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/2023_AMC_10B_Problems/Problem_20 | A | 32 | Four congruent semicircles are drawn on the surface of a sphere with radius $2$ , as
shown, creating a close curve that divides the surface into two congruent regions.
The length of the curve is $\pi\sqrt{n}$ . What is $n$
$\textbf{(A) } 32 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 48 \qquad \textbf{(D) } 36 \qquad ... | [
"There are four marked points on the diagram; let us examine the top two points and call them $A$ and $B$ . Similarly, let the bottom two dots be $C$ and $D$ , as shown:\n\nThis is a cross-section of the sphere seen from the side. We know that ${AO}={BO}={CO}={DO}=2$ , and by Pythagorean Theorem, length of $\\overl... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_18 | D | 12 | Four distinct circles are drawn in a plane . What is the maximum number of points where at least two of the circles intersect?
$\mathrm{(A)}\ 8 \qquad\mathrm{(B)}\ 9 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 16$ | [
"For any given pair of circles, they can intersect at most $2$ times. Since there are ${4\\choose 2} = 6$ pairs of circles, the maximum number of possible intersections is $6 \\cdot 2 = 12$ . We can construct such a situation as below, so the answer is $\\boxed{12}$",
"Because a pair or circles can intersect at m... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_14 | D | 12 | Four distinct circles are drawn in a plane . What is the maximum number of points where at least two of the circles intersect?
$\mathrm{(A)}\ 8 \qquad\mathrm{(B)}\ 9 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 16$ | [
"For any given pair of circles, they can intersect at most $2$ times. Since there are ${4\\choose 2} = 6$ pairs of circles, the maximum number of possible intersections is $6 \\cdot 2 = 12$ . We can construct such a situation as below, so the answer is $\\boxed{12}$",
"Because a pair or circles can intersect at m... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_21 | A | 3 | Four distinct points are arranged on a plane so that the segments connecting them have lengths $a$ $a$ $a$ $a$ $2a$ , and $b$ . What is the ratio of $b$ to $a$
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi$ | [
"For any $4$ non-collinear points with the given requirement, notice that there must be a triangle with side lengths $a$ $a$ $2a$ , which is not possible as $a+a=2a$ . Thus at least $3$ of the $4$ points must be collinear.\nIf all $4$ points are collinear, then there would only be $3$ lines of length $a$ , which wo... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_2 | C | 15 | Four friends do yardwork for their neighbors over the weekend, earning $$15, $20, $25,$ and $$40,$ respectively. They decide to split their earnings equally among themselves. In total, how much will the friend who earned $$40$ give to the others?
$\textbf{(A) }$5 \qquad \textbf{(B) }$10 \qquad \textbf{(C) }$15 \qquad \... | [
"The friends earn $$\\left(15+20+25+40\\right)=$100$ in total. Since they decided to split their earnings equally, it follows that each person will get $$\\left(\\frac{100}{4}\\right)=$25$ . Since the friend who earned $$40$ will need to leave with $$25$ , he will have to give $$\\left(40-25\\right)=\\boxed{15}$ to... |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_14 | A | 7 | Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang $7$ songs, which was more than any other girl, and Mary sang $4$ songs, which was fewer than any other girl. How many songs did these trios sing?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad ... | [
"Alina and Tina must sing more than $4$ , but less than $7$ , songs. Therefore, Alina sang $5$ or $6$ songs, and Tina sang $5$ or $6$ songs, with $4$ possible combinations.\nHowever, since every song is a trio, if you add up all the numbers of songs a person sang for all four singers, it must be divisible by $3$ .... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16 | B | 25 | Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?
$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) ... | [
"Note that the sum of the first two numbers is $21\\cdot2=42,$ the sum of the middle two numbers is $26\\cdot2=52,$ and the sum of the last two numbers is $30\\cdot2=60.$\nIt follows that the sum of the four numbers is $42+60=102,$ so the sum of the first and last numbers is $102-52=50.$ Therefore, the average of t... |
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_16 | null | 3 | Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$ | [
"We assume the side length of the cube is $1$ . The side length of the tetrahedron is $\\sqrt2$ , so the surface area is $4\\times\\frac{2\\sqrt3}{4}=2\\sqrt3$ . The surface area of the cube is $6\\times1\\times1=6$ , so the ratio of the surface area of the cube to the surface area of the tetrahedron is $\\frac{6}{... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_21 | null | 10 | Four positive integers $a$ $b$ $c$ , and $d$ have a product of $8!$ and satisfy:
\[\begin{array}{rl} ab + a + b & = 524 \\ bc + b + c & = 146 \\ cd + c + d & = 104 \end{array}\]
What is $a-d$
$\text{(A) }4 \qquad \text{(B) }6 \qquad \text{(C) }8 \qquad \text{(D) }10 \qquad \text{(E) }12$ | [
"Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:\n\\begin{align*} (a+1)(b+1) & = 525 \\\\ (b+1)(c+1) & = 147 \\\\ (c+1)(d+1) & = 105 \\end{align*}\nLet $(e,f,g,h)=(a+1,b+1,c+1,d+1)$ . We get:\n\\begin{align*} ef & = 3\\cdot 5\\cdot 5\\cdot 7 \\\\ fg & = 3\\cdot 7\\cdot 7 \\... |
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_38 | B | 21 | Four positive integers are given. Select any three of these integers, find their arithmetic average,
and add this result to the fourth integer. Thus the numbers $29, 23, 21$ , and $17$ are obtained. One of the original integers is:
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ ... | [
"Define numbers $a, b, c,$ and $d$ to be the four numbers. In order to satisfy the following conditions, the system of equation should be constructed. (It doesn't matter which variable is which.) \\[\\frac{a+b+c}{3}+d=29\\] \\[\\frac{a+b+d}{3}+c=23\\] \\[\\frac{a+c+d}{3}+b=21\\] \\[\\frac{b+c+d}{3}+a=17\\] Adding a... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_25 | B | 4 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
Wh... | [
"\nRefer to the diagram above.\nLet the origin be at the center of the square, $A$ be the intersection of the top and right hexagons, $B$ be the intersection of the top and left hexagons, and $M$ and $N$ be the top points in the diagram.\nBy symmetry, $A$ lies on the line $y = x$ . The equation of line $AN$ is $y =... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_3 | E | 52 | Four squares of side length $4, 7, 9,$ and $10$ are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units? [asy] size(150); filldraw((0,0)... | [
"We work inwards. The area of the outer shaded square is the area of the whole square minus the area of the second largest square. The area of the inner shaded region is the area of the third largest square minus the area of the smallest square. The sum of these areas is \\[10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \\boxe... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_3 | A | 40 | Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?
$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$ | [
"Let $r$ be the remaining student's score. We know that the average, 70, is equal to $\\frac{70 + 80 + 90 + r}{4}$ . We can use basic algebra to solve for $r$ \\[\\frac{70 + 80 + 90 + r}{4} = 70\\] \\[\\frac{240 + r}{4} = 70\\] \\[240 + r = 280\\] \\[r = 40\\] giving us the answer of $\\boxed{40}$",
"Since $90$... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_13 | B | 6 | Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces
[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; cu... | [
"This is the number cubes that are adjacent to another cube on two sides. The bottom corner cubes are connected on three sides, and the top corner cubes are connected on one. The number we are looking for is the number of middle cubes, which is $\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_9 | B | 137 | Francesca uses $100$ grams of lemon juice, $100$ grams of sugar, and $400$ grams of water to make lemonade. There are $25$ calories in $100$ grams of lemon juice and $386$ calories in $100$ grams of sugar. Water contains no calories. How many calories are in $200$ grams of her lemonade?
$\textbf{(A) } 129\qquad \textbf... | [
"The calorie to gram ratio of Francesca's lemonade is $\\frac{25+386+0}{100+100+400}=\\frac{411\\textrm{ calories}}{600\\textrm{ grams}}=\\frac{137\\textrm{ calories}}{200\\textrm{ grams}}$\nSo in $200\\textrm{ grams}$ of Francesca's lemonade there are $200\\textrm{ grams}\\cdot\\frac{137\\textrm{ calories}}{200\\t... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_6 | B | 137 | Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade?
$\text {(A) } 129 \qquad \text {(B) } 137 \qq... | [
"Francesca makes a total of $100+100+400=600$ grams of lemonade, and in those $600$ grams, there are $25$ calories from the lemon juice and $386$ calories from the sugar, for a total of $25+386=411$ calories per $600$ grams. We want to know how many calories there are in $200=600/3$ grams, so we just divide $411$ b... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_13 | null | 273 | Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line $y = 24$ . A fence is located at the horizontal line $y = 0$ . On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a poi... | [
"Clearly Freddy's $x$ -coordinate is irrelevant, so we let $E(y)$ be the expected value of the number of jumps it will take him to reach the river from a given $y$ -coordinate. Observe that $E(24)=0$ , and \\[E(y)=1+\\frac{E(y+1)+E(y-1)+2E(y)}{4}\\] for all $y$ such that $1\\le y\\le 23$ . Also note that $E(0)=1+\\... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_30 | A | 40 | From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \text{... | [
"Let us represent the number of boys $b$ , and the number of girls $g$\nFrom the first sentence, we get that $2(g-15)=b$\nFrom the second sentence, we get $5(b-45)=g-15$\nExpanding both equations and simplifying, we get $2g-30 = b$ and $5b = g+210$\nSubstituting $b$ for $2g-30$ , we get $5(2g-30)=g+210$ . Solving f... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23 | D | 57 | From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[... | [
"Choose side \"lengths\" $a,b,c$ for the triangle, where \"length\" is how many vertices of the octagon are skipped between vertices of the triangle, starting from the shortest side, and going clockwise, and choosing $a=b$ if the triangle is isosceles: $a+b+c=5$ , where either [ $a\\leq b$ and $a < c$ ] or [ $a=b=c... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_12 | null | 803 | From the set of integers $\{1,2,3,\dots,2009\}$ , choose $k$ pairs $\{a_i,b_i\}$ with $a_i<b_i$ so that no two pairs have a common element. Suppose that all the sums $a_i+b_i$ are distinct and less than or equal to $2009$ . Find the maximum possible value of $k$ | [
"Suppose that we have a valid solution with $k$ pairs. As all $a_i$ and $b_i$ are distinct, their sum is at least $1+2+3+\\cdots+2k=k(2k+1)$ . On the other hand, as the sum of each pair is distinct and at most equal to $2009$ , the sum of all $a_i$ and $b_i$ is at most $2009 + (2009-1) + \\cdots + (2009-(k-1)) = \\... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_18 | E | 2 | Gage skated $1$ hr $15$ min each day for $5$ days and $1$ hr $30$ min each day for $3$ days. How long would he have to skate the ninth day in order to average $85$ minutes of skating each day for the entire time?
$\text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text... | [
"Converting into minutes and adding, we get that she skated $75*5+90*3+x = 375+270+x = 645+x$ minutes total, where $x$ is the amount she skated on day $9$ . Dividing by $9$ to get the average, we get $\\frac{645+x}{9}=85$ . Solving for $x$ \\[645+x=765\\] \\[x=120\\] Now we convert back into hours and minutes to ge... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_1 | null | 37 | Gary purchased a large beverage, but only drank $m/n$ of it, where $m$ and $n$ are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only $2/9$ as much beverage. Find $m+n$ | [
"Let $x$ be the fraction consumed, then $(1-x)$ is the fraction wasted. We have $\\frac{1}{2} - 2x =\\frac{2}{9} (1-x)$ , or $9 - 36x = 4 - 4x$ , or $32x = 5$ or $x = 5/32$ . Therefore, $m + n = 5 + 32 = \\boxed{037}$"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_17 | B | 6 | George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must Geor... | [
"Note that on a normal day, it takes him $1/3$ hour to get to school. However, today it took $\\frac{1/2 \\text{ mile}}{2 \\text{ mph}}=1/4$ hour to walk the first $1/2$ mile. That means that he has $1/3 -1/4 = 1/12$ hours left to get to school, and $1/2$ mile left to go. Therefore, his speed must be $\\frac{1/2 \\... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_8 | E | 54 | Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?
$\textbf{(A) }20\q... | [
"After Gilda gives $20$ % of the marbles to Pedro, she has $80$ % of the marbles left. If she then gives $10$ % of what's left to Ebony, she has $(0.8*0.9)$ $72$ % of what she had at the beginning. Finally, she gives $25$ % of what's left to her brother, so she has $(0.75*0.72)$ $\\boxed{54}$ of what she had in the... |
https://artofproblemsolving.com/wiki/index.php/1993_AHSME_Problems/Problem_30 | D | 31 | Given $0\le x_0<1$ , let \[x_n=\left\{ \begin{array}{ll} 2x_{n-1} &\text{ if }2x_{n-1}<1 \\ 2x_{n-1}-1 &\text{ if }2x_{n-1}\ge 1 \end{array}\right.\] for all integers $n>0$ . For how many $x_0$ is it true that $x_0=x_5$
$\text{(A) 0} \quad \text{(B) 1} \quad \text{(C) 5} \quad \text{(D) 31} \quad \text{(E) }\infty$ | [
"We are going to look at this problem in binary.\n$x_0 = (0.a_1 a_2 \\cdots )_2$\n$2x_0 = (a_1.a_2 a_3 \\cdots)_2$\nIf $2x_0 < 1$ , then $x_0 < \\frac{1}{2}$ which means that $a_1 = 0$ and so $x_1 = (.a_2 a_3 a_4 \\cdots)_2$\nIf $2x_0 \\geq 1$ then $x \\geq \\frac{1}{2}$ which means that $x_1 = 2x_0 - 1 = (.a_2 a_3... |
https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_11 | D | 27 | Given $2^x=8^{y+1}$ and $9^y=3^{x-9}$ , find the value of $x+y$
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$ | [
"Since $8^{y + 1} = 2^{3(y+1)}$ and $9^y = 3^{2y}$ , we have:\n$2^x = 2^{3(y+1)}$ and $3^{2y} = 3^{x - 9}$\nNote that if $a^b = a^c$ , then $b=c$ . Setting the exponents equal gives $x = 3y + 3$ and $2y = x - 9$ . Plugging the first equation into the second equation gives:\n$2y = (3y + 3) - 9$\n$2y = 3y - 6$\n$0 ... |
https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_4 | C | 2 | Given $\frac{\log{a}}{p}=\frac{\log{b}}{q}=\frac{\log{c}}{r}=\log{x}$ , all logarithms to the same base and $x \not= 1$ . If $\frac{b^2}{ac}=x^y$ , then $y$ is:
$\text{(A)}\ \frac{q^2}{p+r}\qquad\text{(B)}\ \frac{p+r}{2q}\qquad\text{(C)}\ 2q-p-r\qquad\text{(D)}\ 2q-pr\qquad\text{(E)}\ q^2-pr$ | [
"We are given: \\[\\frac{b^2}{ac} = x^y\\]\nTaking the logarithm on both sides: \\[\\log{\\left(\\frac{b^2}{ac}\\right)} = \\log{x^y}\\]\nUsing the properties of logarithms: \\[2\\log{b} - \\log{a} - \\log{c} = y \\log{x}\\]\nSubstituting the values given in the problem statement: \\[2q \\log{x} - p \\log{x} - r \\... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_14 | null | 459 | Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive intege... | [
"We now need to solve $a^2+ab+b^2 = 3^2\\cdot 73^2$ . A quick $(\\bmod 9)$ check gives that $3\\mid a$ and $3\\mid b$ . Thus, it's equivalent to solve $x^2+xy+y^2 = 73^2$\nLet $\\omega$ be one root of $\\omega^2+\\omega+1=0$ . Then, recall that $\\mathbb Z[\\omega]$ is the ring of integers of $\\mathbb Q[\\sqrt{-3}... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12 | null | 230 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ | [
"Notice that we must have \\[\\frac{f(f(z))-f(z)}{f(z)-z}=-\\frac{f(f(z))-f(z)}{z-f(z)}\\in i\\mathbb R .\\] However, $f(t)-t=t(t-20)$ , so \\begin{align*} \\frac{f(f(z))-f(z)}{f(z)-z}&=\\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\\\ &=\\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\\\ &=(z-19)(z+1)\\\\ &=(z-9)^2-100. \\end{align*... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_10 | null | 146 | Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be wri... | [
"\nNow we put the figure in the Cartesian plane, let the center of the circle $O (0,0)$ , then $B (\\sqrt{13},0)$ , and $A(4+\\sqrt{13},0)$\nThe equation for Circle O is $x^2+y^2=13$ , and let the slope of the line $AKL$ be $k$ , then the equation for line $AKL$ is $y=k(x-4-\\sqrt{13})$\nThen we get $(k^2+1)x^2-2k^... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_12 | null | 177 | Given a function $f$ for which \[f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)\] holds for all real $x,$ what is the largest number of different values that can appear in the list $f(0),f(1),f(2),\ldots,f(999)?$ | [
"\\begin{align*}f(2158 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\\\ f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\\end{align*}\nSince $\\mathrm{gcd}(1056, 1760) = 352$ we can conclude that (by the Euclidean algorithm\n\\[f(x) = f(352 + x)\\]\nSo we need only to consider one period $f(0), f(1), .... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_9 | null | 233 | Given a nonnegative real number $x$ , let $\langle x\rangle$ denote the fractional part of $x$ ; that is, $\langle x\rangle=x-\lfloor x\rfloor$ , where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle$ , and $2<a^2<3$ . F... | [
"Looking at the properties of the number, it is immediately guess-able that $a = \\phi = \\frac{1+\\sqrt{5}}2$ (the golden ratio ) is the answer. The following is the way to derive that:\nSince $\\sqrt{2} < a < \\sqrt{3}$ $0 < \\frac{1}{\\sqrt{3}} < a^{-1} < \\frac{1}{\\sqrt{2}} < 1$ . Thus $\\langle a^2 \\rangle =... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_15 | null | 597 | Given a point $P^{}_{}$ on a triangular piece of paper $ABC,\,$ consider the creases that are formed in the paper when $A, B,\,$ and $C\,$ are folded onto $P.\,$ Let us call $P_{}^{}$ a fold point of $\triangle ABC\,$ if these creases, which number three unless $P^{}_{}$ is one of the vertices, do not intersect. Suppo... | [
"Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $\\overline{PA}$ and $\\overline{PB}$ , and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $\\triangle PAB, PBC, PCA$ . According to the problem statement, the circ... |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_14 | null | 490 | Given a positive integer $n$ , it can be shown that every complex number of the form $r+si$ , where $r$ and $s$ are integers, can be uniquely expressed in the base $-n+i$ using the integers $0,1,2,\ldots,n^2$ as digits. That is, the equation
is true for a unique choice of non-negative integer $m$ and digits $a_0,a_1,\l... | [
"First, we find the first three powers of $-3+i$\n$(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i$\nSo we solve the diophantine equation $a_1-6a_2+26a_3=0 \\Longrightarrow a_1-6a_2=-26a_3$\nThe minimum the left-hand side can go is -54, so $1\\leq a_3 \\leq 2$ since $a_3$ can't equal 0, so we try cases:\nSo we hav... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_5 | null | 103 | Given a positive integer $n\,$ , let $p(n)\,$ be the product of the non-zero digits of $n\,$ . (If $n\,$ has only one digits, then $p(n)\,$ is equal to that digit.) Let
What is the largest prime factor of $S\,$ | [
"Suppose we write each number in the form of a three-digit number (so $5 \\equiv 005$ ), and since our $p(n)$ ignores all of the zero-digits, replace all of the $0$ s with $1$ s. Now note that in the expansion of\nwe cover every permutation of every product of $3$ digits, including the case where that first $1$ rep... |
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_5 | B | 68 | Given a quadrilateral $ABCD$ inscribed in a circle with side $AB$ extended beyond $B$ to point $E$ , if $\measuredangle BAD=92^\circ$ and $\measuredangle ADC=68^\circ$ , find $\measuredangle EBC$
$\mathrm{(A)\ } 66^\circ \qquad \mathrm{(B) \ }68^\circ \qquad \mathrm{(C) \ } 70^\circ \qquad \mathrm{(D) \ } 88^\circ \q... | [
"Since $ABCD$ is cyclic, opposite angles must sum to $180^\\circ$ . Therefore, $\\angle ADC+\\angle ABC=180^\\circ$ , and $\\angle ABC=180^\\circ-\\angle ADC=180^\\circ-68^\\circ=112^\\circ$ . Notice also that $\\angle ABC$ and $\\angle CBE$ form a linear pair, and so they sum to $180^\\circ$ . Therefore, $\\angle ... |
https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_5 | null | 128 | Given a rational number , write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator . For how many rational numbers between $0$ and $1$ will $20_{}^{}!$ be the resulting product | [
"If the fraction is in the form $\\frac{a}{b}$ , then $a < b$ and $\\gcd(a,b) = 1$ . There are 8 prime numbers less than 20 ( $2, 3, 5, 7, 11, 13, 17, 19$ ), and each can only be a factor of one of $a$ or $b$ . There are $2^8$ ways of selecting some combination of numbers for $a$ ; however, since $a<b$ , only half ... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_7 | null | 553 | Given a real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ For a certain integer $k,$ there are exactly $70$ positive integers $n_{1}, n_{2}, \ldots, n_{70}$ such that $k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{2}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$... | [
"Obviously $k$ is positive. Then, we can let $n_1$ equal $k^3$ and similarly let $n_i$ equal $k^3 + (i - 1)k$\nThe wording of this problem (which uses \"exactly\") tells us that $k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \\leq k^3+70k$ . Taking away $k^3$ from our inequality results in $69k<3k^2+3k+1\\leq 70k$ . Since $6... |
https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_10 | A | 2 | Given a square side of length $s$ . On a diagonal as base a triangle with three unequal sides is constructed so that its area
equals that of the square. The length of the altitude drawn to the base is:
$\textbf{(A)}\ s\sqrt{2} \qquad \textbf{(B)}\ s/\sqrt{2} \qquad \textbf{(C)}\ 2s \qquad \textbf{(D)}\ 2\sqrt{s} \qquad... | [
"The area of the square is $s^2$ . The diagonal of a square with side $s$ bisects the square into two $45-45-90$ right triangles, so the diagonal has length $s\\sqrt{2}$\nThe area of the triangle is $\\frac{1}{2}bh$ . The base $b$ of the triangle is the diagonal of the square, which is $b = s\\sqrt{2}$ . If the ... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_12 | null | 101 | Given a triangle , its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra $P_{i}$ is defined recursively as follows: $P_{0}$ is a regular tetrahedron whose volume is 1. To obtain $P_{i + 1}$ , replace the midpoint triangle of every face of $P_{i}$ by an outward-pointing regular... | [
"On the first construction, $P_1$ , four new tetrahedra will be constructed with side lengths $\\frac 12$ of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follows that each of these new tetrahedra will have volume $\\left(\\frac 12\\r... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_13 | B | 12 | Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 15$ | [
"This is a Pythagorean triple (a $3-4-5$ actually) with legs $15$ and $20$ . The area is then $\\frac{(15)(20)}{2}=150$ . Now, consider an altitude drawn to any side. Since the area remains constant, the altitude and side to which it is drawn are inversely proportional. To get the smallest altitude, it must be draw... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_5 | null | 376 | Given eight distinguishable rings, let $n$ be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost three nonzero digits of $n$ | [
"There are $\\binom{8}{5}$ ways to choose the rings, and there are $5!$ distinct arrangements to order the rings [we order them so that the first ring is the bottom-most on the first finger that actually has a ring, and so forth]. The number of ways to distribute the rings among the fingers is equivalent the number... |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_25 | C | 5 | Given square $ABCD$ with side $8$ feet. A circle is drawn through vertices $A$ and $D$ and tangent to side $BC$ . The radius of the circle, in feet, is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4\sqrt{2}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5\sqrt{2}\qquad\textbf{(E)}\ 6$ | [
"Let $O$ be the center of the circle and $E$ be the point of tangency of the circle and $BC$ and let $F$ be the point of intersection of lines $OE$ and $AD$ Because of the symmetry, $BE=EC=AF=FD=4$ feet. Let the length of $OF$ be $x$ . The length of $OE$ is $EF-OF=-x+8$ . By Pythagorean Theorem, $OA=OD=\\sqrt{x^2+4... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_3 | null | 898 | Given that
\begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
find the value of $x_{531}+x_{753}+x_{975}$ | [
"We find that $x_5 = 267$ by the recursive formula. Summing the recursions\n\\begin{align*} x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \\\\ x_{n-1}&=x_{n-2}-x_{n-3}+x_{n-4}-x_{n-5} \\end{align*}\nyields $x_{n} = -x_{n-5}$ . Thus $x_n = (-1)^k x_{n-5k}$ . Since $531 = 106 \\cdot 5 + 1,\\ 753 = 150 \\cdot 5 + 3,\\ 975 = ... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_7 | null | 137 | Given that
find the greatest integer that is less than $\frac N{100}$ | [
"Multiplying both sides by $19!$ yields:\n\\[\\frac {19!}{2!17!}+\\frac {19!}{3!16!}+\\frac {19!}{4!15!}+\\frac {19!}{5!14!}+\\frac {19!}{6!13!}+\\frac {19!}{7!12!}+\\frac {19!}{8!11!}+\\frac {19!}{9!10!}=\\frac {19!N}{1!18!}.\\]\n\\[\\binom{19}{2}+\\binom{19}{3}+\\binom{19}{4}+\\binom{19}{5}+\\binom{19}{6}+\\binom... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_1 | null | 839 | Given that
where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k + n.$ | [
"Note that \\[{{\\left((3!)!\\right)!}\\over{3!}}= {{(6!)!}\\over{6}}={{720!}\\over6}={{720\\cdot719!}\\over6}=120\\cdot719!.\\] Because $120\\cdot719!<720!$ , we can conclude that $n < 720$ . Thus, the maximum value of $n$ is $719$ . The requested value of $k+n$ is therefore $120+719=\\boxed{839}$"
] |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_7 | null | 27 | Given that $(1+\sin t)(1+\cos t)=5/4$ and
where $k, m,$ and $n_{}$ are positive integers with $m_{}$ and $n_{}$ relatively prime , find $k+m+n.$ | [
"From the givens, $2\\sin t \\cos t + 2 \\sin t + 2 \\cos t = \\frac{1}{2}$ , and adding $\\sin^2 t + \\cos^2t = 1$ to both sides gives $(\\sin t + \\cos t)^2 + 2(\\sin t + \\cos t) = \\frac{3}{2}$ . Completing the square on the left in the variable $(\\sin t + \\cos t)$ gives $\\sin t + \\cos t = -1 \\pm \\sqrt{\... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_15 | D | 12 | Given that $-4\leq x\leq-2$ and $2\leq y\leq4$ , what is the largest possible value of $\frac{x+y}{x}$
$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1$ | [
"Rewrite $\\frac{(x+y)}x$ as $\\frac{x}x+\\frac{y}x=1+\\frac{y}x$\nWe also know that $\\frac{y}x<0$ because $x$ and $y$ are of opposite sign.\nTherefore, $1+\\frac{y}x$ is maximized when $|\\frac{y}x|$ is minimized, which occurs when $|x|$ is the largest and $|y|$ is the smallest.\nThis occurs at $(-4,2)$ , so $\\f... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_25 | B | 195 | Given that $2^{2004}$ is a $604$ digit number whose first digit is $1$ , how many elements of the set $S = \{2^0,2^1,2^2,\ldots ,2^{2003}\}$ have a first digit of $4$
$\mathrm{(A)}\ 194 \qquad \mathrm{(B)}\ 195 \qquad \mathrm{(C)}\ 196 \qquad \mathrm{(D)}\ 197 \qquad \mathrm{(E)}\ 198$ | [
"Given $n$ digits, there must be exactly one power of $2$ with $n$ digits such that the first digit is $1$ . Thus $S$ contains $603$ elements with a first digit of $1$ . For each number in the form of $2^k$ such that its first digit is $1$ , then $2^{k+1}$ must either have a first digit of $2$ or $3$ , and $2^{k+2}... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_14 | D | 407 | Given that $3^8\cdot5^2=a^b,$ where both $a$ and $b$ are positive integers, find the smallest possible value for $a+b$
$\textbf{(A) } 25 \qquad\textbf{(B) } 34 \qquad\textbf{(C) } 351 \qquad\textbf{(D) } 407 \qquad\textbf{(E) } 900$ | [
"\\[3^8\\cdot5^2 = (3^4)^2\\cdot5^2 = (3^4\\cdot5)^2 = 405^2\\]\n$405$ is not a perfect power, so the smallest possible value of $a+b$ is $405+2=\\boxed{407}$"
] |
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_5 | null | 40 | Given that $A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,$ find $|A_{19} + A_{20} + \cdots + A_{98}|.$ | [
"Though the problem may appear to be quite daunting, it is actually not that difficult. $\\frac {k(k-1)}2$ always evaluates to an integer ( triangular number ), and the cosine of $n\\pi$ where $n \\in \\mathbb{Z}$ is 1 if $n$ is even and -1 if $n$ is odd. $\\frac {k(k-1)}2$ will be even if $4|k$ or $4|k-1$ , and od... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_10 | null | 11 | Given that $O$ is a regular octahedron , that $C$ is the cube whose vertices are the centers of the faces of $O,$ and that the ratio of the volume of $O$ to that of $C$ is $\frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$ | [
"Let the side of the octahedron be of length $s$ . Let the vertices of the octahedron be $A, B, C, D, E, F$ so that $A$ and $F$ are opposite each other and $AF = s\\sqrt2$ . The height of the square pyramid $ABCDE$ is $\\frac{AF}2 = \\frac s{\\sqrt2}$ and so it has volume $\\frac 13 s^2 \\cdot \\frac s{\\sqrt2} =... |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_4 | null | 12 | Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$ | [
"Using the properties of logarithms , we can simplify the first equation to $\\log_{10} \\sin x + \\log_{10} \\cos x = \\log_{10}(\\sin x \\cos x) = -1$ . Therefore, \\[\\sin x \\cos x = \\frac{1}{10}.\\qquad (*)\\]\nNow, manipulate the second equation. \\begin{align*} \\log_{10} (\\sin x + \\cos x) &= \\frac{1}{2}... |
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_11 | null | 177 | Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$ | [
"Let $s = \\sum_{k=1}^{35}\\sin 5k = \\sin 5 + \\sin 10 + \\ldots + \\sin 175$ . We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to telescope the sum. Using ... |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_5 | B | 320 | Given that $\text{1 mile} = \text{8 furlongs}$ and $\text{1 furlong} = \text{40 rods}$ , the number of rods in one mile is
$\text{(A)}\ 5 \qquad \text{(B)}\ 320 \qquad \text{(C)}\ 660 \qquad \text{(D)}\ 1760 \qquad \text{(E)}\ 5280$ | [
"\\[(1\\ \\text{mile}) \\left( \\frac{8\\ \\text{furlongs}}{1\\ \\text{mile}} \\right) \\left( \\frac{40\\ \\text{rods}}{1\\ \\text{furlong}} \\right) = \\boxed{320}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_17 | null | 3 | Given that $i^2=-1$ , for how many integers $n$ is $(n+i)^4$ an integer?
$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$ | [
"$(n+i)^4=n^4+4in^3-6n^2-4in+1$ , and this has to be an integer, so the sum of the imaginary parts must be $0$ \\[4in^3-4in=0\\] \\[4in^3=4in\\] \\[n^3=n\\] Since $n^3=n$ , there are $\\boxed{3}$ solutions for $n$ $0$ and $\\pm1$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_8 | D | 2 | Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$ | [
"$x+\\tfrac{2}{x}= y+\\tfrac{2}{y}$\nSince $x\\not=y$ , we may assume that $x=\\frac{2}{y}$ and/or, equivalently, $y=\\frac{2}{x}$\nCross multiply in either equation, giving us $xy=2$\n$\\boxed{2}$",
"\\[x + \\frac{2}{x} = y + \\frac{2}{y}.\\]\nMultiply both sides by xy to get\n\\[x^2y + 2y = y^2x +2x\\]\nRearran... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_8 | null | 2 | Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$ | [
"Let $A = x + \\frac{2}{x} = y + \\frac{2}{y}.$ Consider the equation \\[u + \\frac{2}{u} = A.\\] Reorganizing, we see that $u$ satisfies \\[u^2 - Au + 2 = 0.\\] Notice that there can be at most two distinct values of $u$ which satisfy this equation, and $x$ and $y$ are two distinct possible values for $u.$ Therefo... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_15 | null | 9 | Given that $x, y,$ and $z$ are real numbers that satisfy: \begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*} and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are posit... | [
"Let $\\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$ , and suppose this triangle is acute (so all altitudes are in the interior of the triangle).\nLet the altitude to the side of length $x$ be of length $h_x$ , and similarly for $y$ and $z$ . Then we have by two applications of the Pythagorean Th... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | B | 73 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | [
"Complete the square to get \\[(x-7)^2 + (y-3)^2 = 64.\\] Applying Cauchy-Schwarz directly, \\[64\\cdot25=(3^2+4^2)((x-7)^2 + (y-3)^2) \\ge (3(x-7)+4(y-3))^2.\\] \\[40 \\ge 3x+4y-33\\] \\[3x+4y \\le 73.\\] Thus our answer is $\\boxed{73}$",
"The first equation is a circle , so we find its center and radius by com... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | null | 73 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | [
"Completing the square gives the equation of a circle, $(x-7)^2+(y-3)^2=64.$ Seeing that we would like to maximize $3x+4y,$ we parameterize the circle using polar coordinates: \\begin{align*} x&=7+8\\cos\\theta&y&=3+8\\sin\\theta. \\end{align*} Then, we have $3x+4y=3(7+8\\cos\\theta)+4(3+8\\sin\\theta)=33+24\\cos\\... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_9 | null | 0 | Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$ | [
"Using the quadratic equation on $z^2 - (2 \\cos 3 )z + 1 = 0$ , we have $z = \\frac{2\\cos 3 \\pm \\sqrt{4\\cos^2 3 - 4}}{2} = \\cos 3 \\pm i\\sin 3 = \\text{cis}\\,3^{\\circ}$\nThere are other ways we can come to this conclusion. Note that if $z$ is on the unit circle in the complex plane, then $z = e^{i\\theta} ... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_1 | null | 9 | Given that \begin{eqnarray*}&(1)& x\text{ and }y\text{ are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\ &(2)& y\text{ is the number formed by reversing the digits of }x\text{; and}\\ &(3)& z=|x-y|. \end{eqnarray*}
How many distinct values of $z$ are possible? | [
"We express the numbers as $x=100a+10b+c$ and $y=100c+10b+a$ . From this, we have \\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\\\&=&|99a-99c|\\\\&=&99|a-c|\\\\ \\end{eqnarray*} Because $a$ and $c$ are digits, and $a$ and $c$ are both between 1 and 9 (from condition 1), there are $\\boxed{009}$ possible values (s... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_15 | null | 27 | Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$ | [
"Suppose $b_{i} = \\frac {x_{i}}3$ .\nWe have \\[\\sum_{i = 1}^{2006}b_{i}^{2} = \\sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \\sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)\\] So \\[\\sum_{i = 0}^{2005}b_{i} = \\frac {b_{2006}^{2} - 2006}2\\] Now \\[\\sum_{i = 1}^{2006}b_{i} = \\frac {b_{2006}^{2} + 2b_{2006} - 2006}2\\] Th... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_2 | C | 6 | Given that a, b, and c are non-zero real numbers, define $(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ , find $(2, 12, 9)$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$ | [
"$(2, 12, 9)=\\frac{2}{12}+\\frac{12}{9}+\\frac{9}{2}=\\frac{1}{6}+\\frac{4}{3}+\\frac{9}{2}=\\frac{1}{6}+\\frac{8}{6}+\\frac{27}{6}=\\frac{36}{6}=6$ . Our answer is then $\\boxed{6}$",
"Without computing the answer exactly, we see that $2/12=\\text{a little}$ $12/9=\\text{more than }1$ , and $9/2=4.5$ .\nThe sum... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_6 | B | 30 | Given the areas of the three squares in the figure, what is the area of the interior triangle?
[asy] draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1)); label("$25$",(14.5,1),N); label("$144$",(6,-7.5),N); label("$169$",(3.5,7),N); [/asy]
$\mathrm{... | [
"The sides of the squares are $5, 12$ and $13$ for the square with area $25, 144$ and $169$ , respectively. The legs of the interior triangle are $5$ and $12$ , so the area is $\\frac{5 \\times 12}{2}=\\boxed{30}$"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_14 | null | 66 | Given the nine-sided regular polygon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$ , how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set $\{A_1,A_2,\dots,A_9\}$
$\text{(A) }30 \qquad \text{(B) }36 \qquad \text{(C) }63 \qquad \text{(D) }66 \qquad \text{(E) }72$ | [
"Each of the $\\binom{9}{2} = 36$ pairs of vertices determines two equilateral triangles, for a total of 72 triangles. However, the three triangles $A_1A_4A_7$ $A_2A_5A_8$ , and $A_3A_6A_9$ are each counted 3 times, resulting in an overcount of 6. Thus, there are $\\boxed{66}$ distinct equilateral triangles."
] |
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_48 | B | 5 | Given the polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$ , where $n$ is a positive integer or zero, and $a_0$ is a positive integer. The remaining $a$ 's are integers or zero. Set $h=n+a_0+|a_1|+|a_2|+\cdots+|a_n|$ . [See example 25 for the meaning of $|x|$ .] The number of polynomials with $h=3$ is: $\textbf{(A)}\... | [
"We perform casework by the value of $n$ , the degree of our polynomial $a_0x^n+a_1x^{n-1}+\\cdots+a_{n-1}x+a_n$\nCase $n = 0$ : In this case we are forced to set $a_0 = 3$ . This contributes $1$ possibility.\nCase $n = 1$ : In this case we must have $a_0 + |a_1| = 2$ , so our polynomial could be $1 + 1x, 0 + 2x, -... |
https://artofproblemsolving.com/wiki/index.php/1971_AHSME_Problems/Problem_29 | E | 11 | Given the progression $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$ .
The least positive integer $n$ such that the product of the first $n$ terms of the progression exceeds $100,000$ is
$\textbf{(A) }7\qquad \textbf{(B) }8\qquad \textbf{(C) }9\qquad \textbf... | [
"The product of the sequence $10^{\\dfrac{1}{11}}, 10^{\\dfrac{2}{11}}, 10^{\\dfrac{3}{11}}, 10^{\\dfrac{4}{11}},\\dots , 10^{\\dfrac{n}{11}}$ is equal to $10^{\\dfrac{1}{11}+\\frac{2}{11}\\dots\\frac{n}{11}}$ since we are looking for the smallest value $n$ that will create $100,000$ , or $10^5$ . From there, we ca... |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_13 | C | 100 | Given two positive integers $x$ and $y$ with $x < y$ . The percent that $x$ is less than $y$ is:
$\textbf{(A)}\ \frac{100(y-x)}{x}\qquad \textbf{(B)}\ \frac{100(x-y)}{x}\qquad \textbf{(C)}\ \frac{100(y-x)}{y}\qquad \\ \textbf{(D)}\ 100(y-x)\qquad \textbf{(E)}\ 100(x - y)$ | [
"Suppose that $x$ is $p$ percent less than $y$ . Then $x = \\frac{100 - p}{100}y$ , so that $y - x = \\frac{p}{100}y$ . Solving for $p$ , we get $p = \\frac{100(y-x)}{y}$ , or $\\boxed{100}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_11 | D | 60 | Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perim... | [
"Let the side lengths of the rectangular pan be $m$ and $n$ . It follows that $(m-2)(n-2) = \\frac{mn}{2}$ , since half of the brownie pieces are in the interior. This gives $2(m-2)(n-2) = mn \\iff mn - 4m - 4n + 8 = 0$ . Adding 8 to both sides and applying Simon's Favorite Factoring Trick , we obtain $(m-4)(n-4) =... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_3 | E | 23 | Granny Smith has $63. Elberta has $2 more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have?
$\text{(A)}\ 17 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 23$ | [
"Since Anjou has $\\frac{1}{3}$ the amount of money as Granny Smith and Granny Smith has $ $63$ , Anjou has $\\frac{1}{3}\\times63=21$ dollars. Elberta has $ $2$ more than this, so she has $ $23$ , or $\\boxed{23}$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_18 | D | 411 | Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position?
$\textbf{(A) } 405 \qquad \textbf{(B) } 407 \... | [
"We have $2$ directions going $5$ right or $3$ left. We can assign a variable to each of these directions. We can call going right $1$ direction $\\text{X}$ and we can call going $1$ left $\\text{Y}$ . We can build a equation of $5\\text{X}-3\\text{Y}=2023$ , where we have to limit the number of moves we do. We can... |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_18 | C | 36 | Half the people in a room left. One third of those remaining started to dance. There were then $12$ people who were not dancing. The original number of people in the room was
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 72$ | [
"Let the original number of people in the room be $x$ . Half of them left, so $\\frac{x}{2}$ of them are left in the room.\nAfter that, one third of this group is dancing, so $\\frac{x}{2}-\\frac{1}{3}\\left( \\frac{x}{2}\\right) =\\frac{x}{3}$ people are not dancing.\nThis is given to be $12$ , so \\[\\frac{x}{3}... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_5 | C | 42 | Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next $n$ shots are bullseyes she will be guaranteed victory. What is the minimu... | [
"Let $k$ be the number of points Chelsea currently has. In order to guarantee victory, we must consider the possibility that the opponent scores the maximum amount of points by getting only bullseyes.\n\\begin{align*}k+ 10n + 4(50-n) &> (k-50) + 50\\cdot{10}\\\\ 6n &> 250\\end{align*}\nThe lowest integer value tha... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_5 | E | 20 | Hammie is in $6^\text{th}$ grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?
$\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \... | [
"Listing the elements from least to greatest, we have $(5, 5, 6, 8, 106)$ , we see that the median weight is 6 pounds.\nThe average weight of the five kids is $\\frac{5+5+6+8+106}{5} = \\frac{130}{5} = 26$\nHence, \\[26-6=\\boxed{20}.\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_10 | E | 15 | Handy Aaron helped a neighbor $1 \frac14$ hours on Monday, $50$ minutes on Tuesday, from 8:20 to 10:45 on Wednesday morning, and a half-hour on Friday. He is paid $\textdollar 3$ per hour. How much did he earn for the week?
$\textbf{(A)}\ \textdollar 8 \qquad \textbf{(B)}\ \textdollar 9 \qquad \textbf{(C)}\ \textdollar... | [
"Let's convert everything to minutes and add them together. On Monday he worked for $\\frac54 \\cdot 60 = 75$ minutes. On Tuesday he worked $50$ minutes. On Wednesday he worked for $2$ hours $25$ minutes, or $2(60)+25=145$ minutes. On Friday he worked $\\frac{60}{2}=30$ minutes. This adds up to $75+50+145+30=300$ m... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_9 | null | 757 | Harold, Tanya, and Ulysses paint a very long picket fence.
Call the positive integer $100h+10t+u$ paintable when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers. | [
"Note that it is impossible for any of $h,t,u$ to be $1$ , since then each picket will have been painted one time, and then some will be painted more than once.\n$h$ cannot be $2$ , or that will result in painting the third picket twice. If $h=3$ , then $t$ may not equal anything not divisible by $3$ , and the same... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_1 | A | 10 | Harry and Terry are each told to calculate $8-(2+5)$ . Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5$ . If Harry's answer is $H$ and Terry's answer is $T$ , what is $H-T$
$\textbf{(A) }-10\qquad\textbf{(B) }-6\qquad\textbf{(C) }0\qquad\textbf{(D) }6\qquad \textbf{(E) }10$ | [
"We have $H=8-7=1$ and $T=8-2+5=11$ . Clearly $1-11=-10$ , so our answer is $\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_11 | C | 12 | Harry has 3 sisters and 5 brothers. His sister Harriet has $\text{S}$ sisters and $\text{B}$ brothers. What is the product of $\text{S}$ and $\text{B}$
$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18$ | [
"Harry has 3 sisters and 5 brothers. His sister, being a girl, would have 1 less sister and 1 more brother.\n$S = 3-1=2$\n$B = 5+1=6$\n$S\\cdot B = 2\\times6=12=\\boxed{12}$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_11 | D | 44 | Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating $3$ inches of pasta from the middle of one piece. In the end, he has $10$ pieces of pasta whose total length is $17$ inches. How long, in inches, was the piece of pasta he started with?
$\textbf{(A) } 34\qquad\textbf{... | [
"If there are $10$ pieces of pasta, Henry took $10-1=9$ bites. Each of these $9$ bites took $3$ inches of pasta out, and thus his bites in total took away $9\\cdot 3 = 27$ inches of pasta. Thus, the original piece of pasta was $27+17=\\boxed{44}$ inches long."
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_12 | C | 768 | Henry's Hamburger Haven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two,or three meat patties and any collection of condiments. How many different kinds of hamburgers can be ordered?
$\text{(A) \ } 24 \qqua... | [
"For each condiment, a customer may either choose to order it or not. There are $8$ total condiments to choose from. Therefore, there are $2^8=256$ ways to order the condiments. There are also $3$ choices for the meat, making a total of $256\\times3=768$ possible hamburgers. $\\boxed{768}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_18 | E | 21 | Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$ . Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$
$\textbf{(A) ... | [
"Notice that $1000$ is $3E8$ when converted to hexadecimal ( $3 \\cdot 16^2 + 14 \\cdot 16^1 + 8 \\cdot 16^0$ ). We will proceed by constructing numbers that consist of only numeric digits in hexadecimal.\nThe first digit could be $0,$ $1,$ $2,$ or $3,$ and the second two could be any digit $0 - 9$ , giving $4 \\cd... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_8 | null | 89 | Hexagon $ABCDEF$ is divided into five rhombuses $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent , and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$ . Given that $K$ is a positive integer , find the number of possible values... | [
"Let $x$ denote the common side length of the rhombi.\nLet $y$ denote one of the smaller interior angles of rhombus $\\mathcal{P}$ . Then $x^2\\sin(y)=\\sqrt{2006}$ . We also see that $K=x^2\\sin(2y) \\Longrightarrow K=2x^2\\sin y \\cdot \\cos y \\Longrightarrow K = 2\\sqrt{2006}\\cdot \\cos y$ . Thus $K$ can be ... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | B | 13 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the... | [
"Suppose the roommate took sheets $a$ through $b$ , or equivalently, page numbers $2a-1$ through $2b$ . Because there are $(2b-2a+2)$ numbers taken, \\[\\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\\frac{50\\cdot51}{2} \\implies (2a+2b-39)(b-a+1)=\\frac{50\\cdot13}{2}=25\\cdot13.\\] The first possible solution th... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_15 | A | 20 | Homer began peeling a pile of 44 potatoes at the rate of 3 potatoes per minute. Four minutes later Christen joined him and peeled at the rate of 5 potatoes per minute. When they finished, how many potatoes had Christen peeled?
$\text{(A)}\ 20 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 33 \qquad \tex... | [
"After the $4$ minutes of Homer peeling alone, he had peeled $4\\times3=12$ potatoes. This means that there are $44-12=32$ potatoes left. Once Christen joins him, the two are peeling potatoes at a rate of $3+5=8$ potatoes per minute. So, they finish peeling after another $\\frac{32}{8}=4$ minutes. In these $4$ minu... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_16 | D | 21 | How many $3$ -digit positive integers have digits whose product equals $24$
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24$ | [
"With the digits listed from least to greatest, the $3$ -digit integers are $138,146,226,234$ $226$ can be arranged in $\\frac{3!}{2!} = 3$ ways, and the other three can be arranged in $3!=6$ ways. There are $3+6(3) = \\boxed{21}$ $3$ -digit positive integers."
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17 | D | 576 | How many $4 \times 4$ arrays whose entries are $0$ s and $1$ s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array \[\left[ \begin{arr... | [
"Note that the arrays and the sum configurations have one-to-one correspondence. Furthermore, the row sum configuration and the column sum configuration are independent of each other. Therefore, the answer is $(4!)^2=\\boxed{576}.$",
"In this problem, we call a matrix that satisfies all constraints given in the p... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_6 | B | 100 | How many $4$ -digit positive integers (that is, integers between $1000$ and $9999$ , inclusive) having only even digits are divisible by $5?$
$\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500$ | [
"The units digit, for all numbers divisible by 5, must be either $0$ or $5$ . However, since all digits are even, the units digit must be $0$ . The middle two digits can be 0, 2, 4, 6, or 8, but the thousands digit can only be 2, 4, 6, or 8 since it cannot be zero. There is one choice for the units digit, 5 choices... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_4 | B | 100 | How many $4$ -digit positive integers (that is, integers between $1000$ and $9999$ , inclusive) having only even digits are divisible by $5?$
$\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500$ | [
"The units digit, for all numbers divisible by 5, must be either $0$ or $5$ . However, since all digits are even, the units digit must be $0$ . The middle two digits can be 0, 2, 4, 6, or 8, but the thousands digit can only be 2, 4, 6, or 8 since it cannot be zero. There is one choice for the units digit, 5 choices... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_11 | A | 6 | How many $7$ -digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$ $2$ $3$ $3$ $5$ $5$ $5$
$\text{(A) } 6 \qquad \text{(B) } 12 \qquad \text{(C) } 24 \qquad \text{(D) } 36 \qquad \text{(E) } 48$ | [
"A seven-digit palindrome is a number of the form $\\overline{abcdcba}$ . Clearly, $d$ must be $5$ , as we have an odd number of fives. We are then left with $\\{a,b,c\\} = \\{2,3,5\\}$ . There are $3!$ permutations of these three numbers, since each is reflected over the midpoint we only have to count the first th... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_10 | D | 9 | How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$ | [
"For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of $2012$ , since all of the valid 4-digit number will always be greater than $1000$ . The best way to solve this problem is by using casework.\nThere can be only two leading digits, namely $1$ or $2$\n... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_23 | D | 84 | How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108$ | [
"We can separate this into two cases. If an integer is a multiple of $5,$ the last digit must be either $0$ or $5.$\nCase 1: The last digit is $5.$ The leading digit can be $1,2,3,$ or $4.$ Because the second digit can be $0$ but not the leading digit, there are also $4$ choices. The third digit cannot be the leadi... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_14 | D | 7 | How many complex numbers satisfy the equation $z^5=\overline{z}$ , where $\overline{z}$ is the conjugate of the complex number $z$
$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~5\qquad\textbf{(D)} ~6\qquad\textbf{(E)} ~7$ | [
"Let $z = re^{i\\theta}.$ We now have $\\overline{z} = re^{-i\\theta},$ and want to solve\n\\[r^5e^{5i\\theta} = re^{-i\\theta}.\\]\nFrom this, we have $r = 0$ as a solution, which gives $z = 0$ . If $r\\neq 0$ , then we divide by it, yielding\n\\[r^4e^{5i\\theta} = e^{-i\\theta}.\\]\nDividing both sides by $e^{-i\... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_14 | E | 7 | How many complex numbers satisfy the equation $z^5=\overline{z}$ , where $\overline{z}$ is the conjugate of the complex number $z$
$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~5\qquad\textbf{(D)} ~6\qquad\textbf{(E)} ~7$ | [
"When $z^5=\\overline{z}$ , there are two conditions: either $z=0$ or $z\\neq 0$ . When $z\\neq 0$ , since $|z^5|=|\\overline{z}|$ $|z|=1$ $z^5\\cdot z=z^6=\\overline{z}\\cdot z=|z|^2=1$ . Consider the $r(\\cos \\theta +i\\sin \\theta)$ form, when $z^6=1$ , there are 6 different solutions for $z$ . Therefore, the n... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_8 | null | 90 | How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0? | [
"The problem is asking us for all configurations of $4\\times 4$ grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:\nAdding these cases up, we get $36 + 48 + 6 = \\boxed{090}$",
"Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_2 | A | 2 | How many different combinations of $5 bills and $2 bills can be used to make a total of $17? Order does not matter in this problem.
$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad \text {(C)}\ 4 \qquad \text {(D)}\ 5 \qquad \text {(E)}\ 6$ | [
"You cannot use more than 4 $<dollar>5$ bills, but if you use 3 $<dollar>5$ bills, you can add another $<dollar>2$ bill to make a combination. You cannot use 2 $<dollar>5$ bills since you have an odd number of dollars that need to be paid with $<dollar>2$ bills. You can also use 1 $<dollar>5$ bill and 6 $<dollar>2$... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_2 | B | 6 | How many different four-digit numbers can be formed by rearranging the four digits in $2004$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 81$ | [
"We can solve this problem easily, just by calculating how many choices there are for each of the four digits.\nFirst off, we know there are only $2$ choices for the first digit, because $0$ isn't a valid choice, or the number would a 3-digit number, which is not what we want. \nWe have $3$ choices for the second d... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_10 | A | 13 | How many different integers can be expressed as the sum of three distinct members of the set $\{1,4,7,10,13,16,19\}$
$\text{(A)}\ 13 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$ | [
"Subtracting 10 from each number in the set, and dividing the results by 3, we obtain the set $\\{-3, -2, -1, 0, 1, 2, 3\\}$ . It is easy to see that we can get any integer between $-6$ and $6$ inclusive as the sum of three elements from this set, for the total of $\\boxed{13}$ integers.",
"The set is an arithme... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_15 | C | 6 | How many different isosceles triangles have integer side lengths and perimeter 23?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$ | [
"Let $b$ be the base of the isosceles triangles, and let $a$ be the lengths of the other legs. From this, $2a+b=23$ and $b=23-2a$ . From triangle inequality, $2a>b$ , then plug in the value from the previous equation to get $2a>23-2a$ or $a>5.75$ . The maximum value of $a$ occurs when $b=1$ , in which from the firs... |
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