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/1989_AJHSME_Problems/Problem_20 | D | 14 | The figure may be folded along the lines shown to form a number cube. Three number faces come together at each corner of the cube. What is the largest sum of three numbers whose faces come together at a corner?
[asy] draw((0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,1)--(4,1)--(4,0)--(2,0)--(2,-1)--(1,-1)--(1,0)--cycle); dr... | [
"It is clear that $6$ $5$ , and $4$ will not come together to get a sum of $15$\nThe faces $6$ $5$ , and $3$ come together at a common vertex, making the maximal sum $6+5+3=14\\rightarrow \\boxed{14}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_15 | E | 761 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13... | [
"The sequence $\\{ d_n\\}$ goes $1, 5, 13, 25, 41,\\dots$ . The first finite differences go $4, 8, 12, 16, \\dots$ . The second finite differences go $4, 4, 4, \\dots$ , so we see that the second finite difference is constant. Thus, $d_n$ can be represented as a quadratic, $d_n = an^2 + bn + c$ . However, we alread... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_15 | null | 761 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13... | [
"Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$ th triangular number (i.e. the diamonds within the triangles or between the diagonals) and $4(n-1)+1$ (the diamonds on sides of the t... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_11 | E | 761 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13... | [
"The sequence $\\{ d_n\\}$ goes $1, 5, 13, 25, 41,\\dots$ . The first finite differences go $4, 8, 12, 16, \\dots$ . The second finite differences go $4, 4, 4, \\dots$ , so we see that the second finite difference is constant. Thus, $d_n$ can be represented as a quadratic, $d_n = an^2 + bn + c$ . However, we alread... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_11 | null | 761 | The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13... | [
"Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$ th triangular number (i.e. the diamonds within the triangles or between the diagonals) and $4(n-1)+1$ (the diamonds on sides of the t... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_10 | A | 1,979 | The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8$ . In what year was Samantha born?
$\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$ | [
"The seventh AMC 8 would have been given in $1991$ . If Samantha was 12 then, that means she was born 12 years ago, so she was born in $1991-12=1979$\nOur answer is $\\boxed{1979}$ corrections made by DrDominic",
"Since she was 12 when she took the seventh AMC 8, she should be $12-6=6$ years old when the first AM... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_10 | A | 8,041 | The first four terms of an arithmetic sequence are $p$ $9$ $3p-q$ , and $3p+q$ . What is the $2010^\text{th}$ term of this sequence?
$\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049$ | [
"$3p-q$ and $3p+q$ are consecutive terms, so the common difference is $(3p+q)-(3p-q) = 2q$\n\\begin{align*}p+2q &= 9\\\\ 9+2q &= 3p-q\\\\ q&=2\\\\ p&=5\\end{align*}\nThe common difference is $4$ . The first term is $5$ and the $2010^\\text{th}$ term is\n\\[5+4(2009) = \\boxed{8041}\\]",
"Since all the answer choi... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_11 | E | 250 | The first term of a sequence is $2005$ . Each succeeding term is the sum of the cubes of the digits of the previous term. What is the ${2005}^{\text{th}}$ term of the sequence?
$\textbf{(A) } 29 \qquad \textbf{(B) } 55 \qquad \textbf{(C) } 85 \qquad \textbf{(D) } 133 \qquad \textbf{(E) } 250$ | [
"Performing this operation several times yields the results of $133$ for the second term, $55$ for the third term, and $250$ for the fourth term. The sum of the cubes of the digits of $250$ equal $133$ , a complete cycle. The cycle is, excluding the first term, the $2^{\\text{nd}}$ $3^{\\text{rd}}$ , and $4^{\\text... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_10 | E | 250 | The first term of a sequence is $2005$ . Each succeeding term is the sum of the cubes of the digits of the previous term. What is the ${2005}^{\text{th}}$ term of the sequence?
$\textbf{(A) } 29 \qquad \textbf{(B) } 55 \qquad \textbf{(C) } 85 \qquad \textbf{(D) } 133 \qquad \textbf{(E) } 250$ | [
"Performing this operation several times yields the results of $133$ for the second term, $55$ for the third term, and $250$ for the fourth term. The sum of the cubes of the digits of $250$ equal $133$ , a complete cycle. The cycle is, excluding the first term, the $2^{\\text{nd}}$ $3^{\\text{rd}}$ , and $4^{\\text... |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_3 | B | 0 | The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$ , in the order shown. The value of $x$ is:
$\textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined}$ | [
"Let $y$ represent the common difference between the terms. We have $(x+1)-y=(x-1)\\implies y=2$\nSubstituting gives us $(2x+3)-2=(x+1)\\implies 2x+1=x+1\\implies x=0$\nTherefore, our answer is $\\boxed{0}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_7 | null | 502 | The first three terms of an arithmetic sequence are $2x - 3$ $5x - 11$ , and $3x + 1$ respectively. The $n$ th term of the sequence is $2009$ . What is $n$
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 502 \qquad \textbf{(C)}\ 1004 \qquad \textbf{(D)}\ 1506 \qquad \textbf{(E)}\ 8037$ | [
"As this is an arithmetic sequence, the difference must be constant: $(5x-11) - (2x-3) = (3x+1) - (5x-11)$ . This solves to $x=4$ . The first three terms then are $5$ $9$ , and $13$ . In general, the $n$ th term is $1+4n$ . Solving $1+4n=2009$ , we get $n=\\boxed{502}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_25 | A | 0 | The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$ . For $n\ge1$
What is $|a_{2009}|$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3$ | [
"Consider another sequence $\\{\\theta_1, \\theta_2, \\theta_3...\\}$ such that $a_n = \\tan{\\theta_n}$ , and $0 \\leq \\theta_n < 180$\nThe given recurrence becomes\nIt follows that $\\theta_{n + 2} \\equiv \\theta_{n + 1} + \\theta_{n} \\pmod{180}$ . Since $\\theta_1 = 45, \\theta_2 = 30$ , all terms in the sequ... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_9 | E | 11 | The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\tfrac{a-\sqrt{2}}{b}$ , where $a$ and $b$ are positive integers. What i... | [
"Let $s$ be the side length of the small squares.\nThe diagonal of the big square can be written in two ways: $\\sqrt{2}$ and $s \\sqrt{2} + s + s \\sqrt{2}$\nSolving for $s$ , we get $s = \\frac{4 - \\sqrt{2}}{7}$ , so our answer is $4 + 7 \\Rightarrow \\boxed{11}$",
"The diagonal of the small square can be writ... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_21 | A | 7 | The five solutions to the equation \[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$ . The eccentricity of $\mathcal E$ ... | [
"The solutions to this equation are $z = 1$ $z = -1 \\pm i\\sqrt 3$ , and $z = -2\\pm i\\sqrt 2$ . Consider the five points $(1,0)$ $\\left(-1,\\pm\\sqrt 3\\right)$ , and $\\left(-2,\\pm\\sqrt 2\\right)$ ; these are the five points which lie on $\\mathcal E$ . Note that since these five points are symmetric about t... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_12 | C | 24,000 | The five tires of a car (four road tires and a full-sized spare) were rotated so that each tire was used the same number of miles during the first $30,000$ miles the car traveled. For how many miles was each tire used?
$\text{(A)}\ 6000 \qquad \text{(B)}\ 7500 \qquad \text{(C)}\ 24,000 \qquad \text{(D)}\ 30,000 \qquad... | [
"In the $30,000$ miles, four tires were always used at one time, so the amount of miles the five tires were used in total is $30,000 \\times 4=120,000$ . Five tires were used and each was used equally, so each tire was used for $\\frac{120,000}{5}=\\boxed{24,000}$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_14 | null | 608 | The following analog clock has two hands that can move independently of each other. [asy] unitsize(2cm); draw(unitcircle,black+linewidth(2)); for (int i = 0; i < 12; ++i) { draw(0.9*dir(30*i)--dir(30*i)); } for (int i = 0; i < 4; ++i) { ... | [
"This problem is, in essence, the following: A $12\\times12$ coordinate grid is placed on a (flat) torus; how many loops are there that pass through each point while only moving up or right? In other words, Felix the frog starts his journey at $(0,0)$ in the coordinate plane. Every second, he jumps either to the... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_6 | B | 4 | The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? [asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.... | [
"We first notice that the median name will be the $(19+1)/2=10^{\\mbox{th}}$ name. The $10^{\\mbox{th}}$ name is $\\boxed{4}$",
"To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us $7 + 3 + 1 + 4 + 4 = 19$ . Thus the index of the median length would be the 1... |
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_22 | D | 3 | The following four statements, and only these are found on a card: [asy] pair A,B,C,D,E,F,G; A=(0,1); B=(0,5); C=(11,5); D=(11,1); E=(0,4); F=(0,3); G=(0,2); draw(A--B--C--D--cycle); label("On this card exactly one statement is false.", B, SE); label("On this card exactly two statements are false.", E, SE); label("On t... | [
"There can be at most one true statement on the card, eliminating $\\textbf{(A)}, \\textbf{(B)},$ and $\\textbf{(C)}$ . If there are $0$ true on the card, statement $4$ (\"On this card exactly four statements are false\") will be correct, causing a contradiction. Therefore, the answer is $\\boxed{3}$ , since $3$ ar... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_5 | null | 539 | The formula for converting a Fahrenheit temperature $F$ to the corresponding Celsius temperature $C$ is $C = \frac{5}{9}(F-32).$ An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer
For how many integer Fahrenhe... | [
"Examine $F - 32$ modulo 9.\nGeneralizing this, we define that $9x + k = F - 32$ . Thus, $F = \\left[\\frac{9}{5}\\left[\\frac{5}{9}(9x + k)\\right] + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5}(5x + \\left[\\frac{5}{9}k\\right]) + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5} \\left[\\frac{5}{9}k \\r... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_23 | B | 883 | The fraction
\[\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},\]
where $n$ is the length of the period of the repeating decimal expansion. What is the sum $b_0+b_1+\cdots+b_{n-1}$
$\textbf{(A) }874\qquad \textbf{(B) }883\qquad \textbf{(C) }887\qquad \textbf{(D) }891\qquad \textbf{(E) }892\qquad$ | [
"$\\frac{1}{99^2}\\\\\\\\ =\\frac{1}{99} \\cdot \\frac{1}{99}\\\\\\\\ =\\frac{0.\\overline{01}}{99}\\\\\\\\ =0.\\overline{00010203...9799}$\nSo, the answer is $0+0+0+1+0+2+0+3+...+9+7+9+9=2\\cdot10\\cdot\\frac{9\\cdot10}{2}-(9+8)$ or $\\boxed{883}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10A_Problems/Problem_7 | E | 9 | The fraction
\[\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}\] simplifies to which of the following?
$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9$ | [
"Using Difference of Squares, $\\frac{(3^{2008})^{2}-(3^{2006})^{2}}{(3^{2007})^{2}-(3^{2005}){^2}}$ becomes\n$\\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}$\n$= \\frac{3^{2006}(9+1) \\cdot 3^{2006}(9-1)}{3^{2005}(9+1) \\cdot 3^{2005}(9-1)}$\n$= \\boxed{9}$"
] |
https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_8 | null | 364 | The function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following properties: \[f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).\] Calculate $f(14,52)$ | [
"Let $z = x+y$ . By the substitution $z=x+y,$ we rewrite the third property in terms of $x$ and $z,$ then solve for $f(x,z):$ \\begin{align*} zf(x,z-x) &= (z-x)f(x,z) \\\\ f(x,z) &= \\frac{z}{z-x} \\cdot f(x,z-x). \\end{align*} Using the properties of $f,$ we have \\begin{align*} f(14,52) &= \\frac{52}{38} \\cdot f... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_12 | null | 58 | The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ $b$ $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ | [
"From $f(f(x))=x$ , it is obvious that $\\frac{-d}{c}$ is the value not in the range. First notice that since $f(0)=\\frac{b}{d}$ $f(\\frac{b}{d})=0$ which means $a(\\frac{b}{d})+b=0$ so $a=-d$ . Using $f(19)=19$ , we have that $b=361c+38d$ ; on $f(97)=97$ we obtain $b=9409c+194d$ . Solving for $d$ in terms of $c$ ... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_18 | E | 11 | The function $f$ has the property that for each real number $x$ in its domain, $1/x$ is also in its domain and
$f(x)+f\left(\frac{1}{x}\right)=x$
What is the largest set of real numbers that can be in the domain of $f$
$\mathrm{(A) \ } \{x|x\ne 0\}\qquad \mathrm{(B) \ } \{x|x<0\}$
$\mathrm{(C) \ } \{x|x>0\}$ $\mathrm{(... | [
"We know that $f(x) + f \\left(\\frac{1}{x}\\right) = x.$ Plugging in $x = \\frac{1}{x}$ we get \\[f \\left(\\frac{1}{x}\\right) + f \\left(\\frac{1}{\\frac{1}{x}}\\right) = \\frac{1}{x}\\] \\[f \\left(\\frac{1}{x}\\right) + f(x) = \\frac{1}{x}.\\]\nAlso notice \\[f \\left(\\frac{1}{x}\\right) + f(x) = x\\] by the ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_2 | B | 2 | The function $f$ is given by the table
\[\begin{tabular}{|c||c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 4 & 1 & 3 & 5 & 2 \\ \hline \end{tabular}\]
If $u_0=4$ and $u_{n+1} = f(u_n)$ for $n \ge 0$ , find $u_{2002}$
$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }3 \qquad \text{(D) }4 \qquad \te... | [
"We can guess that the series given by the problem is periodic in some way. Starting off, $u_0=4$ is given. $u_1=u_{0+1}=f(u_0)=f(4)=5,$ so $u_1=5.$ $u_2=u_{1+1}=f(u_1)=f(5)=2,$ so $u_2=2.$ $u_3=u_{2+1}=f(u_2)=f(2)=1,$ so $u_3=1.$ $u_4=u_{3+1}=f(u_3)=f(1)=4,$ so $u_4=4.$ Plugging in $4$ will give us $5$ as found be... |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_16 | E | 8 | The function $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x$ . If the equation $f(x)=0$ has exactly four distinct real roots , then the sum of these roots is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$ | [
"Let one of the roots be $r_1$ . Also, define $x$ such that $2+x=r_1$ . Thus, we have $f(2+x)=f(r_1)=0$ and $f(2+x)=f(2-x)$ . Therefore, we have $f(2-x)=0$ , and $2-x$ is also a root. Let this root be $r_2$ . The sum $r_1+r_2=2+x+2-x=4$ . Similarly, we can let $r_3$ be a root and define $y$ such that $2+y=r_3$ , an... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_3 | null | 561 | The function $f_{}^{}$ has the property that, for each real number $x,\,$
If $f(19)=94,\,$ what is the remainder when $f(94)\,$ is divided by $1000$ | [
"\\begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\\cdots \\\\ &= (94^2-93^2) + (92^2-91^2) +\\cdots+ (22^2-21^2)+ 20^2-f(19) \\\\ &= 94+93+\\cdots+21+400-94 \\\\ &= 4561 \\end{align*}\nSo, the remainder is $\\boxed{561}$",
"Those familiar with triangular numbers and some of their properties... |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_13 | E | 2 | The function $x^2+px+q$ with $p$ and $q$ greater than zero has its minimum value when:
$\textbf{(A) \ }x=-p \qquad \textbf{(B) \ }x=\frac{p}{2} \qquad \textbf{(C) \ }x=-2p \qquad \textbf{(D) \ }x=\frac{p^2}{4q} \qquad$
$\textbf{(E) \ }x=\frac{-p}{2}$ | [
"The minimum value of this parabola is found at its turning point, on the line $\\boxed{2}$ .\nIndeed, the turning point of any function of the form $ax^2+bx+c$ has an x-coordinate of $\\frac{-b}{2a}$ . This can be seen at the average of the quadratic's two roots (whose sum is $\\frac{-b}{a}$ ) or (using calculus) ... |
https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_7 | null | 997 | The function f is defined on the set of integers and satisfies $f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}$
Find $f(84)$ | [
"Define $f^{h} = f(f(\\cdots f(f(x))\\cdots))$ , where the function $f$ is performed $h$ times. We find that $f(84) = f(f(89)) = f^2(89) = f^3(94) = \\ldots f^{y}(1004)$ $1004 = 84 + 5(y - 1) \\Longrightarrow y = 185$ . So we now need to reduce $f^{185}(1004)$\nLet’s write out a couple more iterations of this funct... |
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_20 | C | 125 | The glass gauge on a cylindrical coffee maker shows that there are $45$ cups left when the coffee maker is $36\%$ full. How many cups of coffee does it hold when it is full?
$\text{(A)}\ 80 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 130 \qquad \text{(E)}\ 262$
[asy] draw((5,0)..(0,-1.3)..(-5,0));... | [
"Let the amount of coffee the maker will hold when full be $x$ . Then, \\[.36x=45 \\Rightarrow x=125 \\rightarrow \\boxed{125}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_19 | B | 2.5 | The graph below shows the total accumulated dollars (in millions) spent by the Surf City government during $1988$ . For example, about $.5$ million had been spent by the beginning of February and approximately $2$ million by the end of April. Approximately how many millions of dollars were spent during the summer mon... | [
"Since we want to know how much money is spent in June, July and August, we need the difference between the amount of money spent by the beginning of June and the amount of money spent by the end of August.\nWe estimate these to be about $2.2$ million and $4.8$ million, respectively. The difference is \\[4.8-2.2=2... |
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_23 | C | 2 | The graph of $x^2 + y = 10$ and the graph of $x + y = 10$ meet in two points. The distance between these two points is:
$\textbf{(A)}\ \text{less than 1} \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ \sqrt{2}\qquad \textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{more than 2}$ | [
"We can merge the two equations to create $x^2+y=x+y$ . Using either the quadratic equation or factoring, we get two solutions with $x$ -coordinates $0$ and $1$\nPlugging this into either of the original equations, we get $(0,10)$ and $(1,9)$ . The distance between those two points is $\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_21 | A | 4 | The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$ , where the graph and the line intersect. What is the largest of these values?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$ | [
"The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$\nSince there are $3$ zeros and the function is never negative, all $3$ zeros must be double roots because the function's degree is $6$\nSuppose we let $p$ $q$ , and $r$ be the... |
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_3 | null | 800 | The graph of $y^2 + 2xy + 40|x|= 400$ partitions the plane into several regions. What is the area of the bounded region? | [
"The equation given can be rewritten as:\nWe can split the equation into a piecewise equation by breaking up the absolute value\nFactoring the first one: (alternatively, it is also possible to complete the square\nHence, either $y = -20$ , or $2x = 20 - y \\Longrightarrow y = -2x + 20$\nSimilarily, for the second o... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_19 | D | 6 | The graph of the function $f$ is shown below. How many solutions does the equation $f(f(x))=6$ have?
[asy] size(200); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6); real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; real[] yticks={-6,-5,-4,-3,-2,-1,1,... | [
"First of all, note that the equation $f(t)=6$ has two solutions: $t=-2$ and $t=1$\nGiven an $x$ , let $f(x)=t$ . Obviously, to have $f(f(x))=6$ , we need to have $f(t)=6$ , and we already know when that happens. In other words, the solutions to $f(f(x))=6$ are precisely the solutions to ( $f(x)=-2$ or $f(x)=1$ ).\... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_3 | C | 6 | The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?
[asy] import graph; /* this is a label */ Label f; f.p=fontsize(0); xaxis(-0.9,20,Ticks(f, 5.0, 5.0)); yaxis(-0.9,20, Ticks(f, 22.0,5.0)); // real f(real x... | [
"Suzanna's speed is $\\frac{1}{5}$ . This means she runs $\\frac{1}{5} \\cdot 30 = \\boxed{6}$",
"From the graph, we can see that every $5$ minutes Suzanna goes, her distance increases by $1$ . Since half an hour is $10$ minutes away, she would go $2$ more miles. $4+2$ is $6$ , so the answer is $\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_11 | A | 6 | The graph shows the number of minutes studied by both Asha (black bar) and Sasha (grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?
[asy] size(300); real i; defaultpen(linewidth(0.8)); draw((0,140)--origin--(220,0)); for(i=1;i<13;i=i+1) { draw((0,10*i)--(220,10*i)); } label... | [
"Average the differences between each day. We get $10, -10,\\text{ } 20,\\text{ } 30,-20$ . We find the average of this list to get $\\boxed{6}$",
"This solution may take longer to do than the first solution. In total, Asha studied for 400 minutes a week (80 minutes per day) and Sasha studied for 430 minutes a we... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_3 | C | 70 | The graph shows the price of five gallons of gasoline during the first ten months of the year. By what percent is the highest price more than the lowest price?
[asy] import graph; size(16.38cm); real lsf=2; pathpen=linewidth(0.7); pointpen=black; pen fp = fontsize(10); pointfontpen=fp; real xmin=-1.33,xmax=11.05,ymin=-... | [
"The highest price was in Month 1, which was $17. The lowest price was in Month 3, which was $10. 17 is $\\frac{17}{10}\\cdot100=170\\%$ of 10, and is $170-100=70\\%$ more than 10.\nTherefore, the answer is $\\boxed{70}$"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6 | null | 36 | The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ | [
"Begin by setting $x$ to 0, then set both equations to $h^2=\\frac{2013-j}{3}$ and $h^2=\\frac{2014-k}{2}$ , respectively. Notice that because the two parabolas have to have positive x-intercepts, $h\\ge32$\nWe see that $h^2=\\frac{2014-k}{2}$ , so we now need to find a positive integer $h$ which has positive integ... |
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_46 | B | 1 | The graphs of $2x+3y-6=0, 4x-3y-6=0, x=2$ , and $y=\frac{2}{3}$ intersect in:
$\textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points}$ | [
"We first convert each of the lines into slope-intercept form ( $y = mx + b$ ):\n$2x+3y-6=0 ==> 3y = -2x + 6 ==> y = -\\frac{2}{3}x + 2$\n$4x - 3y - 6 = 0 ==> 4x - 6 = 3y ==> y = \\frac{4}{3}x - 2$\n$x = 2$ stays as is.\n$y = \\frac{2}{3}$ stays as is\nWe can graph the four lines here: [1]\nWhen we do that, the ans... |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_23 | null | 140 | The graphs of $x^2 + y^2 = 4 + 12x + 6y$ and $x^2 + y^2 = k + 4x + 12y$ intersect when $k$ satisfies $a \le k \le b$ , and for no other values of $k$ . Find $b-a$
$\mathrm{(A) \ }5 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ }104 \qquad \mathrm{(D) \ }140 \qquad \mathrm{(E) \ }144$ | [
"Both sets of points are quite obviously circles. To show this, we can rewrite each of them in the form $(x-x_0)^2 + (y-y_0)^2 = r^2$\nThe first curve becomes $(x-6)^2 + (y-3)^2 = 7^2$ , which is a circle centered at $(6,3)$ with radius $7$\nThe second curve becomes $(x-2)^2 + (y-6)^2 = 40+k$ , which is a circle ce... |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_22 | null | 10 | The graphs of $y = -|x-a| + b$ and $y = |x-c| + d$ intersect at points $(2,5)$ and $(8,3)$ . Find $a+c$
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18$ | [
"Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a $45^\\circ$ angle, in the second graph they point upwards. One can easily find out that the only way how to get these graphs to intersect in two points is the one depicted below:\nObviously, the maximum of the fi... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_16 | D | 5 | The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ ... | [
"Setting the first two equations equal to each other, $\\log_3 x = \\log_x 3$\nSolving this, we get $\\left(3, 1\\right)$ and $\\left(\\frac{1}{3}, -1\\right)$\nSimilarly with the last two equations, we get $\\left(3, -1\\right)$ and $\\left(\\frac{1}{3}, 1\\right)$\nNow, by setting the first and third equations eq... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_6 | null | 660 | The graphs of the equations
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed? | [
"We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.\n\nSolving the above equations for $k=\\pm 10$ , we see that the hexagon in question is regular, with side length $\\frac{20}{\\sq... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_8 | C | 10 | The greatest prime number that is a divisor of $16384$ is $2$ because $16384 = 2^{14}$ . What is the sum of the digits of the greatest prime number that is a divisor of $16383$
$\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22$ | [
"We have \\begin{align*} 16383 & = 2^{14} - 1 \\\\ & = \\left( 2^7 + 1 \\right) \\left( 2^7 - 1 \\right) \\\\ & = 129 \\cdot 127 \\\\ \\end{align*}\nSince $129$ is composite, $127$ is the largest prime which can divide $16383$ . The sum of $127$ 's digits is $1+2+7=\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_6 | C | 10 | The greatest prime number that is a divisor of $16384$ is $2$ because $16384 = 2^{14}$ . What is the sum of the digits of the greatest prime number that is a divisor of $16383$
$\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22$ | [
"We have \\begin{align*} 16383 & = 2^{14} - 1 \\\\ & = \\left( 2^7 + 1 \\right) \\left( 2^7 - 1 \\right) \\\\ & = 129 \\cdot 127 \\\\ \\end{align*}\nSince $129$ is composite, $127$ is the largest prime which can divide $16383$ . The sum of $127$ 's digits is $1+2+7=\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20 | D | 8 | The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] un... | [
"The sum of the numbers in each row is $12$ . Consider the second row. In order for the sum of the numbers in this row to equal $12$ , the two shaded numbers must add up to $13$ If two numbers add up to $13$ , one of them must be at least $7$ : If both shaded numbers are no more than $6$ , their sum can be at most... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_2 | A | 2 | The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of $1$ and $2016$ is closest to which integer?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 504 \qquad \textbf{(D)}\ 1008 \qquad \textbf{(E)}\ 2015$ | [
"Since the harmonic mean is $2$ times their product divided by their sum, we get the equation\n$\\frac{2\\times1\\times2016}{1+2016}$\nwhich is then\n$\\frac{4032}{2017}$\nwhich is finally closest to $\\boxed{2}$ .\n-dragonfly"
] |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_8 | null | 799 | The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers $(x,y)$ with $x<y$ is the harmonic mean of $x$ and $y$ equal to $6^{20}$ | [
"The harmonic mean of $x$ and $y$ is equal to $\\frac{1}{\\frac{\\frac{1}{x}+\\frac{1}{y}}2} = \\frac{2xy}{x+y}$ , so we have $xy=(x+y)(3^{20}\\cdot2^{19})$ , and by SFFT $(x-3^{20}\\cdot2^{19})(y-3^{20}\\cdot2^{19})=3^{40}\\cdot2^{38}$ . Now, $3^{40}\\cdot2^{38}$ has $41\\cdot39=1599$ factors, one of which is the ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_22 | E | 8 | The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad... | [
"Let the hundreds, tens, and units digits of the original three-digit number be $a$ $b$ , and $c$ , respectively. We are given that $a=c+2$ . The original three-digit number is equal to $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$ . The hundreds, tens, and units digits of the reversed three-digit number are $c$ $b$... |
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_43 | B | 22 | The hypotenuse of a right triangle is $10$ inches and the radius of the inscribed circle is $1$ inch. The perimeter of the triangle in inches is:
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30$ | [
"To begin, let's notice that the inscribed circle of the right triangle is its incircle , and that the radius of the incircle is the right triangle's inradius . In this case, the hypotenuse is 10, and the inradius is 1. The formula for the inradius of a right triangle is $r = (a+b-c)/2$ , where $r$ is the length of... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_14 | null | 227 | The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ $PB = 4$ $... | [
"Let the sides $\\overline{AB}$ and $\\overline{AC}$ be tangent to $\\omega$ at $Z$ and $W$ , respectively. Let $\\alpha = \\angle BAX$ and $\\beta = \\angle AXC$ . Because $\\overline{PQ}$ and $\\overline{BC}$ are both tangent to $\\omega$ and $\\angle YXC$ and $\\angle QYX$ subtend the same arc of $\\omega$ , it ... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_12 | null | 91 | The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$ | [
"Suppose that $x_0 = a$ , and that the common ratio between the terms is $r$\nThe first conditions tells us that $\\log_3 a + \\log_3 ar + \\ldots + \\log_3 ar^7 = 308$ . Using the rules of logarithms , we can simplify that to $\\log_3 a^8r^{1 + 2 + \\ldots + 7} = 308$ . Thus, $a^8r^{28} = 3^{308}$ . Since all of t... |
https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | null | 981 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | [
"Rewrite all of the terms in base 3. Since the numbers are sums of distinct powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. $100$ is equal... |
https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_1 | null | 528 | The increasing sequence $2,3,5,6,7,10,11,\ldots$ consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence. | [
"Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than $500$ . This happens to be $23^2=529$ . Notice that there are $23$ squares and $8$ cubes less than or equal to $529$ , but $1$ and $2^6$ are both squares and cubes. Thus, there are $529-23-8+2=500$ numb... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_1 | null | 63 | The increasing sequence $3, 15, 24, 48, \ldots\,$ consists of those positive multiples of 3 that are one less than a perfect square . What is the remainder when the 1994th term of the sequence is divided by 1000? | [
"One less than a perfect square can be represented by $n^2 - 1 = (n+1)(n-1)$ . Either $n+1$ or $n-1$ must be divisible by 3. This is true when $n \\equiv -1,\\ 1 \\equiv 2,\\ 1 \\pmod{3}$ . Since 1994 is even, $n$ must be congruent to $1 \\pmod{3}$ . It will be the $\\frac{1994}{2} = 997$ th such term, so $n = 4 + ... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_8 | A | 10 | The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?
$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$ | [
"We can write $\\sqrt[3]{10}$ as $10 ^ \\frac{1}{3}$ . Similarly, $\\sqrt[3]{\\sqrt[3]{10}} = (10 ^ \\frac{1}{3}) ^ \\frac{1}{3} = 10 ^ \\frac{1}{3^2}$\nBy continuing this, we get the form \\[10 ^ \\frac{1}{3} \\cdot 10 ^ \\frac{1}{3^2} \\cdot 10 ^ \\frac{1}{3^3} \\cdots,\\] which is \\[10 ^ {\\frac{1}{3} + \\frac{... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_8 | null | 10 | The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?
$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$ | [
"Move the first term inside the second radical. We get \\[\\sqrt[3]{10} \\cdot \\sqrt[3]{\\sqrt[3]{10}} \\cdot \\sqrt[3]{\\sqrt[3]{\\sqrt[3]{10}}} \\cdots = \\sqrt[3]{10\\sqrt[3]{10}} \\cdot \\sqrt[3]{\\sqrt[3]{\\sqrt[3]{10}}} \\cdots.\\] Do this for the third radical as well: \\[\\sqrt[3]{10\\sqrt[3]{10}} \\cdot \... |
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_12 | null | 345 | The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is $21$ . Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle. | [
"Let $Q$ be the tangency point on $\\overline{AC}$ , and $R$ on $\\overline{BC}$ . By the Two Tangent Theorem $AP = AQ = 23$ $BP = BR = 27$ , and $CQ = CR = x$ . Using $rs = A$ , where $s = \\frac{27 \\cdot 2 + 23 \\cdot 2 + x \\cdot 2}{2} = 50 + x$ , we get $(21)(50 + x) = A$ . By Heron's formula $A = \\sqrt{s(s-a... |
https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_2 | null | 592 | The integer $n$ is the smallest positive multiple of $15$ such that every digit of $n$ is either $8$ or $0$ . Compute $\frac{n}{15}$ | [
"Any multiple of 15 is a multiple of 5 and a multiple of 3.\nAny multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0.\nThe sum of the digits of any multiple of 3 must be divisible by 3. If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$ . ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_13 | D | 240 | The internal angles of quadrilateral $ABCD$ form an arithmetic progression. Triangles $ABD$ and $DCB$ are similar with $\angle DBA = \angle DCB$ and $\angle ADB = \angle CBD$ . Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the tw... | [
"Since the angles of Quadrilateral $ABCD$ form an arithmetic sequence, we can assign each angle with the value $a$ $a+d$ $a+2d$ , and $a+3d$ . Also, since these angles form an arithmetic progression, we can reason out that $(a)+(a+3d)=(a+d)+(a+2d)=180$\nFor the sake of simplicity, lets rename the angles of each sim... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_24 | A | 100 | The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior a... | [
"Extend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below.\n\nEach of the angles at $X$ is $\\frac{180^\\circ}9 = 20^\\circ$ . From $\\triangle XYZ$ , the degree measure of the smaller interior angle of the trapezoid is $\\frac... |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_23 | B | 1 | The large circle has diameter $\text{AC}$ . The two small circles have their centers on $\text{AC}$ and just touch at $\text{O}$ , the center of the large circle. If each small circle has radius $1$ , what is the value of the ratio of the area of the shaded region to the area of one of the small circles?
[asy] pair A... | [
"The small circle has radius $1$ , thus its area is $\\pi$\nThe large circle has radius $2$ , thus its area is $4\\pi$\nThe area of the semicircle above $AC$ is then $2\\pi$\nThe part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is $\\pi$ . This means t... |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_7 | C | 20 | The large cube shown is made up of $27$ identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is
$\text{(A)}\ 10 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 22 \qquad \text{... | [
"Clearly, no unit cube has more than one face painted, so the number of unit cubes with at least one face painted is equal to the number of painted unit squares.\nThere are $10$ painted unit squares on the half of the cube shown, so there are $10\\cdot 2=20$ unit cubes with at least one face painted, thus our answe... |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_5 | D | 11 | The largest integer $n$ for which $n^{200}<5^{300}$ is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }9 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }11 \qquad \mathrm{(E) \ } 12$ | [
"Since both sides are positive, we can take the $100th$ root of both sides to find the largest integer $n$ such that $n^2<5^3$ . Fortunately, this is simple to evaluate: $5^3=125$ , and the largest square less than $125$ is $11^2=121$ , so the largest $n$ is $11, \\boxed{11}$"
] |
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_15 | E | 6 | The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$ , is:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$ | [
"Factoring the polynomial gives $(n+1)(n)(n-1)$ According to the factorization, one of those factors must be a multiple of two because there are more than 2 consecutive integers. In addition, because there are three consecutive integers, one of the integers must be a multiple of 3. Therefore $6$ must divide the gi... |
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_1 | C | 14 | The largest whole number such that seven times the number is less than 100 is
$\text{(A)} \ 12 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ 16$ | [
"We want to find the smallest integer $x$ so that $7x < 100$ . Dividing by 7 gets $x < 14\\dfrac{2}{7}$ , so the answer is 14. $\\boxed{14}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_20 | A | 20 | The least common multiple of $a$ and $b$ is $12$ , and the least common multiple of $b$ and $c$ is $15$ . What is the least possible value of the least common multiple of $a$ and $c$
$\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180$ | [
"We wish to find possible values of $a$ $b$ , and $c$ . By finding the greatest common factor of $12$ and $15$ , we can find that $b$ is 3. Moving on to $a$ and $c$ , in order to minimize them, we wish to find the least such that the least common multiple of $a$ and $3$ is $12$ $\\rightarrow 4$ . Similarly, with $3... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_7 | B | 6 | The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | [
"Note that \\begin{align*} 18 &= 2\\cdot3^2, \\\\ 180 &= 2^2\\cdot3^2\\cdot5, \\\\ 45 &= 3^2\\cdot5 \\\\ 15 &= 3\\cdot5. \\end{align*} Let $n = 2^a\\cdot3^b\\cdot5^c.$ It follows that:\nTogether, we conclude that $n=2^2\\cdot3\\cdot5=60.$ The sum of its digits is $6+0=\\boxed{6}.$",
"The options for $\\text{lcm}(... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_4 | B | 6 | The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | [
"Note that \\begin{align*} 18 &= 2\\cdot3^2, \\\\ 180 &= 2^2\\cdot3^2\\cdot5, \\\\ 45 &= 3^2\\cdot5 \\\\ 15 &= 3\\cdot5. \\end{align*} Let $n = 2^a\\cdot3^b\\cdot5^c.$ It follows that:\nTogether, we conclude that $n=2^2\\cdot3\\cdot5=60.$ The sum of its digits is $6+0=\\boxed{6}.$",
"The options for $\\text{lcm}(... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_6 | B | 58 | The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$ , where $m$ and $k$ are integers and $6$ is not a divisor of $m$ . What is $m+k?$
$(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90$ | [
"Let this positive integer be written as $p_1^{e_1}\\cdot p_2^{e_2}$ . The number of factors of this number is therefore $(e_1+1) \\cdot (e_2+1)$ , and this must equal 2021. The prime factorization of 2021 is $43 \\cdot 47$ , so $e_1+1 = 43 \\implies e_1=42$ and $e_2+1=47\\implies e_2=46$ . To minimize this integer... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_8 | B | 99 | The length of a rectangle is increased by $10\%$ percent and the width is decreased by $10\%$ percent. What percent of the old area is the new area?
$\textbf{(A)}\ 90 \qquad \textbf{(B)}\ 99 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 101 \qquad \textbf{(E)}\ 110$ | [
"In a rectangle with dimensions $10 \\times 10$ , the new rectangle would have dimensions $11 \\times 9$ . The ratio of the new area to the old area is $99/100 = \\boxed{99}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_11 | D | 20 | The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10$ . What is $b - a$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 20 \qquad \mathrm{(E)}\ 30$ | [
"Since we are given the range of the solutions, we must re-write the inequalities so that we have $x$ in terms of $a$ and $b$\n$a\\le 2x+3\\le b$\nSubtract $3$ from all of the quantities:\n$a-3\\le 2x\\le b-3$\nDivide all of the quantities by $2$\n$\\frac{a-3}{2}\\le x\\le \\frac{b-3}{2}$\nSince we have the range o... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_35 | B | 10 | The lengths of the sides of a triangle are integers, and its area is also an integer.
One side is $21$ and the perimeter is $48$ . The shortest side is:
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 16$ | [
"Let $b$ and $c$ be the other two sides of the triangle. The perimeter of the triangle is $48$ units, so $c = 27-b$ and the semiperimeter equals $24$ units.\nBy Heron's Formula , the area of the triangle is $\\sqrt{24 \\cdot 3(24-b)(b-3)}$ . Plug in the answer choices for $b$ and write the prime factorization of ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_13 | E | 11 | The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | [
"Let $n$ $n+1$ , and $n+2$ be the lengths of the sides of the triangle. Then the perimeter of the triangle is $n + (n+1) + (n+2) = 3n+3$ . Using the fact that the length of the smallest side is $30\\%$ of the perimeter, it follows that:\n$n = 0.3(3n+3) \\Rightarrow n = 0.9n+0.9 \\Rightarrow 0.1n = 0.9 \\Rightarrow ... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_2 | null | 893 | The lengths of the sides of a triangle with positive area are $\log_{10} 12$ $\log_{10} 75$ , and $\log_{10} n$ , where $n$ is a positive integer. Find the number of possible values for $n$ | [
"By the Triangle Inequality and applying the well-known logarithmic property $\\log_{c} a + \\log_{c} b = \\log_{c} ab$ , we have that\nAlso,\nCombining these two inequalities:\n\\[6.25 < n < 900\\]\nThus $n$ is in the set $(6.25 , 900)$ ; the number of positive integer $n$ which satisfies this requirement is $\\bo... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_45 | E | 2 | The lengths of two line segments are $a$ units and $b$ units respectively. Then the correct relation between them is:
$\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad \textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad \textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad \textbf{(E)}\ \frac{... | [
"Since both lengths are positive, the AM-GM Inequality is satisfied. The correct relationship between $a$ and $b$ is $\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_6 | C | 20 | The letter T is formed by placing two $2 \times 4$ inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?
[asy] size(150); draw((0,6)--(4,6)--(4,4)--(3,4)--(3,0)--(1,0)--(1,4)--(0,4)--cycle, linewidth(1));[/asy]
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 20\qquad\textbf... | [
"If the two rectangles were seperate, the perimeter would be $2(2(2+4)=24$ . It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is $24-2 \\times 2 = \\boxed{20}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_24 | E | 9 | The letters $A$ $B$ $C$ and $D$ represent digits. If $\begin{tabular}{ccc}&A&B\\ +&C&A\\ \hline &D&A\end{tabular}$ and $\begin{tabular}{ccc}&A&B\\ -&C&A\\ \hline &&A\end{tabular}$ ,what digit does $D$ represent?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | [
"Because $B+A=A$ $B$ must be $0$ .\nNext, because $B-A=A\\implies0-A=A,$ we get $A=5$ as the \"0\" mentioned above is actually 10 in this case.\nNow we can rewrite $\\begin{tabular}{ccc}&A&0\\\\ +&C&A\\\\ \\hline &D&A\\end{tabular}$ as $\\begin{tabular}{ccc}&5&0\\\\ +&C&5\\\\ \\hline &D&5\\end{tabular}$ . Therefore... |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_22 | C | 12 | The letters $\text{A}$ $\text{J}$ $\text{H}$ $\text{S}$ $\text{M}$ $\text{E}$ and the digits $1$ $9$ $8$ $9$ are "cycled" separately as follows and put together in a numbered list:
\[\begin{tabular}[t]{lccc} & & AJHSME & 1989 \\ & & & \\ 1. & & JHSMEA & 9891 \\ 2. & & HSMEAJ & 8919 \\ 3. & & SMEAJH & 9198 \\ & & ...... | [
"Every $4\\text{th}$ line has $1989$ as part of it and every $6\\text{th}$ line has $\\text{AJHSME}$ as part of it. In order for both to be part of line $n$ $n$ must be a multiple of $4$ and $6$ , the least of which is $\\text{lcm}(4,6)=12\\rightarrow \\boxed{12}$"
] |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_40 | D | 2 | The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$ | [
"Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$ , using the L'Hôpital's rule, we have $\\lim \\limits_{x\\to 1}\\frac{x^2-1}{x-1} = \\lim \\limits_{x\\to 1}\\frac{2x}{1} = 2$ .\nThus, the answer is $\\boxed{2}$",
"The numerator of $\\frac {x^2-1}{x-1}$ can be factored as $(x+1)(x-1)$ . The $x-1$ terms in... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_19 | E | 1 | The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$ | [
"Due to rotations preserving an equal distance, we can bash the answer with the distance formula. $D(A, P) = D(A', P)$ , and $D(B, P) = D(B',P)$ .\nThus we will square our equations to yield: $(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2$ , and $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$ .\nCanceling $(3-s)^2$ from the second equation ma... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_10 | E | 13 | The lines with equations $ax-2y=c$ and $2x+by=-c$ are perpendicular and intersect at $(1, -5)$ . What is $c$
$\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$ | [
"Writing each equation in slope-intercept form, we get $y=\\frac{a}{2}x-\\frac{1}{2}c$ and $y=-\\frac{2}{b}x-\\frac{c}{b}$ . We observe the slope of each equation is $\\frac{a}{2}$ and $-\\frac{2}{b}$ , respectively. Because the slope of a line perpendicular to a line with slope $m$ is $-\\frac{1}{m}$ , we see that... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1 | C | 665 | The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?
$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$ | [
"It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes. We know that there is $60$ minutes in a hour. Therefore, there are $11 \\cdot 60 = 660$ minutes in 11 hours. Adding the second part(the 5 minutes) we get $660 + 5 = \\boxed{665}$",
"The best method comes when yo... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_23 | E | 995 | The manager of a company planned to distribute a $$50$ bonus to each employee from the company fund, but the fund contained $$5$ less than what was needed. Instead the manager gave each employee a $$45$ bonus and kept the remaining $$95$ in the company fund. The amount of money in the company fund before any bonuses ... | [
"Let $p$ be the number of people in the company, and $f$ be the amount of money in the fund.\nThe first sentence states that $50p = f + 5$\nThe second sentence states that $45p = f - 95$\nSubtracing the second equation from the first, we get $5p = 100$ , leading to $p = 20$\nPlugging that number into the first equa... |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_5 | C | 35 | The marked price of a book was 30% less than the suggested retail price. Alice purchased the book for half the marked price at a Fiftieth Anniversary sale. What percent of the suggested retail price did Alice pay?
$\mathrm{(A)\ }25\%\qquad\mathrm{(B)\ }30\%\qquad\mathrm{(C)\ }35\%\qquad\mathrm{(D)\ }60\%\qquad\mathrm... | [
"Without loss of generality, let's assume that the retail price was $100$ USD.\nThe marked price of the book is $30 \\%$ off of $100$ which is equal to $100-100(0.3)=70.$\nHalf of that marked price is $0.5(70)=35.$\nTherefore the percent Alice payed of the suggested retail price is $35/100=\\boxed{35}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_5 | null | 22 | The mean age of Amanda's $4$ cousins is $8$ , and their median age is $5$ . What is the sum of the ages of Amanda's youngest and oldest cousins?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25$ | [
"The sum of the ages of the cousins is $4$ times the mean, or $32$ .\nThere are an even number of cousins, so there is no single median, so $5$ must be the mean of the two in the middle.\nTherefore the sum of the ages of the two in the middle is $10$ . Subtracting $10$ from $32$ produces $\\textbf{(D)}\\ \\boxed{22... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_21 | D | 35 | The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
$\text{(A)}\ 19 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 40$ | [
"Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is $18$ , and there are $2$ numbers less than $18$ and $2$ numbers greater than $18$ . The sum of these integers is $5(15)=75$ , since the mean is $15$ . To make the largest possible number with a ... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_7 | D | 90 | The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$ | [
"Since $x$ is the mean, \\begin{align*} x&=\\frac{60+100+x+40+50+200+90}{7}\\\\ &=\\frac{540+x}{7}. \\end{align*}\nTherefore, $7x=540+x$ , so $x=\\boxed{90}.$",
"Note that $x$ must be the median so it must equal either $60$ or $90$ . You can see that the mean is also $x$ , and by intuition $x$ should be the great... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_4 | D | 90 | The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$ | [
"Since $x$ is the mean, \\begin{align*} x&=\\frac{60+100+x+40+50+200+90}{7}\\\\ &=\\frac{540+x}{7}. \\end{align*}\nTherefore, $7x=540+x$ , so $x=\\boxed{90}.$",
"Note that $x$ must be the median so it must equal either $60$ or $90$ . You can see that the mean is also $x$ , and by intuition $x$ should be the great... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_11 | D | 11 | The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$
$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$ | [
"We can eliminate answer choices ${\\textbf{(A)}\\ 5}$ and ${\\textbf{(C)}\\ 7}$ , because of the above statement. Now we need to test the remaining answer choices.\nCase 1: $x = 6$\nMode: $6$\nMedian: $6$\nMean: $\\frac{37}{7}$\nSince the mean does not equal the median or mode, ${\\textbf{(B)}\\ 6}$ can also be el... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_21 | D | 14 | The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$ | [
"As the unique mode is $8$ , there are at least two $8$ s.\nAs the range is $8$ and one of the numbers is $8$ , the largest one can be at most $16$\nIf the largest one is $16$ , then the smallest one is $8$ , and thus the mean is strictly larger than $8$ , which is a contradiction.\nIf we have 2 8's we can add find... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_15 | D | 14 | The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$ | [
"As the unique mode is $8$ , there are at least two $8$ s.\nAs the range is $8$ and one of the numbers is $8$ , the largest one can be at most $16$\nIf the largest one is $16$ , then the smallest one is $8$ , and thus the mean is strictly larger than $8$ , which is a contradiction.\nIf we have 2 8's we can add find... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_24 | C | 115 | The measure of angle $ABC$ is $50^\circ$ $\overline{AD}$ bisects angle $BAC$ , and $\overline{DC}$ bisects angle $BCA$ . The measure of angle $ADC$ is
[asy] pair A,B,C,D; A = (0,0); B = (9,10); C = (10,0); D = (6.66,3); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--cycle); draw(A--D--C); label("$A$",A,SW); label("$B$... | [
"Let $\\angle CAD = \\angle BAD = x$ , and let $\\angle ACD = \\angle BCD = y$\nFrom $\\triangle ABC$ , we know that $50 + 2x + 2y = 180$ , leading to $x + y = 65$\nFrom $\\triangle ADC$ , we know that $x + y + \\angle D = 180$ . Plugging in $x + y = 65$ , we get $\\angle D = 180 - 65 = 115$ , which is answer $\\b... |
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_14 | A | 6 | The measures of the interior angles of a convex polygon are in arithmetic progression.
If the smallest angle is $100^\circ$ , and the largest is $140^\circ$ , then the number of sides the polygon has is
$\textbf{(A) }6\qquad \textbf{(B) }8\qquad \textbf{(C) }10\qquad \textbf{(D) }11\qquad \textbf{(E) }12$ | [
"Let $n$ equal the number of sides the polygon has. The sum of all the interior angles of a polygon is: $180(n-2)$\nThe formula for an arithmetic series is $\\frac{n(a_1 + a_n)}{2}$ . Set this equal to $180(n-2)$ and solve. In this case, $a_1=100$ and $a_n=140$\nOur equation becomes $\\frac{n(100+140)}{2} = 180(n-2... |
https://artofproblemsolving.com/wiki/index.php/1968_AHSME_Problems/Problem_20 | null | 9 | The measures of the interior angles of a convex polygon of $n$ sides are in arithmetic progression. If the common difference is $5^{\circ}$ and the largest angle is $160^{\circ}$ , then $n$ equals:
$\text{(A) } 9\quad \text{(B) } 10\quad \text{(C) } 12\quad \text{(D) } 16\quad \text{(E) } 32$ | [
"The formula for the sum of the angles in any polygon is $180(n-2)$ . Because this particular polygon is convex and has its angles in an arithmetic sequence with its largest angle being $160$ , we can find the sum of the angles.\n$a_{n}=160$\n$a_{1}=160-5(n-1)$\nPlugging this into the formula for finding the sum of... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_1 | E | 11 | The median of the list $n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15$ is $10$ . What is the mean?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | [
"The median of the list is $10$ , and there are $9$ numbers in the list, so the median must be the 5th number from the left, which is $n+6$\nWe substitute the median for $10$ and the equation becomes $n+6=10$\nSubtract both sides by 6 and we get $n=4$\n$n+n+3+n+4+n+5+n+6+n+8+n+10+n+12+n+15=9n+63$\nThe mean of those... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_12 | null | 134 | The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What was the smallest possible number of members of ... | [
"Let $v_i$ be the number of votes candidate $i$ received, and let $s=v_1+\\cdots+v_{27}$ be the total number of votes cast. Our goal is to determine the smallest possible $s$\nCandidate $i$ got $\\frac{v_i}s$ of the votes, hence the percentage of votes they received is $\\frac{100v_i}s$ . The condition in the probl... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_18 | C | 40 | The midpoints of the four sides of a rectangle are $(-3,0), (2,0), (5,4),$ and $(0,4).$ What is the
area of the rectangle?
$\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80$ | [
"The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle: Note that the diagonals of the rhombus have the same lengths as the sides of the rectangle.\nLet $A=(-3,0), B=(2,0), C=(5,4),$ and $D=(0,4).$ Note that $A,B,C,$ and $D$ are the vertices of... |
https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_33 | C | 10.5 | The minimum value of the quotient of a (base ten) number of three different non-zero digits divided by the sum of its digits is
$\textbf{(A) }9.7\qquad \textbf{(B) }10.1\qquad \textbf{(C) }10.5\qquad \textbf{(D) }10.9\qquad \textbf{(E) }20.5$ | [
"The answer we are looking for can be expressed as $\\dfrac{100a+10b+c}{a+b+c}$ . This is equivalent to $1 + \\dfrac{99a+9b}{a+b+c}$ . Because we are trying to minimize our solution, we set $c$ $9$ , so we have $1 + \\dfrac{99a+9b}{a+b+9}$ . This is equal to $1 + \\dfrac{9a+9b+81}{a+b+9} + \\dfrac{90a-81}{a+b+9}$ ,... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.