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/2000_AMC_8_Problems/Problem_25 | B | 27 | The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is
[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$"... | [
"To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension. Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$ , which is the easiest way to avoid fractions. Labelling the right midpoint as $M$ , and the bottom ... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_9 | A | 2 | The area of the largest triangle that can be inscribed in a semi-circle whose radius is $r$ is:
$\textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2}$ | [
"The area of a triangle is $\\frac12 bh.$ To maximize the base, let it be equal to the diameter of the semi circle, which is equal to $2r.$ To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to $r.$ Therefore the area is $\\f... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_19 | E | 54 | The area of the region bounded by the graph of \[x^2+y^2 = 3|x-y| + 3|x+y|\] is $m+n\pi$ , where $m$ and $n$ are integers. What is $m + n$
$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$ | [
"In order to attack this problem, we can use casework on the sign of $|x-y|$ and $|x+y|$\nCase 1: $|x-y|=x-y, |x+y|=x+y$\nSubstituting and simplifying, we have $x^2-6x+y^2=0$ , i.e. $(x-3)^2+y^2=3^2$ , which gives us a circle of radius $3$ centered at $(3,0)$\nCase 2: $|x-y|=y-x, |x+y|=x+y$\nSubstituting and simpli... |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_15 | D | 64 | The area of the shaded region $\text{BEDC}$ in parallelogram $\text{ABCD}$ is
[asy] unitsize(10); pair A,B,C,D,E; A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0); draw(A--B--C--D--cycle); fill(B--E--D--C--cycle,gray); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S); label("$10$",(9,8),... | [
"Let $[ABC]$ denote the area of figure $ABC$\nClearly, $[BEDC]=[ABCD]-[ABE]$ . Using basic area formulas,\nSince $AE+ED=BC=10$ and $ED=6$ $AE=4$ and the area of $\\triangle ABE$ is $4(4)=16$\nFinally, we have $[BEDC]=80-16=64\\rightarrow \\boxed{64}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/Problem_15 | null | 145 | The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any pri... | [
"Lemma: If $x,y$ satisfy $px+qy=1$ , then the minimal value of $\\sqrt{x^2+y^2}$ is $\\frac{1}{\\sqrt{p^2+q^2}}$\nProof: Recall that the distance between the point $(x_0,y_0)$ and the line $px+qy+r = 0$ is given by $\\frac{|px_0+qy_0+r|}{\\sqrt{p^2+q^2}}$ . In particular, the distance between the origin and any poi... |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_7 | D | 64 | The area of the smallest square that will contain a circle of radius 4 is
$\text{(A)}\ 8 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$ | [
"Draw a square circumscribed around the circle. (Alternately, the circle is inscribed in the square.) If the circle has radius $4$ , it has diameter $8$ . Two of the diameters of the circle will run parallel to the sides of the square. Thus, the smallest square that contains it has side length $8$ , and area $8... |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_15 | E | 50 | The area of this figure is $100\text{ cm}^2$ . Its perimeter is
[asy] draw((0,2)--(2,2)--(2,1)--(3,1)--(3,0)--(1,0)--(1,1)--(0,1)--cycle,linewidth(1)); draw((1,2)--(1,1)--(2,1)--(2,0),dashed); [/asy]
$\text{(A)}\ \text{20 cm} \qquad \text{(B)}\ \text{25 cm} \qquad \text{(C)}\ \text{30 cm} \qquad \text{(D)}\ \text{40 c... | [
"Since the area of the whole figure is $100$ , each square has an area of $25$ and the side length is $5$\nThere are $10$ sides of this length, so the perimeter is $10(5)=50\\rightarrow \\boxed{50}$"
] |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_21 | B | 10 | The area of trapezoid $ABCD$ is $164\text{ cm}^2$ . The altitude is 8 cm, $AB$ is 10 cm, and $CD$ is 17 cm. What is $BC$ , in centimeters?
[asy]/* AMC8 2003 #21 Problem */ size(4inch,2inch); draw((0,0)--(31,0)--(16,8)--(6,8)--cycle); draw((11,8)--(11,0), linetype("8 4")); draw((11,1)--(12,1)--(12,0)); label("$A$", (0... | [
"Using the formula for the area of a trapezoid, we have $164=8(\\frac{BC+AD}{2})$ . Thus $BC+AD=41$ . Drop perpendiculars from $B$ to $AD$ and from $C$ to $AD$ and let them hit $AD$ at $E$ and $F$ respectively. Note that each of these perpendiculars has length $8$ . From the Pythagorean Theorem, $AE=6$ and $DF=15$ ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_20 | D | 3 | The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is
[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.... | [
"The shaded region is a right trapezoid. Assume WLOG that $YZ=8$ . Then because the area of $\\triangle XYZ$ is equal to 8, the height of the triangle $XC=2$ . Because the line $AB$ is a midsegment, the top base of the trapezoid is $\\frac12 AB = \\frac14 YZ = 2$ . Also, $AB$ divides $XC$ in two, so the height of t... |
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_13 | D | 37.5 | The arithmetic mean (average) of a set of $50$ numbers is $38$ . If two numbers, namely, $45$ and $55$ , are discarded, the mean of the remaining set of numbers is: $\textbf{(A)}\ 36.5 \qquad\textbf{(B)}\ 37\qquad\textbf{(C)}\ 37.2\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 37.52$ | [
"Since the arithmetic mean of the $50$ numbers is $38$ , their sum must be $50*38 = 1900$ . After $45$ and $55$ are discarded, the sum decreases by $45 + 55 = 100$ , so it must become $1900 - 100 = 1800$ . \nBut this means that the new mean of the remaining $50 - 2 = 48$ numbers must be $\\frac{1800}{48} = 37.5$ . ... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_15 | A | 81 | The arithmetic mean (average) of four numbers is $85$ . If the largest of these numbers is $97$ , then the mean of the remaining three numbers is
$\text{(A)}\ 81.0 \qquad \text{(B)}\ 82.7 \qquad \text{(C)}\ 83.0 \qquad \text{(D)}\ 84.0 \qquad \text{(E)}\ 84.3$ | [
"Say that the four numbers are $a, b, c,$ $97$ . Then $\\frac{a+b+c+97}{4} = 85$ . What we are trying to find is $\\frac{a+b+c}{3}$ . Solving, \\[\\frac{a+b+c+97}{4} = 85\\] \\[a+b+c+97 = 340\\] \\[a+b+c = 243\\] \\[\\frac{a+b+c}{3} = \\boxed{81}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1958_AHSME_Problems/Problem_6 | B | 1 | The arithmetic mean between $\frac {x + a}{x}$ and $\frac {x - a}{x}$ , when $x \not = 0$ , is:
$\textbf{(A)}\ {2}\text{, if }{a \not = 0}\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ {1}\text{, only if }{a = 0}\qquad \textbf{(D)}\ \frac {a}{x}\qquad \textbf{(E)}\ x$ | [
"We have $\\frac{1}{2}\\cdot \\left(\\frac{x + a}{x} + \\frac{x - a}{x}\\right) = \\frac{2}{2} = \\boxed{1}$"
] |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_11 | B | 37.5 | The arithmetic mean of a set of $50$ numbers is $38$ . If two numbers of the set, namely $45$ and $55$ , are discarded,
the arithmetic mean of the remaining set of numbers is:
$\textbf{(A)}\ 38.5 \qquad \textbf{(B)}\ 37.5 \qquad \textbf{(C)}\ 37 \qquad \textbf{(D)}\ 36.5 \qquad \textbf{(E)}\ 36$ | [
"If the arithmetic mean of a set of $50$ numbers is $38$ , then the sum of the $50$ numbers equals $1900$ . Since $45$ and $55$ are being removed, subtract $100$ to get the sum of the remaining $48$ numbers, which is $1800$ . Therefore, the new mean is $37.5$ , which is answer choice $\\boxed{37.5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_3 | A | 0 | The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ doesn't contain the digit
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$ | [
"We wish to find $\\frac{9+99+\\cdots +999999999}{9}$ , or $\\frac{9(1+11+111+\\cdots +111111111)}{9}=123456789$ . This doesn't have the digit 0, so the answer is $\\boxed{0}$",
"Notice that the final number is guaranteed to have the digits $\\{1, 3, 5, 7, 9\\}$ and that each of these digits can be paired with an... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_1 | A | 0 | The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ doesn't contain the digit
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$ | [
"We wish to find $\\frac{9+99+\\cdots +999999999}{9}$ , or $\\frac{9(1+11+111+\\cdots +111111111)}{9}=123456789$ . This doesn't have the digit 0, so the answer is $\\boxed{0}$",
"Notice that the final number is guaranteed to have the digits $\\{1, 3, 5, 7, 9\\}$ and that each of these digits can be paired with an... |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_19 | C | 55 | The average (arithmetic mean) of $10$ different positive whole numbers is $10$ . The largest possible value of any of these numbers is
$\text{(A)}\ 10 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 55 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 91$ | [
"If the average of the numbers is $10$ , then their sum is $10\\times 10=100$\nTo maximize the largest number of the ten, we minimize the other nine. Since they must be distinct, positive whole numbers, we let them be $1,2,3,4,5,6,7,8,9$ . Their sum is $45$\nThe sum of nine of the numbers is $45$ , and the sum of... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_6 | B | 24 | The average (mean) of $20$ numbers is $30$ , and the average of $30$ other numbers is $20$ . What is the average of all $50$ numbers?
$\textbf{(A) } 23\qquad \textbf{(B) } 24\qquad \textbf{(C) } 25\qquad \textbf{(D) } 26\qquad \textbf{(E) } 27$ | [
"Since the average of the first $20$ numbers is $30$ , their sum is $20\\cdot30=600$\nSince the average of $30$ other numbers is $20$ , their sum is $30\\cdot20=600$\nSo the sum of all $50$ numbers is $600+600=1200$\nTherefore, the average of all $50$ numbers is $\\frac{1200}{50}=\\boxed{24}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_6 | C | 24.75 | The average age of $33$ fifth-graders is $11$ . The average age of $55$ of their parents is $33$ . What is the average age of all of these parents and fifth-graders?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28$ | [
"The sum of the ages of the fifth graders is $33 * 11$ , while the sum of the ages of the parents is $55 * 33$ . Therefore, the total sum of all their ages must be $2178$ , and given $33 + 55 = 88$ people in total, their average age is $\\frac{2178}{88} = \\frac{99}{4} = \\boxed{24.75}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_5 | C | 24.75 | The average age of $33$ fifth-graders is $11$ . The average age of $55$ of their parents is $33$ . What is the average age of all of these parents and fifth-graders?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28$ | [
"The sum of the ages of the fifth graders is $33 * 11$ , while the sum of the ages of the parents is $55 * 33$ . Therefore, the total sum of all their ages must be $2178$ , and given $33 + 55 = 88$ people in total, their average age is $\\frac{2178}{88} = \\frac{99}{4} = \\boxed{24.75}$"
] |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_7 | D | 33 | The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves
the room. What is the average age of the four remaining people?
$\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36$ | [
"Let $x$ be the average of the remaining $4$ people.\nThe equation we get is $\\frac{4x + 18}{5} = 30$\nSimplify,\n$4x + 18 = 150$\n$4x = 132$\n$x = 33$\nTherefore, the answer is $\\boxed{33}$",
"Since an $18$ year old left from a group of people averaging $30$ , The remaining people must total $30 - 18 = 12$ yea... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_10 | D | 34 | The average age of the $6$ people in Room A is $40$ . The average age of the $4$ people in Room B is $25$ . If the two groups are combined, what is the average age of all the people?
$\textbf{(A)}\ 32.5 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 33.5 \qquad \textbf{(D)}\ 34\qquad \textbf{(E)}\ 35$ | [
"The total of all their ages over the number of people is\n\\[\\frac{6 \\cdot 40 + 4\\cdot 25}{6+4} = \\frac{340}{10} = \\boxed{34}.\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_13 | C | 28 | The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?
$\text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 29 \qqua... | [
"First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals $(20)(15)+(15)(16)=540$ . The total amount of everyone's ages can be found from the average age, $17\\cdot40=680$ . Then you do $680-540=140$ to find the sum of the adult's ages. The average age of an adult ... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_6 | E | 80 | The average cost of a long-distance call in the USA in $1985$ was $41$ cents per minute, and the average cost of a long-distance
call in the USA in $2005$ was $7$ cents per minute. Find the
approximate percent decrease in the cost per minute of a long-
distance call.
$\mathrm{(A)}\ 7 \qquad\mathrm{(B)}\ 17 \qquad\mathr... | [
"The percent decrease is (the amount of decrease)/(original amount)\nthe amount of decrease is $41 - 7 = 34$\nso the percent decrease is $\\frac{34}{41}$ which is about $\\boxed{80}$"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_9 | D | 58 | The average of the five numbers in a list is $54$ . The average of the first two
numbers is $48$ . What is the average of the last three numbers?
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 57 \qquad \textbf{(D)}\ 58 \qquad \textbf{(E)}\ 59$ | [
"Let the $5$ numbers be $a, b, c, d$ , and $e$ . Thus $\\frac{a+b+c+d+e}{5}=54$ and $a+b+c+d+e=270$ . Since $\\frac{a+b}{2}=48$ $a+b=96$ . Substituting back into our original equation, we have $96+c+d+e=270$ and $c+d+e=174$ . Dividing by $3$ gives the average of $\\boxed{58}$"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_14 | A | 0 | The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$ | [
"Let the total value, in cents, of the coins Paula has originally be $v$ , and the number of coins she has be $n$ . Then $\\frac{v}{n}=20\\Longrightarrow v=20n$ and $\\frac{v+25}{n+1}=21$ . Substituting yields: $20n+25=21(n+1),$ so $n=4$ $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_11 | A | 0 | The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$ | [
"Let the total value, in cents, of the coins Paula has originally be $v$ , and the number of coins she has be $n$ . Then $\\frac{v}{n}=20\\Longrightarrow v=20n$ and $\\frac{v+25}{n+1}=21$ . Substituting yields: $20n+25=21(n+1),$ so $n=4$ $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she... |
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_18 | C | 138 | The average weight of $6$ boys is $150$ pounds and the average weight of $4$ girls is $120$ pounds. The average weight of the $10$ children is
$\text{(A)}\ 135\text{ pounds} \qquad \text{(B)}\ 137\text{ pounds} \qquad \text{(C)}\ 138\text{ pounds} \qquad \text{(D)}\ 140\text{ pounds} \qquad \text{(E)}\ 141\text{ pound... | [
"Let the $6$ boys have total weight $S_B$ and let the $4$ girls have total weight $S_G$ . We are given\n\\begin{align*} \\frac{S_B}{6} &= 150 \\\\ \\frac{S_G}{4} &= 120 \\end{align*}\nWe want the average of the $10$ children, which is \\[\\frac{S_B+S_G}{10}\\] From the first two equations , we can determine that ... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_11 | B | 24 | The bar graph shows the results of a survey on color preferences. What percent preferred blue?
[asy] for (int a = 1; a <= 6; ++a) { draw((-1.5,4*a)--(1.5,4*a)); } draw((0,28)--(0,0)--(32,0)); draw((3,0)--(3,20)--(6,20)--(6,0)); draw((9,0)--(9,24)--(12,24)--(12,0)); draw((15,0)--(15,16)--(18,16)--(18,0)); draw((21,... | [
"The total frequency is $50+60+40+60+40=250$ , with the blue frequency of $60$ . Therefore, the precentage that preferred blue is $\\frac{60}{250}=\\boxed{24}$"
] |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_14 | C | 13 | The base of isosceles $\triangle ABC$ is $24$ and its area is $60$ . What is the length of one
of the congruent sides?
$\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18$ | [
"The area of a triangle is shown by $\\frac{1}{2}bh$ . We set the base equal to $24$ , and the area equal to $60$ , and we get the height, or altitude, of the triangle to be $5$ . In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, $a^2+b^2=c^2$ , we can solve for one of th... |
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_19 | A | 19 | The base of the decimal number system is ten, meaning, for example, that $123 = 1\cdot 10^2 + 2\cdot 10 + 3$ . In the binary system, which has base two, the first five positive integers are $1,\,10,\,11,\,100,\,101$ . The numeral $10011$ in the binary system would then be written in the decimal system as:
$\textbf{(A)}... | [
"Numbers in binary work similar to their decimal counterparts, where the multiplier associated with each place is multiplied by two every single place to the left. For example, $1111_2$ $1111$ in base $2$ ) would equate to $1 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 = 8+4+2+1 = 15$\nUsing this same logic, $10011_2$ woul... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_12 | D | 3 | The base-nine representation of the number $N$ is $27006000052_{\text{nine}}.$ What is the remainder when $N$ is divided by $5?$
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$ | [
"Recall that $9\\equiv-1\\pmod{5}.$ We expand $N$ by the definition of bases: \\begin{align*} N&=27006000052_9 \\\\ &= 2\\cdot9^{10} + 7\\cdot9^9 + 6\\cdot9^6 + 5\\cdot9 + 2 \\\\ &\\equiv 2\\cdot(-1)^{10} + 7\\cdot(-1)^9 + 6\\cdot(-1)^6 + 5\\cdot(-1) + 2 &&\\pmod{5} \\\\ &\\equiv 2-7+6-5+2 &&\\pmod{5} \\\\ &\\equiv... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_10 | D | 3 | The base-nine representation of the number $N$ is $27006000052_{\text{nine}}.$ What is the remainder when $N$ is divided by $5?$
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$ | [
"Recall that $9\\equiv-1\\pmod{5}.$ We expand $N$ by the definition of bases: \\begin{align*} N&=27006000052_9 \\\\ &= 2\\cdot9^{10} + 7\\cdot9^9 + 6\\cdot9^6 + 5\\cdot9 + 2 \\\\ &\\equiv 2\\cdot(-1)^{10} + 7\\cdot(-1)^9 + 6\\cdot(-1)^6 + 5\\cdot(-1) + 2 &&\\pmod{5} \\\\ &\\equiv 2-7+6-5+2 &&\\pmod{5} \\\\ &\\equiv... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_14 | C | 12 | The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$ , where $T$ $M$ , and $H$ denote digits that are not given. What is $T+M+H$
$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$ | [
"We can figure out $H = 0$ by noticing that $19!$ will end with $3$ zeroes, as there are three factors of $5$ in its prime factorization, so there would be 3 powers of 10 meaning it will end in 3 zeros. Next, we use the fact that $19!$ is a multiple of both $11$ and $9$ . Their divisibility rules (see Solution 2) t... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_6 | null | 392 | The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this proc... | [
"Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its orig... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_17 | B | 10 | The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?
[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)... | [
"The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals $4$ and $5$ , for an area $A=10$ . The height $h$ , of one pyramid, is $\\frac{3}{2}$ , so the volume of one pyramid is $\\frac{Ah}{3}=5$ . Thus, the octahedron has volume $2\\cdot5=\\box... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_17 | null | 10 | The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?
[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)... | [
"The \"base\" of the octahedron is half the base of the rectangular prism because it is connected by the midpoints. Additionally, the volume of an octahedron is $\\dfrac{1}{3}$ of its respective prism. Thus, the octahedron's volume is $\\dfrac{1}{2} \\cdot \\dfrac{1}{3} = \\dfrac{1}{6}$ of the rectangular prism's v... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_42 | E | 40 | The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches.
The length of the common internal tangent is:
$\textbf{(A)}\ 41\text{ inches} \qquad \textbf{(B)}\ 39\text{ inches} \qquad \textbf{(C)}\ 39.8\text{ inches} \qquad \textbf{(D)}\... | [
"\nLet $A$ be the center of the circle with radius $5$ , and $B$ be the center of the circle with radius $4$ . Let $\\overline{CD}$ be the common internal tangent of circle $A$ and circle $B$ . Extend $\\overline{BD}$ past $D$ to point $E$ such that $\\overline{BE}\\perp\\overline{AE}$ . Since $\\overline{AC}\\perp... |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_24 | D | 1.05 | The check for a luncheon of 3 sandwiches, 7 cups of coffee and one piece of pie came to $$3.15$ . The check for a luncheon consisting of 4 sandwiches, 10 cups of coffee and one piece of pie came to $$4.20$ at the same place. The cost of a luncheon consisting of one sandwich, one cup of coffee, and one piece of pie at t... | [
"Let $s$ be the cost of one sandwich, $c$ be the cost of one cup of coffee, and $p$ be the price of one piece of pie. With the information,\n\\[3s+7c+p=3.15\\] \\[4s+10c+p=4.20\\]\nSubtract the first equation from the second to get\n\\[s+3c=1.05\\]\nThat means $s=1.05-3c$ . Substituting it back in the second equa... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_11 | null | 23 | The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [
"Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$ , respectively. Let $OB = r$ . Notice that $\\triangle{OMB} \\sim \\triangle{QOB}$ because both are right triangles, and $\\angle{OBQ} \\cong \\angle{OBM}$ . By $\\frac{MB}{BO}=\\frac{BO}{BQ}$ $MB = r\\left(\\frac{r}{4.5}\\right) = \\frac{r^2}{4.... |
https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_20 | D | 14 | The coefficient of $x^7$ in the expansion of $\left(\frac{x^2}{2}-\frac{2}{x}\right)^8$ is:
$\textbf{(A)}\ 56\qquad \textbf{(B)}\ -56\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ -14\qquad \textbf{(E)}\ 0$ | [
"By the Binomial Theorem , each term of the expansion is $\\binom{8}{n}\\left(\\frac{x^2}{2}\\right)^{8-n}\\left(\\frac{-2}{x}\\right)^n$\nWe want the exponent of $x$ to be $7$ , so \\[2(8-n)-n=7\\] \\[16-3n=7\\] \\[n=3\\]\nIf $n=3$ , then the corresponding term is \\[\\binom{8}{3}\\left(\\frac{x^2}{2}\\right)^{5}\... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_3 | null | 15 | The complex number $z$ is equal to $9+bi$ , where $b$ is a positive real number and $i^{2}=-1$ . Given that the imaginary parts of $z^{2}$ and $z^{3}$ are the same, what is $b$ equal to? | [
"Squaring, we find that $(9 + bi)^2 = 81 + 18bi - b^2$ . Cubing and ignoring the real parts of the result, we find that $(81 + 18bi - b^2)(9 + bi) = \\ldots + (9\\cdot 18 + 81)bi - b^3i$\nSetting these two equal, we get that $18bi = 243bi - b^3i$ , so $b(b^2 - 225) = 0$ and $b = -15, 0, 15$ . Since $b > 0$ , the so... |
https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_21 | null | 289 | The complex number $z$ satisfies $z + |z| = 2 + 8i$ . What is $|z|^{2}$ ? Note: if $z = a + bi$ , then $|z| = \sqrt{a^{2} + b^{2}}$
$\textbf{(A)}\ 68\qquad \textbf{(B)}\ 100\qquad \textbf{(C)}\ 169\qquad \textbf{(D)}\ 208\qquad \textbf{(E)}\ 289$ | [
"Let the complex number $z$ equal $a+bi$ . Then the preceding equation can be expressed as \\[a+bi+\\sqrt{a^2+b^2} = 2+8i\\] Because $a$ and $b$ must both be real numbers, we immediately have that $bi = 8i$ , giving $b = 8$ . Plugging this in back to our equation gives us $a+\\sqrt{a^2+64} = 2$ . \nRearranging thi... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_6 | null | 71 | The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$ , for relatively prime positive integers $m$ and $n$ with $m<n.$ Find $n.$ | [
"Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{143} = z.$ We know that $z \\neq 0,$ because we are given the imaginary part of $z,$ so we can divide by $z$ to get $z^{142} = 1.$ So, $z$ must be a $142$ nd root of unity, and thus, by De Moivre's theorem, the imaginary... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_8 | null | 40 | The complex numbers $z$ and $w$ satisfy the system \[z + \frac{20i}w = 5+i\] \[w+\frac{12i}z = -4+10i\] Find the smallest possible value of $\vert zw\vert^2$ | [
"Multiplying the two equations together gives us \\[zw + 32i - \\frac{240}{zw} = -30 + 46i\\] and multiplying by $zw$ then gives us a quadratic in $zw$ \\[(zw)^2 + (30-14i)zw - 240 =0.\\] Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \\pm \\sqrt{(15-7i)^2 + 240}$ $6+2i,$ $12i - 3... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11 | D | 11 | The coordinates of $\triangle ABC$ are $A(5,7)$ $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$
[asy] draw((3,11)--(11,7)--(5,7)--(3,11)); dot((5,7)); label("$A(5,7)$",(5,7),S); dot((11,7)); label("$B(11,7)$",(11,7),S); dot((3,11)); label("$C(3,y)$",(3,11),NW); [... | [
"The triangle has base $6,$ which means its height satisfies \\[\\dfrac{6h}{2}=3h=12.\\] This means that $h=4,$ so the answer is $7+4=\\boxed{11}$",
" Label point $D(3,7)$ as the point at which $CD\\perp DA$ . We now have $[\\triangle ABC] = [\\triangle BCD] - [\\triangle ACD]$ , where the brackets denote areas. ... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_11 | null | 131 | The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$ . The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\overline{AB}$ is $m/n$ ... | [
"For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$ . Suppose $B$ has integer coordinates; then $\\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\\overline{CD}$ , and let $D... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_5 | B | 1 | The data set $[6, 19, 33, 33, 39, 41, 41, 43, 51, 57]$ has median $Q_2 = 40$ , first quartile $Q_1 = 33$ , and third quartile $Q_3 = 43$ . An outlier in a data set is a value that is more than $1.5$ times the interquartile range below the first quartle ( $Q_1$ ) or more than $1.5$ times the interquartile range above th... | [
"The interquartile range is defined as $Q3 - Q1$ , which is $43 - 33 = 10$ $1.5$ times this value is $15$ , so all values more than $15$ below $Q1$ $33 - 15 = 18$ is an outlier. The only one that fits this is $6$ . All values more than $15$ above $Q3 = 43 + 15 = 58$ are also outliers, of which there are none so the... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | D | 26 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \... | [
"We have\n\\[\\dfrac{1}{20^{20}} = \\dfrac{1}{(10\\cdot2)^{20}}=\\dfrac{1}{10^{20}\\cdot2^{20}}\\]\nNow we do some estimation. Notice that $2^{20} = 1024^2$ , which means that $2^{20}$ is a little more than $1000^2=1,000,000$ . Multiplying it with $10^{20}$ , we get that the denominator is about $1\\underbrace{00\\... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_21 | B | 30 | The degree measure of angle $A$ is
[asy] unitsize(12); draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle); draw(arc((1,-10),(1+19/sqrt(461),-10+10/sqrt(461)),(25/17,-155/17),CCW)); draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW)); draw(arc((900/83,-400/83),(900/83+19/sqrt(461),-400/83+10/sqrt(461)),(900/83 - 9/sqrt... | [
"Angle-chasing using the small triangles:\nUse the line below and to the left of the $110^\\circ$ angle to find that the rightmost angle in the small lower-left triangle is $180 - 110 = 70^\\circ$\nThen use the small lower-left triangle to find that the remaining angle in that triangle is $180 - 70 - 40 = 70^\\circ... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_3 | null | 143 | The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle | [
"Another way to solve this problem would be to use exterior angles. Exterior angles of any polygon add up to $360^{\\circ}$ . Since there are $18$ exterior angles in an 18-gon, the average measure of an exterior angles is $\\frac{360}{18}=20^\\circ$ . We know from the problem that since the exterior angles must be ... |
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_2 | D | 17 | The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is
$\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72$ | [
"It becomes $(x^{8}+...)(x^{9}+...)$ with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or $\\boxed{17}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_10 | null | 240 | The diagram below shows a $4\times4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors.
Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the sequence is strictly increasing. Let $m$ be the maximu... | [
"We label our points using coordinates $0 \\le x,y \\le 3$ , with the bottom-left point being $(0,0)$ . By the Pythagorean Theorem , the distance between two points is $\\sqrt{d_x^2 + d_y^2}$ where $0 \\le d_x, d_y \\le 3$ ; these yield the possible distances (in decreasing order) \\[\\sqrt{18},\\ \\sqrt{13},\\ \\s... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | C | 4 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what poi... | [
"The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer ... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_18 | C | 64 | The diagram represents a $7$ -foot-by- $7$ -foot floor that is tiled with $1$ -square-foot black tiles and white tiles. Notice that the corners have white tiles. If a $15$ -foot-by- $15$ -foot floor is to be tiled in the same manner, how many white tiles will be needed?
[asy]unitsize(10); draw((0,0)--(7,0)--(7,7)--(0,7... | [
"In a $1$ -foot-by- $1$ -foot floor, there is $1$ white tile. In a $3$ -by- $3$ , there are $4$ . Continuing on, you can deduce the $n^{th}$ positive odd integer floor has $n^2$ white tiles. $15$ is the $8^{th}$ odd integer, so there are $\\boxed{64}$ white tiles."
] |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_4 | null | 260 | The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33... | [
"Call the squares' side lengths from smallest to largest $a_1,\\ldots,a_9$ , and let $l,w$ represent the dimensions of the rectangle.\nThe picture shows that \\begin{align*} a_1+a_2 &= a_3\\\\ a_1 + a_3 &= a_4\\\\ a_3 + a_4 &= a_5\\\\ a_4 + a_5 &= a_6\\\\ a_2 + a_3 + a_5 &= a_7\\\\ a_2 + a_7 &= a_8\\\\ a_1 + a_4 + ... |
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_1 | D | 10.3 | The diagram shows part of a scale of a measuring device. The arrow indicates an approximate reading of
[asy] draw((-3,0)..(0,3)..(3,0)); draw((-3.5,0)--(-2.5,0)); draw((0,2.5)--(0,3.5)); draw((2.5,0)--(3.5,0)); draw((1.8,1.8)--(2.5,2.5)); draw((-1.8,1.8)--(-2.5,2.5)); draw((0,0)--3*dir(120),EndArrow); label("$10$",(-2... | [
"Clearly the arrow marks a value between $10.25$ and $10.5$ , so only $\\text{C}$ and $\\text{D}$ are possible.\nLooking, we see that the arrow is closer to $10.3$ , so $\\boxed{10.3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_4 | A | 15 | The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?
[asy] for (int a = 0; a < 6; ++a) { for (int b = 0; b < 6; ++b) { dot((4*a,3*b)); } } draw((0,0)--(20,0)--(20,15)--(0,15)--cycle); draw((0,0)--(16,12)); draw((0,0)--(16,9)); label... | [
"After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60. Thus the answer is $60-45=15$ $\\boxed{15}$",
"We see that each dot is $15$ units away from the nearest one above it. So the answer is $\\boxed{15}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_2 | null | 154 | The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where ... | [
"Let the radius of the circles be $r$ . The longer dimension of the rectangle can be written as $14r$ , and by the Pythagorean Theorem , we find that the shorter dimension is $2r\\left(\\sqrt{3}+1\\right)$\nTherefore, $\\frac{14r}{2r\\left(\\sqrt{3}+1\\right)}= \\frac{7}{\\sqrt{3} + 1} \\cdot \\left[\\frac{\\sqrt{3... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_6 | D | 99 | The difference between a two-digit number and the number obtained by reversing its digits is $5$ times the sum of the digits of either number. What is the sum of the two digit number and its reverse?
$\textbf{(A) }44\qquad \textbf{(B) }55\qquad \textbf{(C) }77\qquad \textbf{(D) }99\qquad \textbf{(E) }110$ | [
"Let the two digits be $a$ and $b$ . Then, $5a + 5b = 10a + b - 10b - a = 9a - 9b$ , or $2a = 7b$ . This yields $a = 7$ and $b = 2$ because $a, b < 10$ . Then, $72 + 27 = \\boxed{99}.$",
"We start like above. Let the two digits be $a$ and $b$ . Therefore, $5(a+b) = 10a+b-10b-a=9(a-b)$ . Since we are looking for $... |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_6 | E | 85 | The difference of the roots of $x^2-7x-9=0$ is:
$\textbf{(A) \ }+7 \qquad \textbf{(B) \ }+\frac{7}{2} \qquad \textbf{(C) \ }+9 \qquad \textbf{(D) \ }2\sqrt{85} \qquad \textbf{(E) \ }\sqrt{85}$ | [
"Denote the $2$ roots of this quadratic as $r_1$ and $r_2$ . Note that $(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2$ . By Vieta's Formula's $r_1+r_2=7$ , and $r_1r_2=-9$ . Thus, $r_1-r_2=\\sqrt{49+4\\cdot 9}=\\boxed{85}$"
] |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_7 | A | 1 | The digit-sum of $998$ is $9+9+8=26$ . How many 3-digit whole numbers, whose digit-sum is $26$ , are even?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$ | [
"The highest digit sum for three-digit numbers is $9+9+9=27$ . Therefore, the only possible digit combination is $9, 9, 8$ . Of course, of the three possible numbers, only $998$ works. Thus, the answer is $\\boxed{1}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_24 | A | 1 | The digits $1$ $2$ $3$ $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\... | [
"We see that since $QRS$ is divisible by $5$ $S$ must equal either $0$ or $5$ , but it cannot equal $0$ , so $S=5$ . We notice that since $PQR$ must be even, $R$ must be either $2$ or $4$ . However, when $R=2$ , we see that $T \\equiv 2 \\pmod{3}$ , which cannot happen because $2$ and $5$ are already used up; so $R... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_6 | C | 9 | The digits $2,0,2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?
[asy] // Diagram by TheMathGuyd. I can compress this later size(5cm); real w=2.2; pair O,I,J; O=(0,0);I=(1,0);J=(0,1); path bsqb = O--I; path bsqr = I--I+J; path bsqt = I+J--J; path b... | [
"First, let us consider the case where $0$ is a base: This would result in the entire expression being $0.$ Contrastingly, if $0$ is an exponent, we will get a value greater than $0.$ $3^2\\times2^0=9$ is greater than $2^3\\times2^0=8$ and $2^2\\times3^0=4.$ Therefore, the answer is $\\boxed{9}.$",
"The maximum p... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_4 | E | 9 | The digits 1, 2, 3, 4 and 9 are each used once to form the smallest possible even five-digit number. The digit in the tens place is
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 9$ | [
"Since the number is even, the last digit must be $2$ or $4$ . To make the smallest possible number, the ten-thousands digit must be as small as possible, so the ten-thousands digit is $1$ . Simillarly, the thousands digit has second priority, so it must also be as small as possible once the ten-thousands digit is ... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_1 | null | 217 | The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by $37$ | [
"A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form ${\\underline{(n+3)}}\\,{\\underline{(n+2)}}\\,{\\underline{( n+1)}}\\,{\\underline {(n)}}$ $= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n$ , for $n \\in \\lbrace0, 1, 2, 3,... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_32 | A | 0 | The dimensions of a rectangle $R$ are $a$ and $b$ $a < b$ . It is required to obtain a rectangle with dimensions $x$ and $y$ $x < a, y < a$ ,
so that its perimeter is one-third that of $R$ , and its area is one-third that of $R$ . The number of such (different) rectangles is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qq... | [
"Using the perimeter and area formulas, \\[2(x+y) = \\frac{2}{3}(a+b)\\] \\[x+y = \\frac{a+b}{3}\\] \\[xy = \\frac{ab}{3}\\] Dividing the second equation by the last equation results in \\[\\frac1y + \\frac1x = \\frac1b + \\frac1a\\] Since $x,y < a$ $\\tfrac1a < \\tfrac1x, \\tfrac1y$ . Since $a < b$ $\\tfrac1b < \... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_3 | B | 38 | The dimensions of a rectangular box in inches are all positive integers and the volume of the box is $2002$ in $^3$ . Find the minimum possible sum of the three dimensions.
$\text{(A) }36 \qquad \text{(B) }38 \qquad \text{(C) }42 \qquad \text{(D) }44 \qquad \text{(E) }92$ | [
"Given an arbitrary product and an arbitrary amount of terms to multiply to get that product, to maximize the sum, make all of the terms $1$ with the last one being the number. To minimize the sum, make all of the terms equal to each other. (This is a corollary that follows from the $AM-GM$ proof.) Since $2002$ is ... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4 | null | 294 | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no ... | [
"If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$ . If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$ . Thus, if the number of columns is $n$ , the number of students is... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_19 | C | 18 | The distance between the $5^\text{th}$ and $26^\text{th}$ exits on an interstate highway is $118$ miles. If any two consecutive exits are at least $5$ miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the $5^\text{th}$ and $26^\text{th}$ exits?
$\text{(A)... | [
"There are $21$ pairs of consecutive exits. To find the maximum number of miles of one of these, the other $20$ must be equal to the minimum number yielding a total of $(5)(20)=100$ miles. The longest distance must be $118-100=\\boxed{18}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_8 | null | 371 | The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by 1000. | [
"We start with the same method as above. The domain of the arcsin function is $[-1, 1]$ , so $-1 \\le \\log_{m}(nx) \\le 1$\n\\[\\frac{1}{m} \\le nx \\le m\\] \\[\\frac{1}{mn} \\le x \\le \\frac{m}{n}\\] \\[\\frac{m}{n} - \\frac{1}{mn} = \\frac{1}{2013}\\] \\[n = 2013m - \\frac{2013}{m}\\]\nFor $n$ to be an integer... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_18 | C | 271 | The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A) }19\qquad \textbf{(B) }31\qquad \textbf{(C) }271\qquad \textbf{(D) }319\qquad \textbf{(E) }... | [
"For all real numbers $a,b,$ and $c$ such that $b>0$ and $b\\neq1,$ note that:\nTherefore, we have \\begin{align*} \\log_{\\frac12}(\\log_4(\\log_{\\frac14}(\\log_{16}(\\log_{\\frac1{16}}x)))) \\text{ is defined} &\\implies \\log_4(\\log_{\\frac14}(\\log_{16}(\\log_{\\frac1{16}}x)))>0 \\\\ &\\implies \\log_{\\frac1... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11 | D | 39 | The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign l... | [
"Let $x$ be the number of students taking both a math and a foreign language class.\nBy P-I-E, we get $70 + 54 - x$ $93$\nSolving gives us $x = 31$\nBut we want the number of students taking only a math class,\nwhich is $70 - 31 = 39$\n$\\boxed{39}$",
"We have $70 + 54 = 124$ people taking classes. However, we ov... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11 | null | 39 | The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign l... | [
"\nWe know that the sum of all three areas is $93$ So, we have: \\[93 = 70-x+x+54-x\\] \\[93 = 70+54-x\\] \\[93 = 124 - x\\] \\[-31=-x\\] \\[x=31\\]\nWe are looking for the number of students in only math. This is $70-x$ . Substituting $x$ with $31$ , our answer is $\\boxed{39}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_23 | D | 42 | The entries in a $3 \times 3$ array include all the digits from $1$ through $9$ , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$ | [
"Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right.\n\\[\\begin{tabular}{|c|c|c|} \\hli... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_17 | D | 42 | The entries in a $3 \times 3$ array include all the digits from $1$ through $9$ , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$ | [
"Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right.\n\\[\\begin{tabular}{|c|c|c|} \\hli... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_13 | null | 850 | The equation
has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation. Find the value of | [
"Let $t = 1/x$ . After multiplying the equation by $t^{10}$ $1 + (13 - t)^{10} = 0\\Rightarrow (13 - t)^{10} = - 1$\nUsing DeMoivre, $13 - t = \\text{cis}\\left(\\frac {(2k + 1)\\pi}{10}\\right)$ where $k$ is an integer between $0$ and $9$\n$t = 13 - \\text{cis}\\left(\\frac {(2k + 1)\\pi}{10}\\right) \\Rightarrow ... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_13 | null | 200 | The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$ , where $m$ $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$ . Find $m+n+r$ | [
"We may factor the equation as:\n\\begin{align*} 2000x^6+100x^5+10x^3+x-2&=0\\\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\\\ \\end{align*}\nNow $100x^4+10x^2+1\\ge 1>0$ for real ... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_8 | null | 113 | The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots . Given that their sum is $m/n$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$ | [
"Let $y = 2^{111x}$ . Then our equation reads $\\frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$ . Thus, if this equation has roots $r_1, r_2$ and $r_3$ , by Vieta's formulas we have $r_1\\cdot r_2\\cdot r_3 = 4$ . Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$ . Then the previous stat... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_24 | E | 1 | The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$ | [
"$x + \\sqrt{x-2} = 4$ Original Equation\n$\\sqrt{x-2} = 4 - x$ Subtract x from both sides\n$x-2 = 16 - 8x + x^2$ Square both sides\n$x^2 - 9x + 18 = 0$ Get all terms on one side\n$(x-6)(x-3) = 0$ Factor\n$x = \\{6, 3\\}$\nIf you put down A as your answer, it's wrong. You need to check for extraneous roots.\n$6 + \... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_23 | A | 118 | The equation $z(z+i)(z+3i)=2002i$ has a zero of the form $a+bi$ , where $a$ and $b$ are positive real numbers. Find $a.$
$\text{(A) }\sqrt{118} \qquad \text{(B) }\sqrt{210} \qquad \text{(C) }2 \sqrt{210} \qquad \text{(D) }\sqrt{2002} \qquad \text{(E) }100 \sqrt{2}$ | [
"According to Wolfram-Alpha, the answer is $\\boxed{118}$"
] |
https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_8 | null | 160 | The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane . Determine the degree measure of $\theta$ | [
"We shall introduce another factor to make the equation easier to solve. If $r$ is a root of $z^6+z^3+1$ , then $0=(r^3-1)(r^6+r^3+1)=r^9-1$ . The polynomial $x^9-1$ has all of its roots with absolute value $1$ and argument of the form $40m^\\circ$ for integer $m$ (the ninth degree roots of unity ). Now we simply n... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_3 | B | 4 | The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$
$\textbf {(A)} -8 \qquad \textbf{(B)} -4 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 8$ | [
"$2x + 7 = 3 \\Longrightarrow x = -2, \\quad -2b - 10 = -2 \\Longrightarrow -2b = 8 \\Longrightarrow b = \\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_2 | B | 4 | The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$
$\textbf {(A)} -8 \qquad \textbf{(B)} -4 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 8$ | [
"$2x + 7 = 3 \\Longrightarrow x = -2, \\quad -2b - 10 = -2 \\Longrightarrow -2b = 8 \\Longrightarrow b = \\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_12 | C | 2 | The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$ , respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does $L_2$ , and that $L_1$ has 4 times the slope of $L_2$ . If $L_1$ is not horizontal, then $mn$ is
$\text{(A)} \ \frac{\sq... | [
"Solution by e_power_pi_times_i\n$4n = m$ , as stated in the question. In the line $L_1$ , draw a triangle with the coordinates $(0,0)$ $(1,0)$ , and $(1,m)$ . Then $m = \\tan(\\theta_1)$ . Similarly, $n = \\tan(\\theta_2)$ . Since $4n = m$ and $\\theta_1 = 2\\theta_2$ $\\tan(2\\theta_2) = 4\\tan(\\theta_2)$ . Usin... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_24 | D | 1,008,016 | The expression
\[(x+y+z)^{2006}+(x-y-z)^{2006}\]
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028$ | [
"By the Multinomial Theorem , the summands can be written as\n\\[\\sum_{a+b+c=2006}{\\frac{2006!}{a!b!c!}x^ay^bz^c}\\]\nand\n\\[\\sum_{a+b+c=2006}{\\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},\\]\nrespectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another o... |
https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_3 | C | 3 | The expression $(81)^{-2^{-2}}$ has the same value as:
$\textbf{(A)}\ \frac {1}{81} \qquad \textbf{(B) }\ \frac {1}{3} \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 81\qquad \textbf{(E) }\ 81^4$ | [
"Let us recall $\\text{PEMDAS}$ . We calculate the exponent first. $(-2)^{-2}=\\frac{1}{(-2)^2}=\\frac{1}{4}$ When we substitute, we get $81^{\\frac{1}{4}}=\\sqrt[4]{81}=\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_3 | E | 4,041 | The expression $\frac{2021}{2020} - \frac{2020}{2021}$ is equal to the fraction $\frac{p}{q}$ in which $p$ and $q$ are positive integers whose greatest common divisor is ${ }1$ . What is $p?$
$(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041$ | [
"We write the given expression as a single fraction: \\[\\frac{2021}{2020} - \\frac{2020}{2021} = \\frac{2021\\cdot2021-2020\\cdot2020}{2020\\cdot2021}\\] by cross multiplication. Then by factoring the numerator, we get \\[\\frac{2021\\cdot2021-2020\\cdot2020}{2020\\cdot2021}=\\frac{(2021-2020)(2021+2020)}{2020\\cd... |
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_22 | B | 2 | The expression $\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}$ cannot be evaluated for $x=-1$ or $x=2$ ,
since division by zero is not allowed. For other values of $x$
$\textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \te... | [
"$\\frac{2x^2-x}{(x+1)(x-2)}-\\frac{4+x}{(x+1)(x-2)} = \\frac{2x^2-2x-4}{(x+1)(x-2)}$\nThis can be factored as $\\frac{(2)(x^2-x-2)}{(x+1)(x-2)} \\implies \\frac{(2)(x+1)(x-2)}{(x+1)(x-2)}$ , which cancels out to $2 \\implies \\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_47 | C | 1 | The expressions $a+bc$ and $(a+b)(a+c)$ are:
$\textbf{(A)}\ \text{always equal}\qquad\textbf{(B)}\ \text{never equal}\qquad\textbf{(C)}\ \text{equal whenever }a+b+c=1\\ \textbf{(D)}\ \text{equal when }a+b+c=0\qquad\textbf{(E)}\ \text{equal only when }a=b=c=0$ | [
"Using the FOIL method, we see that $(a+b)(a+c) = a^2 + ab + ac + bc.$ We want to solve \\[a + bc = a^2 + ab + ac + bc\\] \\[a = a^2 + ab + ac\\] \\[a((a + b + c) - 1) = 0\\] These expressions are $\\boxed{1.}$"
] |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_9 | D | 39 | The faces of each of $7$ standard dice are labeled with the integers from $1$ to $6$ . Let $p$ be the probabilities that when all $7$ dice are rolled, the sum of the numbers on the top faces is $10$ . What other sum occurs with the same probability as $p$
$\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \... | [
"It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because ... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_12 | E | 315 | The fifth and eighth terms of a geometric sequence of real numbers are $7!$ and $8!$ respectively. What is the first term?
$\mathrm{(A)}\ 60\qquad \mathrm{(B)}\ 75\qquad \mathrm{(C)}\ 120\qquad \mathrm{(D)}\ 225\qquad \mathrm{(E)}\ 315$ | [
"Let the $n$ th term of the series be $ar^{n-1}$ . Because \\[\\frac {8!}{7!} = \\frac {ar^7}{ar^4} = r^3 = 8,\\] it follows that $r = 2$ and the first term is $a = \\frac {7!}{r^4} = \\frac {7!}{16} = \\boxed{315}$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | C | 147 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | [
"There are $7$ segments whose lengths are $2 \\sin \\frac{\\pi}{7}$ $7$ segments whose lengths are $2 \\sin \\frac{2 \\pi}{7}$ $7$ segments whose lengths are $2 \\sin \\frac{3\\pi}{7}$\nTherefore, the sum of the $4$ th powers of these lengths is \\begin{align*} 7 \\cdot 2^4 \\sin^4 \\frac{\\pi}{7} + 7 \\cdot 2^4 \\... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24 | null | 147 | The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy] import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) { draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5)); } } for(int i = 0; i < 7; ++i) { dot(d... | [
"First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle $\\frac{2\\pi}{7},$ which are \\[\\left(\\cos\\dfrac{2\\pi n}{7}, \\sin\\dfrac{2\\pi ... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_10 | E | 4 | The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$ , without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) [asy] ... | [
"Note that of the $12$ cities, $6$ of them ( $2$ on the top, $2$ on the bottom, and $1$ on each side) have $3$ edges coming into/out of them (i.e., in graph theory terms, they have degree $3$ ). Therefore, at least $1$ edge connecting to each of these cities cannot be used. Additionally, the same applies to the sta... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_10 | null | 4 | The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$ , without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) [asy] ... | [
"Observe that only the two central vertices can be visited twice. Since the path is of length 13, we need to repeat a vertex. Using casework on each vertex, we can find there are two paths that go through each central vertex twice, for an answer of $\\boxed{4}.$",
"Looking at the answer choices we see that it wou... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_24 | C | 192 | The figure below shows a polygon $ABCDEFGH$ , consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that $AH = EF = 8$ and $GH = 14$ . What is the volume of the prism?
[asy] usepackage("mathptmx"); size(275); defaultpen(linewidth(0.8)); ... | [
"While imagining the folding, $\\overline{AB}$ goes on $\\overline{BC},$ $\\overline{AH}$ goes on $\\overline{CI},$ and $\\overline{EF}$ goes on $\\overline{FG}.$ So, $BJ=CI=8$ and $FG=BC=8.$ Also, $\\overline{HJ}$ becomes an edge parallel to $\\overline{FG},$ so that means $HJ=8.$\nSince $GH=14,$ then $JG=14-8=6.$... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | null | 732 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | [
"Choose a section to start coloring. Assume, WLOG, that this section is color $1$ . We proceed coloring clockwise around the ring. Let $f(n,C)$ be the number of ways to color the first $n$ sections (proceeding clockwise) such that the last section has color $C$ . In general (except for when we complete the coloring... |
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_4 | B | 8 | The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$
[asy] unitsize(12); //Force a white background in middle even when transparent fill((3,1)--(1... | [
"It is simple to notice that in each and every row, there is always one more black square than the white squares. Since there are $8$ rows, there are $8$ more black squares than the white squares. $8\\rightarrow \\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_20 | D | 23 | The figure is constructed from $11$ line segments, each of which has length $2$ . The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$ , where $m$ and $n$ are positive integers. What is $m + n ?$ [asy] /* Made by samrocksnature */ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.4713... | [
"\nDraw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$ , they each have area $2\\cdot\\frac{1}{2}\\cdot\\frac{4\\sqrt{3}}{4}=\\sqrt{3}$ . For triangle $ACD$ , we can see that $AC=AD=2\\sqrt{3}$ and $... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_15 | D | 23 | The figure is constructed from $11$ line segments, each of which has length $2$ . The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$ , where $m$ and $n$ are positive integers. What is $m + n ?$ [asy] /* Made by samrocksnature */ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.4713... | [
"\nDraw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$ , they each have area $2\\cdot\\frac{1}{2}\\cdot\\frac{4\\sqrt{3}}{4}=\\sqrt{3}$ . For triangle $ACD$ , we can see that $AC=AD=2\\sqrt{3}$ and $... |
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