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/2018_AMC_10B_Problems/Problem_2 | D | 67 | Sam drove $96$ miles in $90$ minutes. His average speed during the first $30$ minutes was $60$ mph (miles per hour), and his average speed during the second $30$ minutes was $65$ mph. What was his average speed, in mph, during the last $30$ minutes?
$\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qqu... | [
"Suppose that Sam's average speed during the last $30$ minutes was $x$ mph.\nRecall that a half hour is equal to $30$ minutes. Therefore, Sam drove $60\\cdot0.5=30$ miles during the first half hour, $65\\cdot0.5=32.5$ miles during the second half hour, and $x\\cdot0.5$ miles during the last half hour. We have \\beg... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_2 | D | 67 | Sam drove $96$ miles in $90$ minutes. His average speed during the first $30$ minutes was $60$ mph (miles per hour), and his average speed during the second $30$ minutes was $65$ mph. What was his average speed, in mph, during the last $30$ minutes?
$\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qqu... | [
"Suppose that Sam's average speed during the last $30$ minutes was $x$ mph.\nRecall that a half hour is equal to $30$ minutes. Therefore, Sam drove $60\\cdot0.5=30$ miles during the first half hour, $65\\cdot0.5=32.5$ miles during the second half hour, and $x\\cdot0.5$ miles during the last half hour. We have \\beg... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_7 | C | 2.8 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilomet... | [
"Let's call the distance that Samia had to travel in total as $2x$ , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both $\\frac{2x}{2}$ , or $x$ \\[\\] She bikes at a rate of $17$ kph, so she travels the distance she bikes in $\\frac{x}{17}$ h... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_7 | null | 2.8 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilomet... | [
"Since the distance traveled by bicycle and foot are equal, we can substitute the time traveled by bike, or $b$ , as $\\frac{5}{17}$ of the time traveled by foot, or $f$\nAccordingly, we have $\\frac{22}{17}f=44$ This comes out to $f=34$ This means that Samia traveled $34$ minutes on foot, and hence, $44-34=10$ min... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_4 | C | 2.8 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilomet... | [
"Let's call the distance that Samia had to travel in total as $2x$ , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both $\\frac{2x}{2}$ , or $x$ \\[\\] She bikes at a rate of $17$ kph, so she travels the distance she bikes in $\\frac{x}{17}$ h... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_4 | null | 2.8 | Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilomet... | [
"Since the distance traveled by bicycle and foot are equal, we can substitute the time traveled by bike, or $b$ , as $\\frac{5}{17}$ of the time traveled by foot, or $f$\nAccordingly, we have $\\frac{22}{17}f=44$ This comes out to $f=34$ This means that Samia traveled $34$ minutes on foot, and hence, $44-34=10$ min... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_1 | A | 31 | Sandwiches at Joe's Fast Food cost $$3$ each and sodas cost $$2$ each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?
$\textbf{(A)}\ 31\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 34\qquad\textbf{(E)}\ 35$ | [
"The $5$ sandwiches cost $5\\cdot 3=15$ dollars. The $8$ sodas cost $8\\cdot 2=16$ dollars. In total, the purchase costs $15+16=\\boxed{31}$ dollars."
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_1 | A | 31 | Sandwiches at Joe's Fast Food cost $$3$ each and sodas cost $$2$ each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?
$\textbf{(A)}\ 31\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 34\qquad\textbf{(E)}\ 35$ | [
"The $5$ sandwiches cost $5\\cdot 3=15$ dollars. The $8$ sodas cost $8\\cdot 2=16$ dollars. In total, the purchase costs $15+16=\\boxed{31}$ dollars."
] |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6 | B | 300 | Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes c... | [
"If $65\\%$ of the votes were likes, then $35\\%$ of the votes were dislikes. $65\\%-35\\%=30\\%$ , so $90$ votes is $30\\%$ of the total number of votes. Doing quick arithmetic shows that the answer is $\\boxed{300}$",
"Let's consider that Sangho has received 100 votes. This means he has received 65 upvotes and ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_8 | C | 12 | Sara makes a staircase out of toothpicks as shown:
[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45); for(int i=0;i<=2;i=i+1) { for(int j=0;j<=3-i;j=j+1) { filldraw(shift((i,j))*h,black); filldraw(shift((j,i))*v,black); } } [/asy]
This is a 3-step stairca... | [
"A staircase with $n$ steps contains $4 + 6 + 8 + ... + 2n + 2$ toothpicks. This can be rewritten as $(n+1)(n+2) -2$\nSo, $(n+1)(n+2) - 2 = 180$\nSo, $(n+1)(n+2) = 182.$\nInspection could tell us that $13 \\cdot 14 = 182$ , so the answer is $13 - 1 = \\boxed{12}$",
"Layer $1$ $4$ steps\nLayer $1,2$ $10$ steps\nLa... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_3 | null | 126 | Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum ... | [
"Let $x$ be the two-digit number, $y$ be the three-digit number. Putting together the given, we have $1000x+y=9xy \\Longrightarrow 9xy-1000x-y=0$ . Using SFFT , this factorizes to $(9x-1)\\left(y-\\dfrac{1000}{9}\\right)=\\dfrac{1000}{9}$ , and $(9x-1)(9y-1000)=1000$\nSince $89 < 9x-1 < 890$ , we can use trial and ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_17 | D | 25 | Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup ... | [
"We will simulate the process in steps.\nIn the beginning, we have:\nIn the first step we pour $4/2=2$ ounces of coffee from cup $1$ to cup $2$ , getting:\nIn the second step we pour $2/2=1$ ounce of coffee and $4/2=2$ ounces of cream from cup $2$ to cup $1$ , getting:\nHence at the end we have $3+2=5$ ounces of li... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_10 | D | 25 | Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup ... | [
"We will simulate the process in steps.\nIn the beginning, we have:\nIn the first step we pour $4/2=2$ ounces of coffee from cup $1$ to cup $2$ , getting:\nIn the second step we pour $2/2=1$ ounce of coffee and $4/2=2$ ounces of cream from cup $2$ to cup $1$ , getting:\nHence at the end we have $3+2=5$ ounces of li... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_10 | D | 3 | Segment $AB$ is both a diameter of a circle of radius $1$ and a side of an equilateral triangle $ABC$ .
The circle also intersects $AC$ and $BC$ at points $D$ and $E$ , respectively. The length of $AE$ is
$\textbf{(A)} \ \frac{3}{2} \qquad \textbf{(B)} \ \frac{5}{3} \qquad \textbf{(C)} \ \frac{\sqrt 3}{2} \qquad \t... | [
"Note that since $AB$ is a diameter, $\\angle AEB = 90^{\\circ}$ , which means $AB$ is an altitude of equilateral triangle $ABC$ . It follows that $\\triangle ABE$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, and so $AE = AB \\cdot \\frac{\\sqrt{3}}{2} = (2 \\cdot 1) (\\frac{\\sqrt{3}}{2}) = \\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_9 | null | 52.5 | Segment $BD$ and $AE$ intersect at $C$ , as shown, $AB=BC=CD=CE$ , and $\angle A = \frac 52 \angle B$ . What is the degree measure of $\angle D$
[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair C=(0,0), Ep=dir(35), D=dir(-35), B=dir(145); pair A=intersectionpoints(Circle(B,1),C--(-1*Ep... | [
"$\\triangle ABC$ is isosceles, hence $\\angle ACB = \\angle CAB$\nThe sum of internal angles of $\\triangle ABC$ can now be expressed as $\\angle B + \\frac 52 \\angle B + \\frac 52 \\angle B = 6\\angle B$ , hence $\\angle B = 30^\\circ$ , and each of the other two angles is $75^\\circ$\nNow we know that $\\angle ... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6 | null | 192 | Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$ | [
"First scale down the whole cube by $12$ . Let point $P$ have coordinates $(x, y, z)$ , point $A$ have coordinates $(0, 0, 0)$ , and $s$ be the side length. Then we have the equations \\begin{align*} (s-x)^2+y^2+z^2&=\\left(5\\sqrt{10}\\right)^2, \\\\ x^2+(s-y)^2+z^2&=\\left(5\\sqrt{5}\\right)^2, \\\\ x^2+y^2+(s-z)... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_24 | D | 122 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively p... | [
"Let $O=\\Gamma$ be the center of the semicircle and $X=\\Omega$ be the center of the circle.\nApplying the Extended Law of Sines to $\\triangle PQR,$ we find the radius of $\\odot X:$ \\[XP=\\frac{QR}{2\\cdot\\sin \\angle QPR}=\\frac{3\\sqrt3}{2\\cdot\\frac{\\sqrt3}{2}}=3.\\] Alternatively, by the Inscribed Angle ... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | null | 201 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | [
"Note that since set $A$ has $m$ consecutive integers that sum to $2m$ , the middle integer (i.e., the median) must be $2$ . Therefore, the largest element in $A$ is $2 + \\frac{m-1}{2}$\nFurther, we see that the median of set $B$ is $0.5$ , which means that the \"middle two\" integers of set $B$ are $0$ and $1$ .... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_6 | C | 20 | Set $A$ has $20$ elements, and set $B$ has $15$ elements. What is the smallest possible number of elements in $A \cup B$
$\textbf{(A)}5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300$ | [
"$A \\cup B$ will be smallest if $B$ is completely contained in $A$ , in which case all the elements in $B$ would be counted for in $A$ . So the total would be the number of elements in $A$ , which is $\\boxed{20}$",
"Assume WLOG that $A={1, 2, 3, 4, \\cdots , 20}$ , and $B={6, 7, 8, 9, 10, \\cdots , 20}$ . Then,... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_18 | A | 10 | Set $u_0 = \frac{1}{4}$ , and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence \[u_{k+1} = 2u_k - 2u_k^2.\]
This sequence tends to a limit; call it $L$ . What is the least value of $k$ such that \[|u_k-L| \le \frac{1}{2^{1000}}?\]
$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf... | [
"Note that terms of the sequence $(u_k)$ lie in the interval $\\left(0,\\frac12\\right),$ strictly increasing.\nSince the sequence $(u_k)$ tends to the limit $L,$ we set $u_{k+1}=u_k=L>0.$\nThe given equation becomes \\[L=2L-2L^2,\\] from which $L=\\frac12.$\nThe given inequality becomes \\[\\frac12-\\frac{1}{2^{10... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_13 | C | 1,504 | Sets $A$ and $B$ , shown in the Venn diagram, have the same number of elements.
Their union has $2007$ elements and their intersection has $1001$ elements. Find
the number of elements in $A$
[asy] defaultpen(linewidth(0.7)); draw(Circle(origin, 5)); draw(Circle((5,0), 5)); label("$A$", (0,5), N); label("$B$", (5,5), N)... | [
"Let $x$ be the number of elements in $A$ and $B$ which is equal.\nThen we could form equation $2x-1001 = 2007$\n$2x = 3008$\n$x = 1504$\nThe answer is $\\boxed{1504}$",
"Let $x$ be the number of elements in $A$ not including the intersection. $2007-1001=1006$ total elements excluding the intersection. Since we k... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_15 | A | 2 | Seven cookies of radius $1$ inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?
[asy] d... | [
"The big cookie has radius $3$ , since the center of the center cookie is the same as that of the large cookie. The difference in areas of the big cookie and the seven small ones is $3^2\\pi-7\\pi=9\\pi-7\\pi=2 \\pi$ . The scrap cookie has this area, so its radius must be $\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_10 | B | 658 | Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area... | [
"The volume of each cube follows the pattern of $n^3$ , for $n$ is between $1$ and $7$\nWe see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the $7\\times 7\\times 7$ cube (which is just $7 \\times 7 = 49$ ). The sides areas can be meas... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_7 | B | 658 | Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area... | [
"The volume of each cube follows the pattern of $n^3$ , for $n$ is between $1$ and $7$\nWe see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the $7\\times 7\\times 7$ cube (which is just $7 \\times 7 = 49$ ). The sides areas can be meas... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_22 | C | 1,932 | Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?
$\textbf{(A)}\ 1930 \qquad \textbf{(B)}\ 1931 \qquad \textbf{(C)}\ 1932 \qquad \textbf{(D)}\ 1933 \qquad ... | [
"We can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.\nEach candy has three choices; it can go in any of the three bags.\nSince there are seven candies, that makes the total distribution $3^7=2187$\nTo find the overcount, we ... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_23 | C | 365 | Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each ... | [
"First look at the numbers Alice says. $1, 3, 4, 6, 7, 9 \\cdots$ skipping every number that is congruent to $2 \\pmod 3$ . Thus, Barbara says those numbers EXCEPT every second - being $2 + 3^1 \\equiv 5 \\pmod{3^2=9}$ . So Barbara skips every number congruent to $5 \\pmod 9$ . We continue and see:\nAlice skips $2 ... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10 | null | 831 | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulate... | [
"The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$ . We let this probability be $p$ ; then the probability that $A$ and $B$ end with the same score... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_10 | B | 2 | Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way?
$\text {(A) } 1 \qquad \text {(B) } 2 \qquad \text {(C) } 3 \qquad \text {(D) } 4 \qquad \text {(E) } 5$ | [
"Place the first triangle. Now, we can place the second triangle either adjacent to the first, or with one side between them, for a total of $\\boxed{2}$",
"Take ${5 \\choose 2}$ to realize there are 10 ways to choose 2 different triangles. Then divide by 5 for each vertex of a pentagon to get $\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_17 | B | 207 | Several sets of prime numbers, such as $\{7,83,421,659\}$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?
$\text{(A) }193 \qquad \text{(B) }207 \qquad \text{(C) }225 \qquad \text{(D) }252 \qquad \text{(E) }447$ | [
"Neither of the digits $4$ $6$ , and $8$ can be a units digit of a prime. Therefore the sum of the set is at least $40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207$\nWe can indeed create a set of primes with this sum, for example the following sets work: $\\{ 41, 67, 89, 2, 3, 5 \\}$ or $\\{ 43, 61, 89, 2, 5, 7 \\}$\nTh... |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_14 | D | 13 | Several students are competing in a series of three races. A student earns $5$ points for winning a race, $3$ points for finishing second and $1$ point for finishing third. There are no ties. What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points tha... | [
"There are two ways for a student to get $11$ $5+5+1$ and $5+3+3$ . Clearly if someone gets one of these combinations someone else could get the other, so we are not guaranteed the most points with $11$\nThere is only one way to get $13$ points: $5+5+3$ . In this case, the largest score another person could get i... |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_22 | B | 11 | Several students are seated at a large circular table. They pass around a bag containing $100$ pieces of candy. Each person receives the bag, takes one piece of candy and then passes the bag to the next person. If Chris takes the first and last piece of candy, then the number of students at the table could be
$\text... | [
"If this is the case, then if there were only $99$ pieces of candy, the bag would have gone around the table a whole number of times. Thus, the number of students is a divisor of $99$ . The only choice that satisfies this is choice $\\boxed{11}$"
] |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7 | A | 48 | Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }7... | [
"Right now, she scored $76, 94,$ and $87$ points, for a total of $257$ points. She wants her average to be $81$ for her $5$ tests, so she needs to score $405$ points in total. This means she needs to score a total of $405-257= 148$ points in her next $2$ tests. Since the maximum score she can get on one of her $2$... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7 | null | 48 | Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }7... | [
"We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \\[\\frac{76+94+87+x+y}{5} = 81,\\] \\[\\frac{257+x+y}{5} = 81.\\] We can now cross multiply to get rid of the denomin... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_12 | C | 18 | Side $\overline{AB}$ of $\triangle ABC$ has length $10$ . The bisector of angle $A$ meets $\overline{BC}$ at $D$ , and $CD = 3$ . The set of all possible values of $AC$ is an open interval $(m,n)$ . What is $m+n$
$\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }2... | [
"Let $AC=x.$ By Angle Bisector Theorem, we have $\\frac{AB}{AC}=\\frac{BD}{CD},$ from which $BD=CD\\cdot\\frac{AB}{AC}=\\frac{30}{x}.$\nRecall that $x>0.$ We apply the Triangle Inequality to $\\triangle ABC:$\nTaking the intersection of the solutions gives \\[(m,n)=(0,\\infty)\\cap(0,15)\\cap(3,\\infty)=(3,15),\\] ... |
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_7 | B | 36 | Sides $AB,BC,CD$ and $DA$ of convex polygon $ABCD$ have lengths 3, 4, 12, and 13, respectively, and $\angle CBA$ is a right angle. The area of the quadrilateral is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); real r=degrees((12,5)), s=degrees((3,4)); pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s)); dr... | [
"Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is $\\frac{3\\cdot4}{2}+\\frac{5\\cdot12}{2}=36\\Rightarrow\\boxed{36}$"
] |
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_24 | E | 25 | Sides $AB,~ BC$ , and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5$ , and $20$ , respectively.
If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5}$ , then side $AD$ has length
$\textbf{(A) }24\qquad \textbf{(B) }24.5\qquad \textbf{(C) }24.6\qquad \textbf{(D) }24.8\qquad \textbf{(E... | [
"We know that $\\sin(C)=-\\cos(B)=\\frac{3}{5}$ . Since $B$ and $C$ are obtuse, we have $\\sin(180-C)=\\cos(180-B)=\\frac{3}{5}$ . It is known that $\\sin(x)=\\cos(90-x)$ , so $180-C=90-(180-C)=180-B$ . We simplify this as follows:\n\\[-90+C=180-B\\]\n\\[B+C=270^{\\circ}\\]\nSince $B+C=270^{\\circ}$ , we know that ... |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_18 | D | 23 | Six bags of marbles contain $18, 19, 21, 23, 25$ and $34$ marbles, respectively. One bag contains chipped marbles only. The other $5$ bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, h... | [
"Let George's bags contain a total of $x$ marbles, so Jane's bag contains $2x$ marbles. This means the total number of non-chipped marbles is $3x \\equiv 0 \\pmod{3}$ , while the total number of marbles is $18+19+21+23+25+34 = 140 \\equiv 2 \\pmod{3}$ , so the number of chipped marbles must also be congruent to $2-... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_5 | null | 52 | Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order. | [
"Realize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two.\nIf we choose any of the numbers $1$ through $6$ , there are five other spots to put them, so we get $6 \\cdot 5 = 30$ . However, we... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_22 | D | 20 | Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\;... | [
"Consider shifting every person over three seats left after each person has gotten up and sat back down again. Now, instead of each person being seated not in the same chair and not in an adjacent chair, each person will be seated either in the same chair or an adjacent chair. The problem now becomes the number of ... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_1 | null | 942 | Six congruent circles form a ring with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius 30. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $\lfloor K \rfloor$ (the floor function ).
2005 AIM... | [
"Define the radii of the six congruent circles as $r$ . If we draw all of the radii to the points of external tangency, we get a regular hexagon . If we connect the vertices of the hexagon to the center of the circle $C$ , we form several equilateral triangles . The length of each side of the triangle is $2r$ . Not... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_22 | C | 26 | Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides.
[asy] /* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(... | [
"Count the number of sides that are not exposed, where a cube is connected to another cube and subtract it from the total number of faces. There are $5$ places with two adjacent cubes, covering $10$ sides, and $(6)(6)=36$ faces. The exposed surface area is $36-10 = \\boxed{26}$",
"We can count the number of showi... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_14 | B | 29 | Six different digits from the set \[\{ 1,2,3,4,5,6,7,8,9\}\] are placed in the squares in the figure shown so that the sum of the entries in the vertical column is 23 and the sum of the entries in the horizontal row is 12.
The sum of the six digits used is
[asy] unitsize(18); draw((0,0)--(1,0)--(1,1)--(4,1)--(4,2)--(1,... | [
"Looking at the vertical column, the three numbers sum to $23$ . If we make the numbers on either end $9$ and $8$ in some order, the middle number will be $6$ . This is the minimum for the intersection.\nLooking at the horizontal row, the four numbers sum to $12$ . If we minimize the three numbers on the right t... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_20 | E | 1 | Six distinct positive integers are randomly chosen between $1$ and $2006$ , inclusive. What is the probability that some pair of these integers has a difference that is a multiple of $5$
$\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{3}{5}\qquad\textbf{(C) } \frac{2}{3}\qquad\textbf{(D) } \frac{4}{5}\qquad\textbf{... | [
"For two numbers to have a difference that is a multiple of $5$ , the numbers must be congruent $\\bmod{5}$ (their remainders after division by $5$ must be the same).\n$0, 1, 2, 3, 4$ are the possible values of numbers in $\\bmod{5}$ . Since there are only $5$ possible values in $\\bmod{5}$ and we are picking $6$ n... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_12 | null | 594 | Six men and some number of women stand in a line in random order. Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent. | [
"Let $n$ be the number of women present, and let _ be some positive number of women between groups of men. Since the problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to consider, where $(k)$ refers to a consecutive group of $k$ men:\nFor the first ... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_4 | D | 8 | Six numbers from a list of nine integers are $7,8,3,5,9$ and $5$ . The largest possible value of the median of all nine numbers in this list is
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ | [
"First, put the six numbers we have in order, since we are concerned with the median: $3, 5, 5, 7, 8, 9$\nWe have three more numbers to insert into the list, and the median will be the $5^{th}$ highest (and $5^{th}$ lowest) number on the list. If we top-load the list by making all three of the numbers greater than... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_10 | B | 23 | Six pepperoni circles will exactly fit across the diameter of a $12$ -inch pizza when placed. If a total of $24$ circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?
$\textbf{(A)}\ \frac 12 \qquad\textbf{(B)}\ \frac 23 \qquad\textbf{(C)}\ \frac 34 \qquad\tex... | [
"The pepperoni circles' diameter is $2$ , since $\\dfrac{12}{6} = 2$ . From that we see that the area of the $24$ circles of pepperoni is $\\left ( \\frac{2}{2} \\right )^2 (24\\pi) = 24\\pi$ . The large pizza's area is $6^2\\pi$ . Therefore, the ratio is $\\frac{24\\pi}{36\\pi} = \\boxed{23}$"
] |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_6 | D | 182 | Six rectangles each with a common base width of $2$ have lengths of $1, 4, 9, 16, 25$ , and $36$ . What is the sum of the areas of the six rectangles?
$\textbf{(A) }91\qquad\textbf{(B) }93\qquad\textbf{(C) }162\qquad\textbf{(D) }182\qquad \textbf{(E) }202$ | [
"The sum of the areas is equal to $2\\cdot1+2\\cdot4+2\\cdot9+2\\cdot16+2\\cdot25+2\\cdot36$ . This is equal to $2(1+4+9+16+25+36)$ , which is equal to $2\\cdot91$ . This is equal to our final answer of $\\boxed{182}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_18 | E | 32 | Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2$ . The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to th... | [
"It can be seen that the diameter of the eighth sphere is equal to the radius of the seventh sphere by drawing out a diagram of the insides of the seventh sphere. The radius of the seventh sphere is $2+1=3$ , the radius of the eight sphere is $\\boxed{32}$"
] |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_27 | C | 22 | Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$ | [
"The first line divides the plane into two regions. The second line intersects one line, creating two regions. The third line intersects two lines, creating three regions. Similarly, the fourth line intersects three lines and creates four regions, the fifth line intersects four lines and creates five regions, an... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_27 | null | 22 | Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$ | [
"We can use the fact that the number of regions that $n$ lines divide a plane is given by the equation $L_n = \\frac{n^2 + n +2}{2}$ , and in this problems, $n=6$ , from which the answer is $\\boxed{22}$"
] |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_6 | B | 100 | Six trees are equally spaced along one side of a straight road. The distance from the first tree to the fourth is 60 feet. What is the distance in feet between the first and last trees?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 105 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 140$ | [
"There are $3$ spaces between the 1st and 4th trees, so each of these spaces has $\\frac{60}{3}=20$ feet. Between the first and last trees there are $5$ spaces, so the distance between them is $20\\times5=100$ feet, $\\boxed{100}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_7 | E | 22.5 | Small lights are hung on a string $6$ inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of $2$ red lights followed by $3$ green lights. How many feet separate the 3rd red light and the 21st red light?
Note: $1$ foot is equal to $12$ inches.
$\textb... | [
"We know the repeating section is made of $2$ red lights and $3$ green lights. The 3rd red light would appear in the 2nd section of this pattern, and the 21st red light would appear in the 11th section. There would then be a total of $44$ lights in between the 3rd and 21st red light, translating to $45$ $6$ -inch g... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_5 | B | 5 | Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15$ | [
"Start by buying the largest packs first. After three $24$ -packs, $90-3(24)=18$ cans are left. After one $12$ -pack, $18-12=6$ cans are left. Then buy one more $6$ -pack. The total number of packs is $3+1+1=\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_14 | C | 8 | Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C... | [
"If we let $p$ be the number of people initially in the group, then $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p$ , but the number of girls is $0.4p-2$ . Since only $30\\%$ of the group are girls, \\begin{align*} \\frac{0.4p-2}{p}&=\\frac{3}{10... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_10 | C | 8 | Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C... | [
"If we let $p$ be the number of people initially in the group, then $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p$ , but the number of girls is $0.4p-2$ . Since only $30\\%$ of the group are girls, \\begin{align*} \\frac{0.4p-2}{p}&=\\frac{3}{10... |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_16 | B | 22 | Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remaining marbles are red. If two blue marbles are removed instead of one red, then one-fifth of the remaining marbles are red. How many marbles were in the bag originally?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 22 ... | [
"Let $r$ and $b$ be the number of red and blue marbles originally in the bag respectively. After $1$ red marble is removed, there are $r+b-1$ marbles left in the bag and $r-1$ red marbles left. So \\[\\frac{r-1}{r+b-1}=\\frac{1}{7}.\\] When $2$ blue marbles are removed, there are $r$ red marbles and $r+b-2$ total m... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_21 | D | 23 | Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((... | [
"An even number comes from multiplying an even and even, even and odd, or odd and even. Since an odd number only comes from multiplying an odd and odd, there are less cases and it would be easier to find the probability of spinning two odd numbers from $1$ . Multiply the independent probabilities of each spinner ge... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | null | 307 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | [
"Let $G$ be the foot of the perpendicular from $O$ to $AB$ . Denote $x = EG$ and $y = FG$ , and $x > y$ (since $AE < BF$ and $AG = BG$ ). Then $\\tan \\angle EOG = \\frac{x}{450}$ , and $\\tan \\angle FOG = \\frac{y}{450}$\nBy the tangent addition rule $\\left( \\tan (a + b) = \\frac{\\tan a + \\tan b}{1 - \\tan a ... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_3 | null | 578 | Square $ABCD$ has side length $13$ , and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$ . Find $EF^{2}$ [asy]unitsize(0.2 cm); pair A, B, C, D, E, F; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); draw(A--B--C--D--cycle); draw(... | [
"Drawing $EF$ , it clearly passes through the center of $ABCD$ . Letting this point be $P$ , we note that $AEBP$ and $CFDP$ are congruent cyclic quadrilaterals, and that $AP=BP=CP=DP=\\frac{13}{\\sqrt{2}}.$ Now, from Ptolemy's, $13\\cdot EP=\\frac{13}{\\sqrt{2}}(12+5)\\implies EP=\\frac{17\\sqrt{2}}{2}$ . Since $EF... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_13 | C | 200 | Square $ABCD$ has side length $30$ . Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$ . The centroids of $\triangle{ABP}$ $\triangle{BCP}$ $\triangle{CDP}$ , and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
[asy] unitsize(120); pair B = (0, 0), A =... | [
"As shown below, let $M_1,M_2,M_3,M_4$ be the midpoints of $\\overline{AB},\\overline{BC},\\overline{CD},\\overline{DA},$ respectively, and $G_1,G_2,G_3,G_4$ be the centroids of $\\triangle{ABP},\\triangle{BCP},\\triangle{CDP},\\triangle{DAP},$ respectively. By SAS, we conclude that $\\triangle G_1G_2P\\sim\\trian... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6 | null | 12 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqr... | [
" Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$ , and define $s$ to be one of the sides of that square. Since the sides are parallel , by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\\frac{AA'... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_4 | null | 86 | Square $ABCD$ has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$ . Find $100k$ | [
"Without loss of generality, let $(0,0)$ $(2,0)$ $(0,2)$ , and $(2,2)$ be the vertices of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex $(0,0)$ . Let the two endpoints of the segment have coordinates $(x,0)$ and $(0,y)$ . Because the segment has length... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_23 | C | 13 | Square $ABCD$ has sides of length 3. Segments $CM$ and $CN$ divide the square's area into three equal parts. How long is segment $CM$
[asy] pair A,B,C,D,M,N; A = (0,0); B = (0,3); C = (3,3); D = (3,0); M = (0,1); N = (1,0); draw(A--B--C--D--cycle); draw(M--C--N); label("$A$",A,SW); label("$M$",M,W); label("$B$",B,NW); ... | [
"Since the square has side length $3$ , the area of the entire square is $9$\nThe segments divide the square into 3 equal parts, so the area of each part is $9 \\div 3 = 3$\nSince $\\triangle CBM$ has area $3$ and base $CB = 3$ , using the area formula for a triangle:\n$A_{tri} = \\frac{1}{2}bh$\n$3 = \\frac{1}{2}3... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_6 | null | 251 | Square $ABCD$ is inscribed in a circle . Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$ , then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$ . Find $... | [
"Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$ $2b$ be the side length of $EFGH$ . By the Pythagorean Theorem , the radius of $\\odot O = OC = a\\sqrt{2}$\nNow consider right triangle $OGI$ , where $I$ is the midpoint of $\\overline{GH}$ . Then, by the Pythagorean Theorem,\n\\begin{alig... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_7 | B | 35 | Square $ABCD$ is rotated $20^{\circ}$ clockwise about its center to obtain square $EFGH$ , as shown below. IMG 1031.jpeg
What is the degree measure of $\angle EAB$
$\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}$ | [
"First, let's call the center of both squares $I$ . Then, $\\angle{AIE} = 20$ , and since $\\overline{EI} = \\overline{AI}$ $\\angle{AEI} = \\angle{EAI} = 80$ . Then, we know that $AI$ bisects angle $\\angle{DAB}$ , so $\\angle{BAI} = \\angle{DAI} = 45$ . Subtracting $45$ from $80$ , we get $\\boxed{35}$",
"First... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_2 | null | 25 | Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$ , and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\triangle GEM$ | [
"Note that if the altitude of the triangle is at most $10$ , then the maximum area of the intersection of the triangle and the square is $5\\cdot10=50$ .\nThis implies that vertex G must be located outside of square $AIME$\nLet $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$ . Clearly, $XY=6$ since the area of ... |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_1 | null | 255 | Square $S_{1}$ is $1\times 1.$ For $i\ge 1,$ the lengths of the sides of square $S_{i+1}$ are half the lengths of the sides of square $S_{i},$ two adjacent sides of square $S_{i}$ are perpendicular bisectors of two adjacent sides of square $S_{i+1},$ and the other two sides of square $S_{i+1},$ are the perpendicular bi... | [
"The sum of the areas of the squares if they were not interconnected is a geometric sequence\nThen subtract the areas of the intersections, which is $\\left(\\frac{1}{4}\\right)^2 + \\ldots + \\left(\\frac{1}{32}\\right)^2$\nThe majority of the terms cancel, leaving $1 + \\frac{1}{4} - \\frac{1}{1024}$ , which simp... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_17 | B | 500 | Square corners, 5 units on a side, are removed from a $20$ unit by $30$ unit rectangular sheet of cardboard. The sides are then folded to form an open box. The surface area, in square units, of the interior of the box is
[asy] fill((0,0)--(20,0)--(20,5)--(0,5)--cycle,lightgray); fill((20,0)--(20+5*sqrt(2),5*sqrt(2))-... | [
"If the sides of the open box are folded down so that a flat sheet with four corners cut out remains, then the revealed surface would have the same area as the interior of the box. This is equal to the area of the four corners subtracted from the area of the original sheet, which is $((20)(30)-4(5)(5)) = 600-100 = ... |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_7 | E | 175 | Squares $ABCD$ and $EFGH$ are congruent, $AB=10$ , and $G$ is the center of square $ABCD$ . The area of the region in the plane covered by these squares is [asy] draw((0,0)--(10,0)--(10,10)--(0,10)--cycle); draw((5,5)--(12,-2)--(5,-9)--(-2,-2)--cycle); label("A", (0,0), W); label("B", (10,0), E); label("C", (10,10), NE... | [
"The area of the entire region in the plane is the area of the figure. However, we cannot simply add the two areas of the squares. We find the area of $\\triangle ABG$ and subtract this from $200$ , the total area of the two squares.\nSince $G$ is the center of $ABCD$ $BG$ is half of the diagonal of the square. The... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_7 | null | 840 | Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$ . The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$ . Find the differe... | [
"Letting $AI=a$ and $IB=b$ , we have \\[IJ^{2}=a^{2}+b^{2} \\geq 1008\\] by AM-GM inequality . Also, since $EFGH||ABCD$ , the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas... |
https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_15 | null | 462 | Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png | [
"1987 AIME-15a.png\nBecause all the triangles in the figure are similar to triangle $ABC$ , it's a good idea to use area ratios . In the diagram above, $\\frac {T_1}{T_3} = \\frac {T_2}{T_4} = \\frac {441}{440}.$ Hence, $T_3 = \\frac {440}{441}T_1$ and $T_4 = \\frac {440}{441}T_2$ . Additionally, the area of triang... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_6 | D | 103 | Star lists the whole numbers $1$ through $30$ once. Emilio copies Star's numbers, replacing each occurrence of the digit $2$ by the digit $1$ . Star adds her numbers and Emilio adds his numbers. How much larger is Star's sum than Emilio's?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}... | [
"For every tens digit 2, we subtract 10, and for every units digit 2, we subtract 1. Because 2 appears 10 times as a tens digit and 2 appears 3 times as a units digit, the answer is $10\\cdot 10+1\\cdot 3=\\boxed{103.}$"
] |
https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3 | null | 67 | Starting at $(0,0),$ an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let $p$ be the probability that the object reaches $(2,2)$ in six or fewer steps. Given that $p$ can be written in the form $m/n,$ where $m$ and $n... | [
"It takes an even number of steps for the object to reach $(2,2)$ , so the number of steps the object may have taken is either $4$ or $6$\nIf the object took $4$ steps, then it must have gone two steps and two steps , in some permutation. There are $\\frac{4!}{2!2!} = 6$ ways for these four steps of occuring, and t... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_17 | C | 45 | Starting with some gold coins and some empty treasure chests, I tried to put $9$ gold coins in each treasure chest, but that left $2$ treasure chests empty. So instead I put $6$ gold coins in each treasure chest, but then I had $3$ gold coins left over. How many gold coins did I have?
$\textbf{(A) }9\qquad\textbf{(B)... | [
"We can represent the amount of gold with $g$ and the amount of chests with $c$ . We can use the problem to make the following equations: \\[9c-18 = g\\] \\[6c+3 = g\\]\nWe do this because for 9 chests there are 2 empty and if 9 were in each 9 multiplied by 2 is 18 left.\nTherefore, $6c+3 = 9c-18.$ This implies tha... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21 | C | 9 | Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the sam... | [
"Let $x$ be the number of shots that Candace made in the first half, and let $y$ be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have $x+y=10+15=25.$ In addition, we have the following inequalities: \\[\... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_6 | null | 440 | Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve sa... | [
"We call the three roots (some may be equal to one another) $x_1$ $x_2$ , and $x_3$ . Using Vieta's formulas, we get $x_1+x_2+x_3 = a$ $x_1 \\cdot x_2+x_1 \\cdot x_3+x_2 \\cdot x_3 = \\frac{a^2-81}{2}$ , and $x_1 \\cdot x_2 \\cdot x_3 = \\frac{c}{2}$\nSquaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \\cdot ... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | D | 11 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulti... | [
"Note that cycles exist initially and after each round of erasing.\nLet the parentheses denote cycles. It follows that:\nSince $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: \\[(12\\underline{425}3415251).\\] T... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_6 | A | 40 | Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many hours will it take to fill Steve's pool?
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 42 \qquad \textbf{(C)}\ 44 \qquad \textbf{(D)}\ 46 \qqu... | [
"Each of the four hoses hose fills $24,000/4 = 6,000$ gallons of water. At the rate it goes at it will take $6,000/2.5 = 2400$ minutes, or $\\boxed{40}$ hours.",
"If all four hoses fill $2.5$ gallons a minute, every minute $10$ gallons would be added. Since every hour has $60$ minutes, $600$ gallons of water woul... |
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_16 | C | 180 | Students from three middle schools worked on a summer project.
The total amount paid for the students' work was 744. Assuming each student received the same amount for a day's work, how much did the students from Balboa school earn altogether?
$\text{(A)}\ 9.00\text{ dollars} \qquad \text{(B)}\ 48.38\text{ dollars} \q... | [
"Altogether, the summer project totaled $(7)(3)+(4)(5)+(5)(9)=21+20+45=86$ days of work for a single student. This equals $744/86=9$ dollars per day per student. The students from Balboa school earned $9(4)(5)=\\boxed{180.00}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_21 | C | 37 | Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$ , and $49$ . Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?
$\textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48$ | [
"If at least half the guesses are too low, then Norb's age must be greater than $36.$\nIf two of the guesses are off by one, then his age is in between two guesses whose difference is $2.$ It could be $31,37,$ or $48,$ but because his age is greater than $36$ it can only be $37$ or $48.$\nLastly, Norb's age is a pr... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_22 | D | 28 | Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these}$ | [
"Without loss of generality, assume something costs $100$ dollars. Then with each successive discount, it would cost $90$ dollars, then $72$ dollars. This amounts to a total of $28$ dollars off, so the single discount would be $\\boxed{28}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15 | C | 75 | Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$ | [
"Since $20\\% = \\frac{1}{5}$ , multiplying the given condition by $5$ shows that $y$ is $15 \\cdot 5 = \\boxed{75}$ percent of $x$",
"Letting $x=100$ (without loss of generality), the condition becomes $0.15\\cdot 100 = 0.2\\cdot y \\Rightarrow 15 = \\frac{y}{5} \\Rightarrow y=75$ . Clearly, it follows that $y$ ... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_7 | A | 100 | Suppose $A>B>0$ and A is $x$ % greater than $B$ . What is $x$
$\textbf {(A) } 100\left(\frac{A-B}{B}\right) \qquad \textbf {(B) } 100\left(\frac{A+B}{B}\right) \qquad \textbf {(C) } 100\left(\frac{A+B}{A}\right)\qquad \textbf {(D) } 100\left(\frac{A-B}{A}\right) \qquad \textbf {(E) } 100\left(\frac{A}{B}\right)$ | [
"We have that A is $x\\%$ greater than B, so $A=\\frac{100+x}{100}(B)$ . We solve for $x$ . We get\n$\\frac{A}{B}=\\frac{100+x}{100}$\n$100\\frac{A}{B}=100+x$\n$100\\left(\\frac{A}{B}-1\\right)=x$\n$100\\left(\\frac{A-B}{B}\\right)=x$ $\\boxed{100}$",
"The question is basically asking the percentage increase from... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_6 | B | 4 | Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?
$\textbf{(A) } \frac{4DQ}{S} \qquad \textbf{(B) } \frac{4DS}{Q} \qquad \textbf{(C) } \... | [
"Each can of soda costs $\\frac QS$ quarters, or $\\frac{Q}{4S}$ dollars. Therefore, $D$ dollars can purchase $\\frac{D}{\\left(\\tfrac{Q}{4S}\\right)}=\\boxed{4}$ cans of soda.",
"Note that $S$ is in the unit of $\\text{can}.$ On the other hand, $Q$ and $D$ are both in the unit of $\\text{cost}.$\nThe units of $... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_16 | B | 438 | Suppose $a$ $b$ $c$ are positive integers such that \[a+b+c=23\] and \[\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9.\] What is the sum of all possible distinct values of $a^2+b^2+c^2$
$\textbf{(A)}\: 259\qquad\textbf{(B)} \: 438\qquad\textbf{(C)} \: 516\qquad\textbf{(D)} \: 625\qquad\textbf{(E)} \: 687$ | [
"Because $a + b + c$ is odd, $a$ $b$ $c$ are either one odd and two evens or three odds.\n$\\textbf{Case 1}$ $a$ $b$ $c$ have one odd and two evens.\nWithout loss of generality, we assume $a$ is odd and $b$ and $c$ are even.\nHence, ${\\rm gcd} \\left( a , b \\right)$ and ${\\rm gcd} \\left( a , c \\right)$ are odd... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_21 | A | 0 | Suppose $a$ $b$ , and $c$ are nonzero real numbers, and $a+b+c=0$ . What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$
$\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }... | [
"There are $2$ cases to consider:\nCase $1$ $2$ of $a$ $b$ , and $c$ are positive and the other is negative. Without loss of generality (WLOG), we can assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that \\[\\frac{a}{|a|}+\\frac{b}{|b|}+\\frac{c}{|c|}+\\frac{abc}{|abc|}=1+1-1-1=0.\\]... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_22 | null | 492 | Suppose $a$ $b$ and $c$ are positive integers with $a+b+c=2006$ , and $a!b!c!=m\cdot 10^n$ , where $m$ and $n$ are integers and $m$ is not divisible by $10$ . What is the smallest possible value of $n$
$\mathrm{(A)}\ 489 \qquad \mathrm{(B)}\ 492 \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 498 \qquad \mathrm{(E)}\ 50... | [
"The power of $10$ for any factorial is given by the well-known algorithm \\[\\left\\lfloor \\frac n{5}\\right\\rfloor + \\left\\lfloor \\frac n{25}\\right\\rfloor + \\left\\lfloor \\frac n{125}\\right\\rfloor + \\cdots\\] It is rational to guess numbers right before powers of $5$ because we won't have any extra nu... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_2 | D | 200 | Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$ | [
"Since $a=1.5b$ , that means $b=\\frac{a}{1.5}$ . We multiply by $3$ to get a $3b$ term, yielding $3b=2a$ , and $2a$ is $\\boxed{200}$ of $a$",
"Without loss of generality, let $b=100$ . Then, we have $a=150$ and $3b=300$ . Thus, $\\frac{3b}{a}=\\frac{300}{150}=2$ , so $3b$ is $200\\%$ of $a$ . Hence the answer i... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_17 | A | 4 | Suppose $a$ is a real number such that the equation \[a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}\] has more than one solution in the interval $(0, \pi)$ . The set of all such $a$ that can be written
in the form \[(p,q) \cup (q,r),\] where $p, q,$ and $r$ are real numbers with $p < q< r$ . What is $p+q+r$
$\textbf{(A) } {-... | [
"We are given that $a\\cdot(\\sin{x}+\\sin{(2x)})=\\sin{(3x)}$\nUsing the sine double angle formula combine with the fact that $\\sin{(3x)} = \\sin{x}\\cdot(4\\cos^2{x}-1)$ , which can be derived using sine angle addition with $\\sin{(2x + x)}$ , we have \\[a\\cdot(\\sin{x}+2\\sin{x}\\cos{x})=\\sin{x}\\cdot(4\\cos^... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_6 | C | 5 | Suppose $d$ is a digit. For how many values of $d$ is $2.00d5 > 2.005$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10$ | [
"We see that $2.0055$ works but $2.0045$ does not. The digit $d$ can be from $5$ through $9$ , which is $\\boxed{5}$ values."
] |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3 | null | 750 | Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if | [
"Repeating decimals represent rational numbers . To figure out which rational number, we sum an infinite geometric series $0.d25d25d25\\ldots = \\sum_{n = 1}^\\infty \\frac{d25}{1000^n} = \\frac{100d + 25}{999}$ . Thus $\\frac{n}{810} = \\frac{100d + 25}{999}$ so $n = 30\\frac{100d + 25}{37} =750\\frac{4d + 1}{37... |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_21 | C | 2 | Suppose $n^{*}$ means $\frac{1}{n}$ , the reciprocal of $n$ . For example, $5^{*}=\frac{1}{5}$ . How many of the following statements are true?
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | [
"We can just test all of these statements: \\begin{align*} 3^*+6^* &= \\frac{1}{3}+\\frac{1}{6} \\\\ &= \\frac{1}{2} \\neq 9^* \\\\ 6^*-4^* &= \\frac{1}{6}-\\frac{1}{4} \\\\ &= \\frac{-1}{12} \\neq 2^* \\\\ 2^*\\cdot 6^* &= \\frac{1}{2}\\cdot \\frac{1}{6} \\\\ &= \\frac{1}{12} = 12^* \\\\ 10^* \\div 2^* &= \\frac{1... |
https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_6 | null | 743 | Suppose $r^{}_{}$ is a real number for which
Find $\lfloor 100r \rfloor$ . (For real $x^{}_{}$ $\lfloor x \rfloor$ is the greatest integer less than or equal to $x^{}_{}$ .) | [
"There are $91 - 19 + 1 = 73$ numbers in the sequence . Since the terms of the sequence can be at most $1$ apart, all of the numbers in the sequence can take one of two possible values. Since $\\frac{546}{73} = 7 R 35$ , the values of each of the terms of the sequence must be either $7$ or $8$ . As the remainder is... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_9 | null | 192 | Suppose $x$ is in the interval $[0, \pi/2]$ and $\log_{24\sin x} (24\cos x)=\frac{3}{2}$ . Find $24\cot^2 x$ | [
"We can rewrite the given expression as \\[\\sqrt{24^3\\sin^3 x}=24\\cos x\\] Square both sides and divide by $24^2$ to get \\[24\\sin ^3 x=\\cos ^2 x\\] Rewrite $\\cos ^2 x$ as $1-\\sin ^2 x$ \\[24\\sin ^3 x=1-\\sin ^2 x\\] \\[24\\sin ^3 x+\\sin ^2 x - 1=0\\] Testing values using the rational root theorem gives $\... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_8 | E | 3 | Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
$\textbf{(A)}\ m+3n\qquad\textbf{(B)}\ 3m-n\qquad\textbf{(C)}\ 3m^2 + 3n^2\qquad\textbf{(D)}\ (nm + 3)^2\qquad\textbf{(E)}\ 3mn$ | [
"Assume WLOG that $m$ and $n$ are both $1$ . Plugging into each of the choices, we get $4, 2, 6, 16,$ and $3$ . The only odd integer is $\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_21 | E | 6 | Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined. How many of the resulting figures can be folded into a topless cubical box?
[asy] draw((1,0)--(2,0)--(2,5)--(1,5)--cycle); draw((0,1)--(3,1)--(3,4)--(0,4)--cycle); draw((0,2)--(4,2)--(4,3)--(0,3)--cyc... | [
"The four squares we already have assemble nicely into four sides of the cube. Let the central one be the bottom, and fold the other three upwards to get the front, right, and back side. Currently, our box is missing its left side and its top side. We have to count the possibilities that would fold to one of these ... |
https://artofproblemsolving.com/wiki/index.php/2020_USOMO_Problems/Problem_4 | null | 197 | Suppose that $(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \le i < j \le 100$ and $|a_ib_j - a_j b_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ... | [
"Let's start off with just $(a_1, b_1), (a_2, b_2)$ and suppose that it satisfies the given condition. We could use $(1, 1), (1, 2)$ for example. We should maximize the number of conditions that the third pair satisfies. We find out that the third pair should equal $(a_1+a_2, b_1+b_2)$\nWe know this must be true: \... |
https://artofproblemsolving.com/wiki/index.php/2020_USOJMO_Problems/Problem_5 | null | 197 | Suppose that $(a_1,b_1),$ $(a_2,b_2),$ $\dots,$ $(a_{100},b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i,j)$ satisfying $1\leq i<j\leq 100$ and $|a_ib_j-a_jb_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ order... | [
"Call the pair $(i, j)$ good if $1\\leq i < j \\leq 100$ and $|a_ib_j-a_jb_i|=1$ . Note that we can reorder the pairs $(a_1, b_1), (a_2, b_2), \\ldots, (a_{100}, b_{100})$ without changing the number of good pairs. Thus, we can reorder them so that $a_1\\leq a_2\\leq\\ldots\\leq a_{100}$ . Furthermore, reorder them... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_11 | B | 53 | Suppose that $(u_n)$ is a sequence of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$
and that $u_3=9$ and $u_6=128$ . What is $u_5$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 53\qquad\mathrm{(C)}\ 68\qquad\mathrm{(D)}\ 88\qquad\mathrm{(E)}\ 104$ | [
"If we plug in $n=4$ , we get\nBy plugging in $n=3$ , we get\nThis is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives $128=5u_4+18 \\Longrightarrow u_4=22$ , therefore $u_5=\\frac{128-22}{2}=53 \\longrightarrow \\textbf{\\boxed{53}$ .\... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22 | D | 8,178 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on... | [
"For $1\\leq k\\leq 12,$ suppose that cards $1, 2, \\ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\\ldots,13$ are picked up on the second pass.\nOnce we pick the spots for the cards on the first pass, there is only one way to arrange all $\\boldsymbol{13}$ cards.\nFor each value of $k,$... |
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