $x$ be the least real number greater than $1$ such that $\sin(x)= \sin(x^2)$ , where the arguments are in degrees. What is $x$ rounded up to the closest integer?
$\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$ | [
"The smallest $x$ to make $\\sin(x) = \\sin(x^2)$ would require $x=x^2$ , but since $x$ needs to be greater than $1$ , these solutions are not valid.\nThe next smallest $x$ would require $x=180-x^2$ , or $x^2+x=180$\nAfter a bit of guessing and checking, we find that $12^2+12=156$ , and $13^2+13=182$ , so the solut... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_11 | D | 3 | Let $x$ be the number \[0.\underbrace{0000...0000}_{1996\text{ zeros}}1,\] where there are 1996 zeros after the decimal point. Which of the following expressions represents the largest number?
$\text{(A)}\ 3+x \qquad \text{(B)}\ 3-x \qquad \text{(C)}\ 3\cdot x \qquad \text{(D)}\ 3/x \qquad \text{(E)}\ x/3$ | [
"Estimate each of the options.\n$A$ will give a number that is just over $3$\n$B$ will give a number that is just under $3$ . This eliminates $B$ , because $A$ is bigger.\n$C$ will give a number that is barely over $0$ , since it is three times a tiny number. This eliminates $C$ , because $A$ is bigger.\n$D$ will... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_15 | null | 33 | Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ ar... | [
"First, let define a triangle with side lengths $\\sqrt{2x}$ $\\sqrt{2z}$ , and $l$ , with altitude from $l$ 's equal to $\\sqrt{xz}$ $l = \\sqrt{2x - xz} + \\sqrt{2z - xz}$ , the left side of one equation in the problem.\nLet $\\theta$ be angle opposite the side with length $\\sqrt{2x}$ . Then the altitude has len... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_9 | null | 49 | Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | [
"Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value (that isn't 0, of course), so to simplify the problem let us assume without loss of generality that \\[2\\log_{x}(2y) = 2\\log_{2x}(4z) = \\log_{2x^4}(8yz) = 2.\\] Then \\begin{align*} ... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_4 | null | 33 | Let $x,y$ and $z$ be positive real numbers that satisfy the following system of equations: \[\log_2\left({x \over yz}\right) = {1 \over 2}\] \[\log_2\left({y \over xz}\right) = {1 \over 3}\] \[\log_2\left({z \over xy}\right) = {1 \over 4}\] Then the value of $\left|\log_2(x^4y^3z^2)\right|$ is $\tfrac{m}{n}$ where $m$ ... | [
"Denote $\\log_2(x) = a$ $\\log_2(y) = b$ , and $\\log_2(z) = c$\nThen, we have: $a-b-c = \\frac{1}{2}$ $-a+b-c = \\frac{1}{3}$ $-a-b+c = \\frac{1}{4}$\nNow, we can solve to get $a = \\frac{-7}{24}, b = \\frac{-9}{24}, c = \\frac{-5}{12}$ . Plugging these values in, we obtain $|4a + 3b + 2c| = \\frac{25}{8} \\impl... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_3 | null | 265 | Let $x,y,$ and $z$ be real numbers satisfying the system \begin{align*} \log_2(xyz-3+\log_5 x)&=5,\\ \log_3(xyz-3+\log_5 y)&=4,\\ \log_4(xyz-3+\log_5 z)&=4.\\ \end{align*} Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$ | [
"First, we get rid of logs by taking powers: $xyz-3+\\log_5 x=2^{5}=32$ $xyz-3+\\log_5 y=3^{4}=81$ , and $(xyz-3+\\log_5 z)=4^{4}=256$ . Adding all the equations up and using the $\\log {xy}=\\log {x}+\\log{y}$ property, we have $3xyz+\\log_5{xyz} = 378$ , so we have $xyz=125$ . Solving for $x,y,z$ by substituting ... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 | null | 273 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | [
"We first subtract the second equation from the first, noting that they both equal $60$ \\begin{align*} xy+4z-yz-4x&=0 \\\\ 4(z-x)-y(z-x)&=0 \\\\ (z-x)(4-y)&=0 \\end{align*}\nCase 1: Let $y=4$\nThe first and third equations simplify to: \\begin{align*} x+z&=15 \\\\ xz&=44 \\end{align*} from which it is apparent tha... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_3 | D | 4,032 | Let $x=-2016$ . What is the value of $\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$ | [
"Substituting carefully, $\\Bigg\\vert\\Big\\vert 2016-(-2016)\\Big\\vert-2016\\Bigg\\vert-(-2016)$\nbecomes $|4032-2016|+2016=2016+2016=4032$ which is $\\boxed{4032}$",
"Solution by e_power_pi_times_i\nSubstitute $-y = x = -2016$ into the equation. Now, it is $\\Bigg\\vert\\Big\\vert |y|+y\\Big\\vert-|y|\\Bigg\\... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_3 | D | 4,032 | Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$ | [
"By: dragonfly\nFirst of all, lets plug in all of the $x$ 's into the equation.\n$\\bigg|$ $||-2016|-(-2016)|-|-2016|$ $\\bigg|$ $-(-2016)$\nThen we simplify to get\n$\\bigg|$ $|2016+2016|-2016$ $\\bigg|$ $+2016$\nwhich simplifies into\n$\\bigg|$ $2016$ $\\bigg|$ $+2016$\nand finally we get $\\boxed{4032}$",
"Con... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_7 | null | 125 | Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $(x+1)^{48}$ | [
"We note that in general,\nIt now becomes apparent that if we multiply the numerator and denominator of $\\frac{4}{ (\\sqrt{5}+1) (\\sqrt[4]{5}+1) (\\sqrt[8]{5}+1) (\\sqrt[16]{5}+1) }$ by $(\\sqrt[16]{5} - 1)$ , the denominator will telescope to $\\sqrt[1]{5} - 1 = 4$ , so\nIt follows that $(x + 1)^{48} = (\\sqrt[1... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | null | 241 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | [
"Note that $\\frac{\\sum_{n=1}^{44} \\cos n}{\\sum_{n=1}^{44} \\sin n} = \\frac{\\sum_{n=1}^{44} \\cos n}{\\sum_{n=46}^{89} \\cos n} = \\frac {\\cos 1 + \\cos 2 + \\dots + \\cos 44}{\\cos 89 + \\cos 88 + \\dots + \\cos 46}$ by the cofunction identities.(We could have also written it as $\\frac{\\sum_{n=1}^{44} \\co... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_25 | A | 6 | Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qq... | [
"In binary numbers, we have \\[S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \\ldots x_{2} x_{1} x_{0})_2.\\] It follows that \\[8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \\ldots x_{2} x_{1} x_{0}000)_2.\\] We obtain $7S_n$ by subtracting the equations: \\[\\begin{array}{clccrccccccr} & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_23 | A | 6 | Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qq... | [
"In binary numbers, we have \\[S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \\ldots x_{2} x_{1} x_{0})_2.\\] It follows that \\[8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \\ldots x_{2} x_{1} x_{0}000)_2.\\] We obtain $7S_n$ by subtracting the equations: \\[\\begin{array}{clccrccccccr} & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n... |
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_2 | B | 3 | Let $x_1$ and $x_2$ be such that $x_1\not=x_2$ and $3x_i^2-hx_i=b$ $i=1, 2$ . Then $x_1+x_2$ equals
$\mathrm{(A)\ } -\frac{h}{3} \qquad \mathrm{(B) \ }\frac{h}{3} \qquad \mathrm{(C) \ } \frac{b}{3} \qquad \mathrm{(D) \ } 2b \qquad \mathrm{(E) \ }-\frac{b}{3}$ | [
"Notice that $x_1$ and $x_2$ are the distinct solutions to the quadratic $3x^2-hx-b=0$ . By Vieta, the sum of the roots of this quadratic is the negation of the coefficient of the linear term divided by the coefficient of the quadratic term, so in this case $-\\frac{-h}{3}=\\frac{h}{3}, \\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_28 | E | 7 | Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$ (ii) $x_1 + \cdots + x_n = 19$ ; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$ .
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$ , respectively. Then $\frac Mm =$
$\ma... | [
"Let $a =$ number of $2$ s, $b =$ number of $1$ s, $c =$ number of $-1$\n\\[4a+b+c=99\\] \\[2a+b-c=19\\]\nMultiplying the second equation by $2$ gives $4a+2b-2c=38$ . By subtracting this equation from the first equation we get $3c-b=61$ $3c=61+b$ . As we need to minimize the value of $c$ $c = 21$ $b = 2$ $a = 19$\n... |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_28 | null | 7 | Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$ (ii) $x_1 + \cdots + x_n = 19$ ; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$ .
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$ , respectively. Then $\frac Mm =$
$\ma... | [
"Clearly, we can ignore the possibility that some $x_i$ are zero, as adding/removing such variables does not change the truth value of any condition, nor does it change the value of the sum of cubes. Thus we'll only consider $x_i\\in\\{-1,1,2\\}$\nAlso, order of the $x_i$ does not matter, so we are only interested ... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9 | null | 2 | Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$ | [
"Substituting $n$ for $2014$ , we get \\[\\sqrt{n}x^3 - (1+2n)x^2 + 2 = \\sqrt{n}x^3 - x^2 - 2nx^2 + 2\\] \\[= x^2(\\sqrt{n}x - 1) - 2(nx^2 - 1) = 0\\] Noting that $nx^2 - 1$ factors as a difference of squares to \\[(\\sqrt{n}x - 1)(\\sqrt{n}x+1)\\] we can factor the left side as \\[(\\sqrt{n}x - 1)(x^2 - 2(\\sqrt{... |
https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_1 | null | 384 | Let $x_1=97$ , and for $n>1$ , let $x_n=\frac{n}{x_{n-1}}$ . Calculate the product $x_1x_2x_3x_4x_5x_6x_7x_8$ | [
"Since $x_n=\\frac{n}{x_{n-1}}$ $x_n \\cdot x_{n - 1} = n$ . Setting $n = 2, 4, 6$ and $8$ in this equation gives us respectively $x_1x_2 = 2$ $x_3x_4 = 4$ $x_5x_6 = 6$ and $x_7x_8 = 8$ so \\[x_1x_2x_3x_4x_5x_6x_7x_8 = 2\\cdot4\\cdot6\\cdot8 = \\boxed{384}.\\] Notice that the value of $x_1$ was completely unneeded... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_6 | null | 841 | Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. F... | [
"To find the greatest value of $x_{76} - x_{16}$ $x_{76}$ must be as large as possible, and $x_{16}$ must be as small as possible. If $x_{76}$ is as large as possible, $x_{76} = x_{77} = x_{78} = \\dots = x_{100} > 0$ . If $x_{16}$ is as small as possible, $x_{16} = x_{15} = x_{14} = \\dots = x_{1} < 0$ . The other... |
https://artofproblemsolving.com/wiki/index.php/1992_AHSME_Problems/Problem_12 | null | 4 | Let $y=mx+b$ be the image when the line $x-3y+11=0$ is reflected across the $x$ -axis. The value of $m+b$ is
$\text{(A) -6} \quad \text{(B) } -5\quad \text{(C) } -4\quad \text{(D) } -3\quad \text{(E) } -2$ | [
"$\\fbox{C}$ First we want to put this is slope-intercept form, so we get $y=\\dfrac{1}{3}x+\\dfrac{11}{3}$ . When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since $m+b$ is the sum of the slope and the y... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_18 | A | 2 | Let $z$ be a complex number satisfying $12|z|^2=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\frac 6z?$
$\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4$ | [
"Using the fact $z\\bar{z}=|z|^2$ , the equation rewrites itself as \\begin{align*} 12z\\bar{z}&=2(z+2)(\\bar{z}+2)+(z^2+1)(\\bar{z}^2+1)+31 \\\\ -12z\\bar{z}+2z\\bar{z}+4(z+\\bar{z})+8+z^2\\bar{z}^2+(z^2+\\bar{z}^2)+32&=0 \\\\ \\left((z^2+2z\\bar{z}+\\bar{z}^2)+4(z+\\bar{z})+4\\right)+\\left(z^2\\bar{z}^2-12z\\bar... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_10 | null | 147 | Let $z$ be a complex number with $|z|=2014$ . Let $P$ be the polygon in the complex plane whose vertices are $z$ and every $w$ such that $\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}$ . Then the area enclosed by $P$ can be written in the form $n\sqrt{3}$ , where $n$ is an integer. Find the remainder when $n$ is divided by $10... | [
"I find that generally, the trick to these kinds of AIME problems is to interpret the problem geometrically, and that is just what I did here. Looking at the initial equation, this seems like a difficult task, but rearranging yields a nicer equation: \\[\\frac1{z+w}=\\frac1z+\\frac1w\\] \\[\\frac w{z+w}=\\frac wz+... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_6 | null | 125 | Let $z=a+bi$ be the complex number with $\vert z \vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$ . Find $c+d$ | [
"Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have\n$|z^3| * |z^2 - (1+2i)|$\n$=|z|^3 * |z^2 - (1+2i)|$\n$=125|z^2 - (1+2i)|$\nThus we only need to maximize the value of $|z^2 - (1+2i)|$\nTo maximize this value, we must have that $z^2$ is in ... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_8 | null | 784 | Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ | [
"The twelve dots above represent the $12$ roots of the equation $z^{12}-2^{36}=0$ . If we write $z=a+bi$ , then the real part of $z$ is $a$ and the real part of $iz$ is $-b$ . The blue dots represent those roots $z$ for which the real part of $z$ is greater than the real part of $iz$ , and the red dots represent th... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_10 | null | 56 | Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$ | [
"Now that we’ve proven this fact, we know that all four points lie on a circle (intuitively one can also observe this because $z=z_1$ and $z=z_2$ satisfy the property in the question, and $z=z_3$ techincally gives no imaginary part), so let’s draw that in;\n\nWhile $Z_1, Z_2, Z_3$ are fixed, $Z$ can be anywhere on ... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22 | B | 512 | Let ( $a_1$ $a_2$ , ... $a_{10}$ ) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,... | [
"If we have 1 as the first number, then the only possible list is $(1,2,3,4,5,6,7,8,9,10)$\nIf we have 2 as the first number, then we have 9 ways to choose where the $1$ goes, and the numbers ascend from the first number, $2$ , with the exception of the $1$ .\nFor example, $(2,3,1,4,5,6,7,8,9,10)$ , or $(2,3,4,1,5,... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_5 | null | 80 | Let ABCDEF be a convex equilateral hexagon in which all pairs of opposite sides are parallel. The triangle whose sides are extensions of segments AB, CD, and EF has side lengths 200, 240, and 300. Find the side length of the hexagon. | [
"(Sorry i have zero idea how to make drawings)\nDraw a good diagram!\nLet $AB \\cap DC$ $CD \\cap FE$ , and $BA \\cap EF$ be P, Q, and R, respectively. Let $QR=200, RP=300, PQ=240$ . Notice that all smaller triangles formed are all similar to the larger $(200,240,300)$ triangle. Let the side length of the hexagon b... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_4 | A | 92 | Let X and Y be the following sums of arithmetic sequences:
\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}
What is the value of $Y - X?$
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$ | [
"We see that both sequences have equal numbers of terms, so reformat the sequence to look like:\n\\begin{align*} Y = \\ &12 + 14 + \\cdots + 100 + 102\\\\ X = 10 \\ + \\ &12 + 14 + \\cdots + 100\\\\ \\end{align*} From here it is obvious that $Y - X = 102 - 10 = \\boxed{92}$",
"\\begin{align*} X&=10+12+14+\\cdots... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_4 | null | 92 | Let X and Y be the following sums of arithmetic sequences:
\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}
What is the value of $Y - X?$
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$ | [
"We see that every number in Y's sequence is two more than every corresponding number in X's sequence. Since there are 46 numbers in each sequence, the difference must be $46\\cdot 2=\\boxed{92}$"
] |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_12 | E | 112 | Let [mathjax]A, M,[/mathjax] and [mathjax]C[/mathjax] be nonnegative integers such that [mathjax]A + M + C=12[/mathjax]. What is the maximum value of [mathjax]A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C[/mathjax]?
[katex] \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D... | [
"It is not hard to see that \\[(A+1)(M+1)(C+1)=\\] \\[AMC+AM+AC+MC+A+M+C+1\\] Since $A+M+C=12$ , we can rewrite this as \\[(A+1)(M+1)(C+1)=\\] \\[AMC+AM+AC+MC+13\\] So we wish to maximize \\[(A+1)(M+1)(C+1)-13\\] Which is largest when all the factors are equal (consequence of AM-GM). Since $A+M+C=12$ , we set $A=M... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_12 | null | 23 | Let \(O=(0,0)\), \(A=\left(\tfrac{1}{2},0\right)\), and \(B=\left(0,\tfrac{\sqrt{3}}{2}\right)\) be points in the coordinate plane. Let \(\mathcal{F}\) be the family of segments \(\overline{PQ}\) of unit length lying in the first quadrant with \(P\) on the \(x\)-axis and \(Q\) on the \(y\)-axis. There is a unique point... | [
"By Furaken \nLet $C = (\\tfrac18,\\tfrac{3\\sqrt3}8)$ This is sus, furaken randomly guessed C and proceeded to prove it works Draw a line through $C$ intersecting the $x$ -axis at $A'$ and the $y$ -axis at $B'$ . We shall show that $A'B' \\ge 1$ , and that equality only holds when $A'=A$ and $B'=B$\nLet $\\theta =... |
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