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/2013_AMC_12B_Problems/Problem_23 | E | 25 | Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers ... | [
"First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.\nSay that $N \\equiv a \\pmod{6}$\nalso that $N \\equiv b \\pmod{5}$\nSubstituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b < 5$... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_6 | E | 26 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C... | [
"Since Bertha has $6$ daughters, she has $30-6=24$ granddaughters, of which none have daughters. Of Bertha's daughters, $\\frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters. Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters. $\\boxed{26}$",
"Draw a tree diagram and ... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_6 | null | 26 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C... | [
"Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \\boxed{26}$"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_4 | E | 26 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C... | [
"Since Bertha has $6$ daughters, she has $30-6=24$ granddaughters, of which none have daughters. Of Bertha's daughters, $\\frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters. Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters. $\\boxed{26}$",
"Draw a tree diagram and ... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_4 | null | 26 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C... | [
"Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \\boxed{26}$"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_15 | A | 40 | Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set a... | [
"Before new letters were added, five different letters could have been chosen for the first position, three for the second, and four for the third. This means that $5\\cdot 3\\cdot 4=60$ plates could have been made.\nIf two letters are added to the second set, then $5\\cdot 5\\cdot 4=100$ plates can be made. If one... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_15 | D | 40 | Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set a... | [
"There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \\cdot 3 \\cdot 4 = 60$ license plates.\nAdding $2$ letters to the start gives $7\\cdot 3 \\cdot 4 = 84$ plates.\nAdding $2$ letters to the middle gives $5 \\cdot 5 \\cdo... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_12 | D | 32 | Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$ | [
"There are $5$ days from May 1 to May 5. The number of bananas he eats each day is an arithmetic sequence. He eats $n$ bananas on May 5, and $n-4(6)=n-24$ bananas on May 1. The sum of this arithmetic sequence is equal to $100$\n\\begin{align*} \\frac{n+n-24}{2} \\cdot 5 &= 100\\\\ n-12&=20\\\\ n&=\\boxed{32}",
"S... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_7 | A | 10 | Blake and Jenny each took four $100$ -point tests. Blake averaged $78$ on the four tests. Jenny scored $10$ points higher than Blake on the first test, $10$ points lower than him on the second test, and $20$ points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's ave... | [
"Blake scored a total of $4 \\times 78=312$ points. Jenny scored $10-10+20+20=40$ points higher than Blake, so her average is $\\frac{312+40}{4}=88$ . \nthe difference is $88-78=\\boxed{10}$",
"The total point difference between Blake's and Jenny's scores is $10-10+20+20=40$ . The average of it is $\\frac{40}{4}=... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_14 | B | 1 | Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$ | [
"Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$ . It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$ . Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to Vieta's For... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_12 | B | 1 | Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$ | [
"Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$ . It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$ . Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to Vieta's For... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_17 | C | 350 | Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
$\... | [
"Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $\\frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a propor... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_15 | C | 350 | Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
$\... | [
"Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $\\frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a propor... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_7 | B | 5 | Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month. If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, then in how many months from that time will they have the same number of goldfish?
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qqua... | [
"Call this month \"Month 0\". Make a table of the fish that Brent and Gretel have each month.\n$\\text{Month / Brent / Gretel}$\n$\\text{0 / 4 / 128}$\n$\\text{1 / 16 / 256}$\n$\\text{2 / 64 / 512}$\n$\\text{3 / 256 / 1024}$\n$\\text{4 / 1024 / 2048}$\n$\\text{5 / 4096 / 4096}$\nYou could create a similar table wi... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_13 | B | 31 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$ . What is the sum of the possible values of $w$
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93$ | [
"Assume that $y-z=a, x-y=b, w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$ . \nThis means $a+b+c=9$ $a,b,c$ must be in the set ${1,3,4,5,6}$ .\nThe only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$ $a+b$ , and $b+c$ then must be the remaining two numbers which ar... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_13 | null | 31 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$ . What is the sum of the possible values of $w$
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93$ | [
"Let the four numbers be $z$ $z+a$ $z+b$ , and $z+c$ . We know that $c$ must be $9$ because that's the greatest difference. So we have $z$ $z+a$ $z+b$ , and $z+9$ . The 6 possible differences are $a$ $b$ $9$ $b-a$ $9-a$ , and $9-b$ . We are given that the differences are 1, 3, 4, 5, 6, 9. $a$ and $9-a$ and $b$ and ... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_21 | B | 31 | Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | [
"The largest difference, $9,$ must be between $w$ and $z.$\nThe smallest difference, $1,$ must be directly between two integers. This also means the differences directly between the other two should add up to $8.$ The only remaining differences that would make this possible are $3$ and $5.$ However, those two diffe... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_5 | C | 25 | Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?
$\textbf{(A) }\ \frac15 \qquad\textbf{(B) }\ \frac13 \qquad\textbf{(C) }\ \frac25 \qq... | [
"Let $m =$ Brianna's money. We have $\\frac15 m = \\frac13 (\\mbox{CDs}) \\Rightarrow \\frac35 m = (\\mbox{CDs})$ . Thus, the money left over is $m-\\frac35m = \\frac25m$ , so the answer is $\\boxed{25}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_3 | C | 25 | Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?
$\textbf{(A) }\ \frac15 \qquad\textbf{(B) }\ \frac13 \qquad\textbf{(C) }\ \frac25 \qq... | [
"Let $m =$ Brianna's money. We have $\\frac15 m = \\frac13 (\\mbox{CDs}) \\Rightarrow \\frac35 m = (\\mbox{CDs})$ . Thus, the money left over is $m-\\frac35m = \\frac25m$ , so the answer is $\\boxed{25}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_18 | B | 900 | Bricklayer Brenda takes $9$ hours to build a chimney alone, and bricklayer Brandon takes $10$ hours to build it alone. When they work together, they talk a lot, and their combined output decreases by $10$ bricks per hour. Working together, they build the chimney in $5$ hours. How many bricks are in the chimney?
$\mathr... | [
"Let $x$ be the number of bricks in the chimney. The work done is the rate multiplied by the time.\nUsing $w = rt$ , we get $x = \\left(\\frac{x}{9} + \\frac{x}{10} - 10\\right)\\cdot(5)$ . Solving for $x$ , we get $\\boxed{900}$"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_3 | E | 52 | Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $\textdollar 2.50$ each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf co... | [
"She first sells one-half of her $48$ loaves, or $\\frac{48}{2}=24$ loaves. Each loaf sells for $\\textdollar 2.50$ , so her total earnings in the morning is equal to \\[24\\cdot \\textdollar 2.50 = \\textdollar 60\\]\nThis leaves 24 loaves left, and Bridget will sell $\\dfrac{2}{3}\\times 24=16$ of them for a pric... |
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_25 | D | 10 | Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes $5$ hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass in the highway (not in the station)?
$\... | [
"Say you are on the Houston-bound bus that left at 12:30 in the afternoon, looking out the window to see how many buses you pass. At 12:45 pm, the Dallas bus that left at 8:00 am is 4:45 away (Note - $a:b$ $a$ is for hrs. and $b$ is for min.) from Dallas, and therefore 15 minutes from Houston. Your bus is also 15 m... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_11 | B | 39.6 | Business is a little slow at Lou's Fine Shoes, so Lou decides to have a
sale. On Friday, Lou increases all of Thursday's prices by $10$ percent. Over the
weekend, Lou advertises the sale: "Ten percent off the listed price. Sale
starts Monday." How much does a pair of shoes cost on Monday that
cost $40$ dollars on Thurs... | [
"On Friday, the shoes would cost $40 \\cdot 1.1= 44$ dollars. Then on Monday, the shoes would cost $44- \\frac{44}{10}=44-4.4=\\boxed{39.60}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_4 | null | 279 | Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their rout... | [
"When they meet at the milepost, Sparky has been ridden for $n$ miles total. Assume Butch rides Sparky for $a$ miles, and Sundance rides for $n-a$ miles. Thus, we can set up an equation, given that Sparky takes $\\frac{1}{6}$ hours per mile, Butch takes $\\frac{1}{4}$ hours per mile, and Sundance takes $\\frac{2}{5... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_13 | B | 5 | Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?
(For example, one sequence of hops is up-up-down-down-up-down.)
2024-AMC8-q13.png
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\... | [
"Looking at the answer choices, you see that you can list them out. \nDoing this gets you:\n$UUDDUD$\n$UDUDUD$\n$UUUDDD$\n$UDUUDD$\n$UUDUDD$\nCounting all the paths listed above gets you $\\boxed{5}$",
"Any combination can be written as some re-arrangement of $UUUDDD$ . Clearly we must end going down, and start g... |
https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_3 | null | 182 | By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called nice if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers? | [
"Let $p(n)$ denote the product of the distinct proper divisors of $n$ . A number $n$ is nice in one of two instances:\nWe now show that the above are the only two cases. Suppose that another nice number existed that does not fall into one of these two categories. Then we can either express it in the form $n = pqr$ ... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_7 | C | 4 | By inserting parentheses, it is possible to give the expression \[2\times3 + 4\times5\] several values. How many different values can be obtained?
$\text{(A) } 2 \qquad \text{(B) } 3 \qquad \text{(C) } 4 \qquad \text{(D) } 5 \qquad \text{(E) } 6$ | [
"The three operations can be performed on any of $3! = 6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions \\begin{align*} (2\\... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_6 | C | 4 | By inserting parentheses, it is possible to give the expression \[2\times3 + 4\times5\] several values. How many different values can be obtained?
$\text{(A) } 2 \qquad \text{(B) } 3 \qquad \text{(C) } 4 \qquad \text{(D) } 5 \qquad \text{(E) } 6$ | [
"The three operations can be performed on any of $3! = 6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions \\begin{align*} (2\\... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_8 | D | 20 | CHIKEN NUGGIEs | [
"Case $1$ : The numbers are separated by $1$\nWe this case with $a=0, b=1,$ and $c=2$ . Following this logic, the last set we can get is $a=7, b=8,$ and $c=9$ . We have $8$ sets of numbers in this case.\nCase $2$ : The numbers are separated by $2$\nThis case starts with $a=0, b=2,$ and $c=2$ . It ends with $a=5, b=... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_1 | D | 25 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | [
"Cagney can frost one in $20$ seconds, and Lacey can frost one in $30$ seconds. Working together, they can frost one in $\\frac{20\\cdot30}{20+30} = \\frac{600}{50} = 12$ seconds. In $300$ seconds ( $5$ minutes), they can frost $\\boxed{25}$ cupcakes.",
"In $300$ seconds ( $5$ minutes), Cagney will frost $\\dfrac... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_2 | D | 25 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | [
"Cagney can frost one in $20$ seconds, and Lacey can frost one in $30$ seconds. Working together, they can frost one in $\\frac{20\\cdot30}{20+30} = \\frac{600}{50} = 12$ seconds. In $300$ seconds ( $5$ minutes), they can frost $\\boxed{25}$ cupcakes.",
"In $300$ seconds ( $5$ minutes), Cagney will frost $\\dfrac... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_1 | null | 840 | Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. | [
"Assume that the largest geometric number starts with a $9$ . We know that the common ratio must be a rational of the form $k/3$ for some integer $k$ , because a whole number should be attained for the 3rd term as well. When $k = 1$ , the number is $931$ . When $k = 2$ , the number is $964$ . When $k = 3$ , we get ... |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_24 | C | 19,990 | Call a $7$ -digit telephone number $d_1d_2d_3-d_4d_5d_6d_7$ memorable if the prefix sequence $d_1d_2d_3$ is exactly the same as either of the sequences $d_4d_5d_6$ or $d_5d_6d_7$ (possibly both). Assuming that each $d_i$ can be any of the ten decimal digits $0,1,2, \ldots, 9$ , the number of different memorable telepho... | [
"In this problem, we only need to consider the digits $\\overline{d_4d_5d_6d_7}$ . Each possibility of $\\overline{d_4d_5d_6d_7}$ gives $2$ possibilities for $\\overline{d_1d_2d_3}$ , which are $\\overline{d_1d_2d_3}=\\overline{d_4d_5d_6}$ and $\\overline{d_1d_2d_3}=\\overline{d_5d_6d_7}$ with the exception of the ... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_7 | C | 11 | Call a fraction $\frac{a}{b}$ , not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$ . How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf... | [
"The special fractions are \\[\\frac{1}{14},\\frac{2}{13},\\frac{3}{12},\\frac{4}{11},\\frac{5}{10},\\frac{6}{9},\\frac{7}{8},\\frac{8}{7},\\frac{9}{6},\\frac{10}{5},\\frac{11}{4},\\frac{12}{3},\\frac{13}{2},\\frac{14}{1}.\\] We rewrite them in the simplest form: \\[\\frac{1}{14},\\frac{2}{13},\\frac{1}{4},\\frac{4... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_5 | C | 11 | Call a fraction $\frac{a}{b}$ , not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$ . How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf... | [
"The special fractions are \\[\\frac{1}{14},\\frac{2}{13},\\frac{3}{12},\\frac{4}{11},\\frac{5}{10},\\frac{6}{9},\\frac{7}{8},\\frac{8}{7},\\frac{9}{6},\\frac{10}{5},\\frac{11}{4},\\frac{12}{3},\\frac{13}{2},\\frac{14}{1}.\\] We rewrite them in the simplest form: \\[\\frac{1}{14},\\frac{2}{13},\\frac{1}{4},\\frac{4... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_10 | null | 486 | Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$ . For example, 53421 and 14253 are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$ , but 45123 is not. Find the number of quasi-increasing permutations of th... | [
"The simple recurrence can be found.\nWhen inserting an integer $n$ into a string with $n - 1$ integers, we notice that the integer $n$ has 3 spots where it can go: before $n - 1$ , before $n - 2$ , and at the very end.\nEXAMPLE: \nPutting 4 into the string 123:\n4 can go before the 2: 1423,\nBefore the 3: 1243,\nA... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_8 | null | 315 | Call a positive integer $N$ 7-10 double if the digits of the base- $7$ representation of $N$ form a base- $10$ number that is twice $N$ . For example, $51$ is a 7-10 double because its base- $7$ representation is $102$ . What is the largest 7-10 double? | [
"Let $A$ be the base $10$ representation of our number, and let $B$ be its base $7$ representation.\nGiven this is an AIME problem, $A<1000$ . If we look at $B$ in base $10$ , it must be equal to $2A$ , so $B<2000$ when $B$ is looked at in base $10.$\nIf $B$ in base $10$ is less than $2000$ , then $B$ as a number i... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_9 | null | 472 | Call a positive integer $n$ $k$ pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$ . For example, $18$ is $6$ -pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$ -pretty. Find $\tfrac{S}{20}$ | [
"Every 20-pretty integer can be written in form $n = 2^a 5^b k$ , where $a \\ge 2$ $b \\ge 1$ $\\gcd(k,10) = 1$ , and $d(n) = 20$ , where $d(n)$ is the number of divisors of $n$ . Thus, we have $20 = (a+1)(b+1)d(k)$ , using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of 2... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_7 | null | 49 | Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$ | [
"$n$ can either be $0$ or $1$ (mod $2$ ).\nCase 1: $n \\equiv 0 \\pmod{2}$\nThen, $n \\equiv 2 \\pmod{4}$ , which implies $n \\equiv 1 \\pmod{3}$ and $n \\equiv 4 \\pmod{6}$ , and therefore $n \\equiv 3 \\pmod{5}$ . Using CRT , we obtain $n \\equiv 58 \\pmod{60}$ , which gives $16$ values for $n$\nCase 2: $n \\equi... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_16 | C | 6 | Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)... | [
"The divisibility rule of $15$ is that the number must be congruent to $0$ mod $3$ and congruent to $0$ mod $5$ . Being divisible by $5$ means that it must end with a $5$ or a $0$ . We can rule out the case when the number ends with a $0$ immediately because the only integer that is uphill and ends with a $0$ is $0... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_17 | B | 1,524 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?... | [
"Case 1: monotonous numbers with digits in ascending order\nThere are $\\sum_{n=1}^{9} \\binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will alwa... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | B | 1,524 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?... | [
"Case 1: monotonous numbers with digits in ascending order\nThere are $\\sum_{n=1}^{9} \\binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will alwa... |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_29 | B | 8 | Call a positive real number special if it has a decimal representation that consists entirely of digits $0$ and $7$ . For example, $\frac{700}{99}= 7.\overline{07}= 7.070707\cdots$ and $77.007$ are special numbers. What is the smallest $n$ such that $1$ can be written as a sum of $n$ special numbers?
$\textbf{(A)}\ 7\q... | [
"Define a super-special number to be a number whose decimal expansion only consists of $0$ 's and $1$ 's. The problem is equivalent to finding the number of super-special numbers necessary to add up to $\\frac{1}{7}=0.142857142857\\hdots$ . This can be done in $8$ numbers if we take \\[0.111111\\hdots, 0.011111\\hd... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_12 | null | 252 | Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$ . For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,... | [
"We shall solve this problem by doing casework on the lowest element of the subset. Note that the number $1$ cannot be in the subset because $1*1=1$ . Let $S$ be a product-free set. If the lowest element of $S$ is $2$ , we consider the set $\\{3, 6, 9\\}$ . We see that 5 of these subsets can be a subset of $S$ $\\{... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | E | 129 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | [
"Let $S_{n}$ denote the number of spacy subsets of $\\{ 1, 2, ... n \\}$ . We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$\nThe spacy subsets of $S_{n + 1}$ can be divided into two groups:\nHence,\nFrom this recursion , we find that\nAnd so the answer is $\\boxed{129}$",
"Let us consider each size of subset individual... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | null | 129 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | [
"Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used.\nFrom the set $\\{1,2,3,4,5,6,7,8,9,10,11,12\\}$ we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_5 | null | 31 | Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences. | [
"Let the terms be $a-b$ $a$ , and $a+b$ . Then we want $(a-b)^2+a^2+(a+b)^2=ab^2$ , or $3a^2+2b^2=ab^2$ . Rearranging, we get $b^2=\\frac{3a^2}{a-2}$ . Simplifying further, $b^2=3a+6+\\frac{12}{a-2}$ . Looking at this second equation, since the right side must be an integer, $a-2$ must equal $\\pm1, 2, 3, 4, 6, 12$... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_10 | D | 11 | Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ ... | [
"Let $M$ be the median. It follows that the two largest integers are both $M+2.$\nLet $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is \\[a,b,M,M+2,M+2.\\] Since the median is $2$ greater than their arithmetic mean, we have $\\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or \\[a+b+14=2M.\\] Note tha... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_7 | D | 11 | Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ ... | [
"Let $M$ be the median. It follows that the two largest integers are both $M+2.$\nLet $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is \\[a,b,M,M+2,M+2.\\] Since the median is $2$ greater than their arithmetic mean, we have $\\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or \\[a+b+14=2M.\\] Note tha... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_5 | D | 40 | Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating 10 pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}... | [
"Denote the number of blueberry and cherry jelly beans as $b$ and $c$ respectively. Then $b = 2c$ and $b-10 = 3(c-10)$ . Substituting, we have $2c-10 = 3c-30$ , so $c=20$ $b=\\boxed{40}$",
"From the problem, we see that 10 less than one of the answer choices must be a multiple of 3 and positive. The only answer c... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_8 | D | 80 | Candy sales from the Boosters Club from January through April are shown. What were the average sales per month in dollars?
[asy] draw((0,0)--(36,0)--(36,24)--(0,24)--cycle); draw((0,4)--(36,4)); draw((0,8)--(36,8)); draw((0,12)--(36,12)); draw((0,16)--(36,16)); draw((0,20)--(36,20)); fill((4,0)--(8,0)--(8,20)--(4,20)--... | [
"There are a total of $100+60+40+120=320$ dollars of sales spread through $4$ months, for an average of $320/4 = \\boxed{80}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_11 | B | 336 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the sho... | [
"If the dimensions are $4a\\times 4b$ , then one side will have $a+1$ posts (including corners) and the other $b+1$ (including corners).\nThe total number of posts is $2(a+b)=20$\nThis diagram represents the number of posts around the garden. \nWe solve the system $\\begin{cases}b+1=2(a+1)\\\\a+b=10\\end{cases}$ to... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_9 | B | 336 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the sho... | [
"By Albert471\nTo start, use algebra to determine the number of posts on each side. You have (the long sides count for $2$ because there are twice as many) $6x = 20 + 4$ (each corner is double counted so you must add $4$ ) Making the shorter end have $4$ , and the longer end have $8$ $((8-1)*4)*((4-1)*4) = 28*12 = ... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_2 | E | 45 | Carl has $5$ cubes each having side length $1$ , and Kate has $5$ cubes each having side length $2$ . What is the total volume of these $10$ cubes?
$\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 45$ | [
"A cube with side length $1$ has volume $1^3=1$ , so $5$ of these will have a total volume of $5\\cdot1=5$\nA cube with side length $2$ has volume $2^3=8$ , so $5$ of these will have a total volume of $5\\cdot8=40$\n$5+40=\\boxed{45}$ ~quacker88",
"The total volume of Carl's cubes is $5$ . This is because to find... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_1 | C | 20 | Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$ | [
"If Carlos took $70\\%$ of the pie, there must be $(100 - 70)\\% = 30\\%$ left. After Maria takes $\\frac{1}{3}$ of the remaining $30\\%, \\ 1 - \\frac{1}{3} = \\frac{2}{3}$ of the remaining $30\\%$ is left.\nTherefore, the answer is $30\\% \\cdot \\frac{2}{3} = \\boxed{20}.$",
"Like solution 1, it is clear that ... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_2 | B | 50 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could a... | [
"The discounted shoe is $20\\%$ off the original price. So that means $1 - 0.2 = 0.8$ . There is also a $7.5\\%$ sales tax charge, so $0.8 * 1.075 = 0.86$ . Now we can set up the equation $0.86x = 43$ , and solving that we get $x=\\boxed{50}$ ~ kabbybear",
"Let the original price be $x$ dollars. \nAfter the disco... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_2 | null | 50 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could a... | [
"We can create the equation: \\[0.8x \\cdot 1.075 = 43\\] using the information given. This is because x, the original price, got reduced by 20%, or multiplied by 0.8, and it also got multiplied by 1.075 on the discounted price. Solving that equation, we get \\[\\frac{4}{5} \\cdot x \\cdot \\frac{43}{40} = 43\\] \\... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | B | 50 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could a... | [
"The discounted shoe is $20\\%$ off the original price. So that means $1 - 0.2 = 0.8$ . There is also a $7.5\\%$ sales tax charge, so $0.8 * 1.075 = 0.86$ . Now we can set up the equation $0.86x = 43$ , and solving that we get $x=\\boxed{50}$ ~ kabbybear",
"Let the original price be $x$ dollars. \nAfter the disco... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | null | 50 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could a... | [
"We can create the equation: \\[0.8x \\cdot 1.075 = 43\\] using the information given. This is because x, the original price, got reduced by 20%, or multiplied by 0.8, and it also got multiplied by 1.075 on the discounted price. Solving that equation, we get \\[\\frac{4}{5} \\cdot x \\cdot \\frac{43}{40} = 43\\] \\... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_9 | E | 5 | Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?
[asy] import graph; size(8.76cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,... | [
"We observe the graph and see that the shape of the graph does not matter. We only want the total time it took Carmen and the total distance she traveled. Based on the graph, Carmen traveled 35 miles for 7 hours. Therefore, her average speed is $\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_3 | D | 1,920 | Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest?
$\textbf{(A) }560 \qquad \t... | [
"Note that the unit of the answer is strawberries , which is the product of\nBy conversion factors, we have \\[\\left(6 \\ \\color{red}\\cancel{\\mathrm{ft}}\\color{black}\\cdot8 \\ \\color{red}\\cancel{\\mathrm{ft}}\\color{black}\\right)\\cdot\\left(4 \\ \\frac{\\color{green}\\cancel{\\mathrm{plants}}}{\\color{red... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_1 | D | 10 | Casey's shop class is making a golf trophy. He has to paint 300 dimples on a golf ball. If it takes him 2 seconds to paint one dimple, how many minutes will he need to do his job?
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$ | [
"It will take him $300\\cdot2=600$ seconds to paint all the dimples.\nThis is equivalent to $\\frac{600}{60}=10$ minutes $\\Rightarrow \\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_14 | null | 574 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$ | [
"First note that $1001 = 143 \\cdot 7$ and $429 = 143 \\cdot 3$ so every point of the form $(7k, 3k)$ is on the line. Then consider the line $l$ from $(7k, 3k)$ to $(7(k + 1), 3(k + 1))$ . Translate the line $l$ so that $(7k, 3k)$ is now the origin. There is one square and one circle that intersect the line ar... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_24 | E | 900 | Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how ... | [
"Let us label the players of the first team $A$ $B$ , and $C$ , and those of the second team, $X$ $Y$ , and $Z$\n$\\textbf{1}$ . One way of scheduling all six distinct rounds could be:\nRound 1: $AX$ $BY$ $CZ$\nRound 2: $AX$ $BZ$ $CY$\nRound 3: $AY$ $BX$ $CZ$\nRound 4: $AY$ $BZ$ $CX$\nRound 5: $AZ$ $BX$ $CY$\nRound... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_5 | B | 25 | Chandler wants to buy a $500$ dollar mountain bike. For his birthday, his grandparents
send him $50$ dollars, his aunt sends him $35$ dollars and his cousin gives him $15$ dollars. He earns $16$ dollars per week for his paper route. He will use all of his birthday money and all
of the money he earns from his paper rout... | [
"Let $x$ be the number of weeks.\nThus, we have the equation $50 + 35 + 15 + 16x = 500$\nSimplify,\n$100 + 16x = 500$\n$16x = 400$\n$x = 25$\nThe answer is $\\boxed{25}$"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_6 | null | 167 | Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ . Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixe... | [
"The probability that he rolls a six twice when using the fair die is $\\frac{1}{6}\\times \\frac{1}{6}=\\frac{1}{36}$ . The probability that he rolls a six twice using the biased die is $\\frac{2}{3}\\times \\frac{2}{3}=\\frac{4}{9}=\\frac{16}{36}$ . Given that Charles rolled two sixes, we can see that it is $16... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | C | 93 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had c... | [
"Let the number of questions that they solved alone be $x$ . Let the percentage of problems they correctly solve together be $a$ %. \nAs given, \\[\\frac{80x}{100} + \\frac{ax}{100} = \\frac{2 \\cdot 88x}{100}\\]\nHence, $a = 96$\nZoe got $\\frac{90x}{100} + \\frac{ax}{100} = \\frac{186x}{100}$ problems right out o... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_14 | B | 481 | Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?
$\textbf{(A)}\ 480 \qquad\textbf{(B)}\ 481 \qquad\textbf{(C)}\ 482 ... | [
"There are 15 rows with 15 black tiles, and 16 rows with 16 black tiles, so the answer is $15^2+16^2 =225+256= \\boxed{481}$",
"We build the $31 \\times 31$ checkerboard starting with a board of $30 \\times 30$ that is exactly half black. There are $15 \\cdot 30$ black tiles in this region.\nAdd to this $30 \\ti... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_6 | A | 15 | Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
$\textbf{(A) } 15\qquad \textbf{(B) } 34\qquad \textbf{(C) } 43\qq... | [
"We work backwards; the number that Cindy started with is $3(43)+9=138$ . Now, the correct result is $\\frac{138-3}{9}=\\frac{135}{9}=15$ . Our answer is $\\boxed{15}$",
"Let the number be $x$ . We transform the problem into an equation: $\\frac{x-9}{3}=43$ . Solve for $x$ gives us $x=138$ . Therefore, the correc... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_2 | A | 15 | Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
$\textbf{(A) } 15\qquad \textbf{(B) } 34\qquad \textbf{(C) } 43\qq... | [
"We work backwards; the number that Cindy started with is $3(43)+9=138$ . Now, the correct result is $\\frac{138-3}{9}=\\frac{135}{9}=15$ . Our answer is $\\boxed{15}$",
"Let the number be $x$ . We transform the problem into an equation: $\\frac{x-9}{3}=43$ . Solve for $x$ gives us $x=138$ . Therefore, the correc... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_19 | null | 8 | Circle $A$ has radius $100$ . Circle $B$ has an integer radius $r<100$ and remains internally tangent to circle $A$ as it rolls once around the circumference of circle $A$ . The two circles have the same points of tangency at the beginning and end of circle $B$ 's trip. How many possible values can $r$ have?
$\mathrm{(... | [
"The circumference of circle $A$ is $200\\pi$ , and the circumference of circle $B$ with radius $r$ is $2r\\pi$ . Since circle $B$ makes a complete revolution and ends up on the same point , the circumference of $A$ must be a multiple of the circumference of $B$ , therefore the quotient must be an integer.\nThus, $... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8 | null | 254 | Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the ... | [
"Using the diagram above, let the radius of $D$ be $3r$ , and the radius of $E$ be $r$ . Then, $EF=r$ , and $CE=2-r$ , so the Pythagorean theorem in $\\triangle CEF$ gives $CF=\\sqrt{4-4r}$ . Also, $CD=CA-AD=2-3r$ , so \\[DF=DC+CF=2-3r+\\sqrt{4-4r}.\\] Noting that $DE=4r$ , we can now use the Pythagorean theorem in... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_12 | null | 110 | Circle $C_0$ has radius $1$ , and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$ . Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$ . Circle $C_2$ has radius $r^2$ and is internally tang... | [
"Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$ . Therefore $A_1=(1-r,r)$ $A_2=(1-r-r^2,r-r^2)$ $A_3=(1-r-r^2+r^3,r-r^2-r^3)$ $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$ , and so on, where the signs alternate in groups of $2$ . The limit of all these points is point $B$ . Using the... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_16 | E | 30 | Circle $C_1$ has its center $O$ lying on circle $C_2$ . The two circles meet at $X$ and $Y$ . Point $Z$ in the exterior of $C_1$ lies on circle $C_2$ and $XZ=13$ $OZ=11$ , and $YZ=7$ . What is the radius of circle $C_1$
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ \sqrt{26}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 2\s... | [
"Let $r$ denote the radius of circle $C_1$ . Note that quadrilateral $ZYOX$ is cyclic. By Ptolemy's Theorem, we have $11XY=13r+7r$ and $XY=20r/11$ . Let $t$ be the measure of angle $YOX$ . Since $YO=OX=r$ , the law of cosines on triangle $YOX$ gives us $\\cos t =-79/121$ . Again since $ZYOX$ is cyclic, the measure ... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_16 | null | 30 | Circle $C_1$ has its center $O$ lying on circle $C_2$ . The two circles meet at $X$ and $Y$ . Point $Z$ in the exterior of $C_1$ lies on circle $C_2$ and $XZ=13$ $OZ=11$ , and $YZ=7$ . What is the radius of circle $C_1$
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ \sqrt{26}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 2\s... | [
"Use the diagram above. Notice that $\\angle YZO=\\angle XZO$ as they subtend arcs of the same length. Let $A$ be the point of intersection of $C_1$ and $XZ$ . We now have $AZ=YZ=7$ and $XA=6$ . Consider the power of point $Z$ with respect to Circle $O,$ we have $13\\cdot 7 = (11 + r)(11 - r) = 11^2 - r^2,$ which g... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_8 | null | 405 | Circles $C_1$ and $C_2$ are externally tangent , and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear . A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the cho... | [
"Let $O_1, O_2, O_3$ be the centers and $r_1 = 4, r_2 = 10,r_3 = 14$ the radii of the circles $C_1, C_2, C_3$ . Let $T_1, T_2$ be the points of tangency from the common external tangent of $C_1, C_2$ , respectively, and let the extension of $\\overline{T_1T_2}$ intersect the extension of $\\overline{O_1O_2}$ at a p... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_9 | null | 27 | Circles $\mathcal{C}_1, \mathcal{C}_2,$ and $\mathcal{C}_3$ have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line $t_1$ is a common internal tangent to $\mathcal{C}_1$ and $\mathcal{C}_2$ and has a positive slope , and line $t_2$ is a common internal tangent to $\mathcal{C}_2$ ... | [
"2006 II AIME-9.png\nCall the centers $O_1, O_2, O_3$ , the points of tangency $r_1, r_2, s_1, s_2$ (with $r$ on $t_1$ and $s$ on $t_2$ , and $s_2$ on $\\mathcal{C}_2$ ), and the intersection of each common internal tangent to the X-axis $r, s$ $\\triangle O_1r_1r \\sim \\triangle O_2r_2r$ since both triangles have... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_15 | null | 282 | Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$ , and the product of the radii is $68$ . The x-axis and the line $y = mx$ , where $m > 0$ , are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$ , where $a$ $b$ , and $c$ are positive ... | [
"Let the smaller angle between the $x$ -axis and the line $y=mx$ be $\\theta$ . Note that the centers of the two circles lie on the angle bisector of the angle between the $x$ -axis and the line $y=mx$ . Also note that if $(x,y)$ is on said angle bisector, we have that $\\frac{y}{x}=\\tan{\\frac{\\theta}{2}}$ . Le... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | null | 129 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and ... | [
"Let $M$ be the intersection of $\\overline{BC}$ and the common internal tangent of $\\mathcal P$ and $\\mathcal Q.$ We claim that $M$ is the circumcenter of right $\\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\\overline{BC},$ imp... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_22 | E | 130 | Circles $\omega$ and $\gamma$ , both centered at $O$ , have radii $20$ and $17$ , respectively. Equilateral triangle $ABC$ , whose interior lies in the interior of $\omega$ but in the exterior of $\gamma$ , has vertex $A$ on $\omega$ , and the line containing side $\overline{BC}$ is tangent to $\gamma$ . Segments $\ove... | [
"\nLet $S$ be the point of tangency between $\\overline{BC}$ and $\\gamma$ , and $M$ be the midpoint of $\\overline{BC}$ . Note that $AM \\perp BS$ and $OS \\perp BS$ . This implies that $\\angle OAM \\cong \\angle AOS$ , and $\\angle AMP \\cong \\angle OSP$ . Thus, $\\triangle PMA \\sim \\triangle PSO$\nIf we let ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | D | 552 | Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\om... | [
"Let $O_1$ and $O_2$ be the centers of $\\omega_1$ and $\\omega_2$ respectively and draw $O_1O_2$ $O_1P_1$ , and $O_2P_2$ . Note that $\\angle{O_1P_1P_2}$ and $\\angle{O_2P_2P_3}$ are both right. Furthermore, since $\\triangle{P_1P_2P_3}$ is equilateral, $m\\angle{P_1P_2P_3} = 60^\\circ$ and $m\\angle{O_2P_2P_1} = ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_25 | null | 552 | Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\om... | [
"Let $O_i$ be the center of circle $\\omega_i$ for $i=1,2,3$ , and let $K$ be the intersection of lines $O_1P_1$ and $O_2P_2$ . Because $\\angle P_1P_2P_3 = 60^\\circ$ , it follows that $\\triangle P_2KP_1$ is a $30-60-90$ triangle. Let $x=P_1K$ ; then $P_2K = 2x$ and $P_1P_2 = x \\sqrt 3$ . The Law of Cosines in $... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | null | 270 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again... | [
"Let $Z = XY \\cap AB$ . By the radical axis theorem $AD, XY, BC$ are concurrent, say at $P$ . Moreover, $\\triangle DXP \\sim \\triangle PXC$ by simple angle chasing. Let $y = PX, x = XZ$ . Then \\[\\frac{y}{37} = \\frac{67}{y} \\qquad \\implies \\qquad y^2 = 37 \\cdot 67.\\] Now, $AZ^2 = \\tfrac 14 AB^2$ , and by... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | null | 672 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $... | [
"Let $O_i$ and $r_i$ be the center and radius of $\\omega_i$ , and let $O$ and $r$ be the center and radius of $\\omega$\nSince $\\overline{AB}$ extends to an arc with arc $120^\\circ$ , the distance from $O$ to $\\overline{AB}$ is $r/2$ . Let $X=\\overline{AB}\\cap \\overline{O_1O_2}$ . Consider $\\triangle OO_1O_... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_4 | D | 8 | Circles of diameter $1$ inch and $3$ inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?
2006amc10b04.gif
$\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad ... | [
"The area painted red is equal to the area of the smaller circle and the area painted blue is equal to the area of the larger circle minus the area of the smaller circle.\nSo we have:\n\\begin{align*} A_{red}&=\\pi\\left(\\frac{1}{2}\\right)^2=\\frac{\\pi}{4}\\\\ A_{blue}&=\\pi\\left(\\frac{3}{2}\\right)^2-\\pi\\le... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_4 | null | 17 | Circles of radii $5, 5, 8,$ and $\frac mn$ are mutually externally tangent, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | [
"1997 AIME-4.png\nIf (in the diagram above) we draw the line going through the centers of the circles with radii $8$ and $\\frac mn = r$ , that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii $5$ . Then we form two right triangles , of lengths $5, x, 5+r$ and $... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_5 | E | 12 | Circles of radius $2$ and $3$ are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.
$\mathrm{(A) \ } 3\pi\qquad \mathrm{(B) \ } 4\pi\qquad \mathrm{(C) \ } 6\pi\qquad \mathrm{(D) \ } 9\pi\qquad \mathrm{(E) \ } 12\pi$ | [
"A line going through the centers of the two smaller circles also goes through the diameter. The length of this line within the circle is $3+3+2+2=10.$ Because this is the length of the larger circle's diameter, the length of its radius is $5.$\nThe area of the large circle is $25\\pi$ , and the area of the two sma... |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_4 | null | 224 | Circles of radius $3$ and $6$ are externally tangent to each other and are internally tangent to a circle of radius $9$ . The circle of radius $9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord. | [
"We label the points as following: the centers of the circles of radii $3,6,9$ are $O_3,O_6,O_9$ respectively, and the endpoints of the chord are $P,Q$ . Let $A_3,A_6,A_9$ be the feet of the perpendiculars from $O_3,O_6,O_9$ to $\\overline{PQ}$ (so $A_3,A_6$ are the points of tangency ). Then we note that $\\overli... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_12 | D | 16 | Cities $A$ $B$ $C$ $D$ , and $E$ are connected by roads $\widetilde{AB}$ $\widetilde{AD}$ $\widetilde{AE}$ $\widetilde{BC}$ $\widetilde{BD}$ $\widetilde{CD}$ , and $\widetilde{DE}$ . How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some cities mor... | [
"Note that cities $C$ and $E$ can be removed when counting paths because if a path goes in to $C$ or $E$ , there is only one possible path to take out of cities $C$ or $E$ .\nSo the diagram is as follows:\n\nNow we proceed with casework. Remember that there are two ways to travel from $A$ to $D$ $D$ to $A$ $B$ to $... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | E | 27 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | [
"This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \\[12x+18x=45\\] Solving gives us $x=1.5$ . The $18x$ is Alicia so $18\\times1.5=\\boxed{27}$",
"The relative speed of the two is $18+12=30$ , so $\\frac{3}{2}$ hours would be required to travel $45$ miles. $d... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_1 | E | 27 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | [
"This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \\[12x+18x=45\\] Solving gives us $x=1.5$ . The $18x$ is Alicia so $18\\times1.5=\\boxed{27}$",
"The relative speed of the two is $18+12=30$ , so $\\frac{3}{2}$ hours would be required to travel $45$ miles. $d... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_11 | D | 143 | Claire adds the degree measures of the interior angles of a convex polygon and arrives at a sum of $2017$ . She then discovers that she forgot to include one angle. What is the degree measure of the forgotten angle?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 63\qquad\textbf{(C)}\ 117\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\... | [
"We know that the sum of the interior angles of the polygon is a multiple of $180$ . Note that $\\left\\lceil\\frac{2017}{180}\\right\\rceil = 12$ and $180\\cdot 12 = 2160$ , so the angle Claire forgot is $\\equiv 2160-2017=143\\mod 180$ . Since the polygon is convex, the angle is $\\leq 180$ , so the answer is $\\... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_13 | C | 5 | Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | [
"Let Claudia have $x$ 5-cent coins and $\\left( 12 - x \\right)$ 10-cent coins. It is easily observed that any multiple of $5$ between $5$ and $5x + 10(12 - x) = 120 - 5x$ inclusive can be obtained by a combination of coins. Thus, $24 - x = 17$ combinations can be made, so $x = 7$ . But the answer is not $7,$ becau... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_11 | null | 341 | Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$ . The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$ , where $m$ and ... | [
"Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the complement principle , the desired probability is half the probability tha... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_14 | null | 375 | Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$ | [
"By Vieta's formula, the sum of the roots is equal to 0, or $a+b+c=0$ . Therefore, $\\frac{(a+b+c)}{3}=0$ . Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be $x$ and the other leg be $y$ . Without the loss of genera... |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | null | 869 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | [
"Note that the four numbers to multiply are symmetric with the center at $29.5$ . \nMultiply the symmetric pairs to get $31\\cdot 28=868$ and $30\\cdot 29=870$ $\\sqrt{868\\cdot 870 + 1} = \\sqrt{(869-1)(869+1) + 1} = \\sqrt{869^2 - 1^2 + 1} = \\sqrt{869^2} = \\boxed{869}$",
"Notice that $(a+1)^2 = a \\cdot (a+2)... |
https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14 | null | 373 | Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\] | [
"The Sophie Germain Identity states that $a^4 + 4b^4$ can be factored as $\\left(a^2 + 2b^2 - 2ab\\right)\\left(a^2 + 2b^2 + 2ab\\right).$ Each of the terms is in the form of $x^4 + 324.$ Using Sophie Germain, we get that \\begin{align*} x^4 + 324 &= x^4 + 4\\cdot 3^4 \\\\ &= \\left(x^2 + 2 \\cdot 3^2 - 2\\cdot 3\\... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_1 | B | 15 | Connie multiplies a number by $2$ and gets $60$ as her answer. However, she should have divided the number by $2$ to get the correct answer. What is the correct answer?
$\textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240$ | [
"If $x$ is the number, then $2x=60$ and $x=30$ . Dividing the number by $2$ yields $\\dfrac{30}{2} = \\boxed{15}$"
] |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_27 | C | 4 | Consider a sequence $x_1,x_2,x_3,\dotsc$ defined by: \begin{align*}&x_1 = \sqrt[3]{3}, \\ &x_2 = \left(\sqrt[3]{3}\right)^{\sqrt[3]{3}},\end{align*} and in general \[x_n = \left(x_{n-1}\right)^{\sqrt[3]{3}} \text{ for } n > 1.\]
What is the smallest value of $n$ for which $x_n$ is an integer?
$\mathrm{(A)\ } 2 \qquad \... | [
"Firstly, we will show by induction that \\[x_n = \\left(\\sqrt[3]{3}\\right)^{\\left(\\left(\\sqrt[3]{3}\\right)^{n-1}\\right)}.\\] For the base case, we indeed have \\begin{align*}x_1 &= \\sqrt[3]{3} \\\\ &= \\left(\\sqrt[3]{3}\\right)^1 \\\\ &= \\left(\\sqrt[3]{3}\\right)^{\\left(\\left(\\sqrt[3]{3}\\right)^0\\r... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_14 | null | 1 | Consider a string of $n$ $7$ 's, $7777\cdots77,$ into which $+$ signs are inserted to produce an arithmetic expression . For example, $7+77+777+7+7=875$ could be obtained from eight $7$ 's in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value $7000$ | [
"It's obvious that we cannot have any number $\\ge 7777$ because $7777 > 7000$ so the max number that an occur is $777$\nLet's say we have $a$ 777's , $b$ 77's and $c$ 7's\nFrom here we get our required equation as $777a + 77b + 7c = 7000$\nNow comes the main problem , one might think that if we find number of $(a,... |
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