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/2015_AMC_12A_Problems/Problem_4 | B | 32 | The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ | [
"Let $a$ be the bigger number and $b$ be the smaller.\n$a + b = 5(a - b)$\nMultiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives\n$\\frac{a}{b} = \\frac32$ , so the answer is $\\boxed{32}$",
"Without loss of generality, let the two numbers be ... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_4 | E | 166 | The sum of two prime numbers is $85$ . What is the product of these two prime numbers?
$\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$ | [
"Since the two prime numbers sum to an odd number, one of them must be even. The only even prime number is $2$ . The other prime number is $85-2=83$ , and the product of these two numbers is $83\\cdot2=\\boxed{166}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_8 | D | 7 | The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | [
"Let the three numbers be equal to $a$ $b$ , and $c$ . We can now write three equations:\n$a+b=12$\n$b+c=17$\n$a+c=19$\nAdding these equations together, we get that\n$2(a+b+c)=48$ and\n$a+b+c=24$\nSubstituting the original equations into this one, we find\n$c+12=24$\n$a+17=24$\n$b+19=24$\nTherefore, our numbers are... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_6 | D | 7 | The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | [
"Let the three numbers be equal to $a$ $b$ , and $c$ . We can now write three equations:\n$a+b=12$\n$b+c=17$\n$a+c=19$\nAdding these equations together, we get that\n$2(a+b+c)=48$ and\n$a+b+c=24$\nSubstituting the original equations into this one, we find\n$c+12=24$\n$a+17=24$\n$b+19=24$\nTherefore, our numbers are... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_7 | B | 38 | The symbolism $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$ . For example, $\lfloor 3 \rfloor = 3,$ and $\lfloor 9/2 \rfloor = 4$ . Compute \[\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor.\]
$\textbf{(A) } 35 \qquad\textbf{(B) } 3... | [
"The first three values in the sum are equal to $1,$ the next five equal to $2,$ the next seven equal to $3,$ and the last one equal to $4.$ For example, since $2^2=4$ any square root of a number less than $4$ must be less than $2.$ Sum them all together to get\n\\[3\\cdot1 + 5\\cdot2 + 7\\cdot3 + 1\\cdot4 = 3+10+2... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_9 | null | 25 | The system of equations \begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*}
has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$ | [
"Since $\\log ab = \\log a + \\log b$ , we can reduce the equations to a more recognizable form:\n\\begin{eqnarray*} -\\log x \\log y + \\log x + \\log y - 1 &=& 3 - \\log 2000\\\\ -\\log y \\log z + \\log y + \\log z - 1 &=& - \\log 2\\\\ -\\log x \\log z + \\log x + \\log z - 1 &=& -1\\\\ \\end{eqnarray*}\nLet $a... |
https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_3 | null | 943 | The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$
In the newspaper story covering the event, it was reported that
What was the total number of fish caught during the festival? | [
"Suppose that the number of fish is $x$ and the number of contestants is $y$ . The $y-(9+5+7)=y-21$ fishers that caught $3$ or more fish caught a total of $x - \\left(0\\cdot(9) + 1\\cdot(5) + 2\\cdot(7)\\right) = x - 19$ fish. Since they averaged $6$ fish,\nSimilarily, those who caught $12$ or fewer fish averaged ... |
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_17 | D | 15 | The table below gives the percent of students in each grade at Annville and Cleona elementary schools:
\[\begin{tabular}{rccccccc}&\textbf{\underline{K}}&\textbf{\underline{1}}&\textbf{\underline{2}}&\textbf{\underline{3}}&\textbf{\underline{4}}&\textbf{\underline{5}}&\textbf{\underline{6}}\\ \textbf{Annville:}& 16\% &... | [
"By the tables, Annville has $11$ 6th graders and Cleona has $34$ . Together they have $45$ 6th graders and $300$ total students, so the percent is $\\frac{45}{300}=\\frac{15}{100}= \\boxed{15}$"
] |
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_15 | B | 62.5 | The table below shows the distance $s$ in feet a ball rolls down an inclined plane in $t$ seconds.
\[\begin{tabular}{|c|c|c|c|c|c|c|}\hline t & 0 & 1 & 2 & 3 & 4 & 5\\ \hline s & 0 & 10 & 40 & 90 & 160 & 250\\ \hline\end{tabular}\] The distance $s$ for $t = 2.5$ is:
$\textbf{(A)}\ 45\qquad \textbf{(B)}\ 62.5\qquad \tex... | [
"Looking at the pattern, we can determine that $t=10s^2$ . Applying the relationship, we can see that $s = \\boxed{62.5}$ when $t=2.5$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_8 | E | 75 | The table shows some of the results of a survey by radiostation KACL. What percentage of the males surveyed listen to the station?
$\begin{tabular}{|c|c|c|c|}\hline & Listen & Don't Listen & Total\\ \hline Males & ? & 26 & ?\\ \hline Females & 58 & ? & 96\\ \hline Total & 136 & 64 & 200\\ \hline\end{tabular}$
$\textbf{... | [
"Filling out the chart, it becomes\n$\\begin{tabular}{|c|c|c|c|}\\hline & Listen & Don't Listen & Total\\\\ \\hline Males & 78 & 26 & 104\\\\ \\hline Females & 58 & 38 & 96\\\\ \\hline Total & 136 & 64 & 200\\\\ \\hline\\end{tabular}$\nThus, the percentage of males surveyed that listen to the station is $100 \\cdot... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_10 | C | 3.3 | The taxi fare in Gotham City is $2.40 for the first $\frac12$ mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?
$\textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(... | [
"Let $x$ be the number of miles you ride. The number of miles you ride after the first half mile is $x-0.5.$ We can write this equation:\n\\begin{align*} 10 &= 2.4 + 0.2 \\times \\frac{x-0.5}{0.1} + 2\\\\ 5.6 &= 2(x-0.5)\\\\ 2.8 &= x-0.5\\\\ x &= \\boxed{3.3}",
"\\begin{array}{|c|c|c|}\nMiles & Money & Remark\\\\... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_2 | null | 781 | The teams $T_1$ $T_2$ $T_3$ , and $T_4$ are in the playoffs. In the semifinal matches, $T_1$ plays $T_4$ , and $T_2$ plays $T_3$ . The winners of those two matches will play each other in the final match to determine the champion. When $T_i$ plays $T_j$ , the probability that $T_i$ wins is $\frac{i}{i+j}$ , and the out... | [
"There are two scenarios in which $T_4$ wins. The first scenario is where $T_4$ beats $T_1$ $T_3$ beats $T_2$ , and $T_4$ beats $T_3$ , and the second scenario is where $T_4$ beats $T_1$ $T_2$ beats $T_3$ , and $T_4$ beats $T_2$ . Consider the first scenario. The probability $T_4$ beats $T_1$ is $\\frac{4}{4+1}$ , ... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_2 | A | 8,671 | The ten-letter code $\text{BEST OF LUCK}$ represents the ten digits $0-9$ , in order. What 4-digit number is represented by the code word $\text{CLUE}$
$\textbf{(A)}\ 8671 \qquad \textbf{(B)}\ 8672 \qquad \textbf{(C)}\ 9781 \qquad \textbf{(D)}\ 9782 \qquad \textbf{(E)}\ 9872$ | [
"We can derive that $C=8$ $L=6$ $U=7$ , and $E=1$ . Therefore, the answer is $\\boxed{8671}$ ~edited by Owencheng"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_2 | null | 195 | The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of... | [
"If the sum of the original sequence is $\\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \\sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \\rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the terms, or $\\... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_13 | null | 90 | The terms of the sequence $(a_i)$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$ | [
"This question is guessable but let's prove our answer\n\\[a_{n + 2} = \\frac {a_n + 2009} {1 + a_{n + 1}}\\]\n\\[a_{n + 2}(1 + a_{n + 1})= a_n + 2009\\]\n\\[a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009\\]\nlets put $n+1$ into $n$ now\n\\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009\\]\nand set them equal now\n\\[a_{n + ... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_7 | E | 130 | The third exit on a highway is located at milepost 40 and the tenth exit is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost would you expect to find this service center?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \te... | [
"There are $160-40=120$ miles between the third and tenth exits, so the service center is at milepost $40+(3/4)(120) = 40+90=\\boxed{130}$"
] |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_16 | D | 4 | The three row sums and the three column sums of the array
\[\left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right]\]
are the same. What is the least number of entries that must be altered to make all six sums different from one another?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\... | [
"If you change $3$ numbers, then you either change one number in each column and row (ie sudoku-style):\n\\[\\left[\\begin{matrix}* & 9 & 2\\\\ 8 & * & 6\\\\ 3 & 5 & *\\end{matrix}\\right]\\]\nOr you leave at least one row and one column unchanged:\n\\[\\left[\\begin{matrix}* & 9 & 2\\\\ * & * & 6\\\\ 3 & 5 & 7\\en... |
https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_1 | C | 6 | The three-digit number $2a3$ is added to the number $326$ to give the three-digit number $5b9$ . If $5b9$ is divisible by 9, then $a+b$ equals
$\text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ | [
"If $5b9$ is divisible by $9$ , this must mean that $5 + b + 9$ is a multiple of $9$ . So, \\[5 + b + 9 = 9, 18, 27, 36...\\]\nBecause $5 + 9 = 14$ and $b$ is in between 0 and 9,\n\\[5 + b + 9 = 18\\] \\[b = 4\\]\nThe question states that \\[2a3 + 326 = 549\\] so \\[2a3 = 549 - 326\\] \\[2a3 = 223\\] \\[a = 2\\]\n\... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_11 | B | 64 | The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$ . In feet, how tall is the taller tree?
$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$ | [
"Let the height of the taller tree be $h$ and let the height of the smaller tree be $h-16$ . Since the ratio of the smaller tree to the larger tree is $\\frac{3}{4}$ , we have $\\frac{h-16}{h}=\\frac{3}{4}$ . Solving for $h$ gives us $h=64 \\Rightarrow \\boxed{64}$",
"To answer this problem, you have to make it s... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_24 | E | 2,013 | The tower function of twos is defined recursively as follows: $T(1) = 2$ and $T(n + 1) = 2^{T(n)}$ for $n\ge1$ . Let $A = (T(2009))^{T(2009)}$ and $B = (T(2009))^A$ . What is the largest integer $k$ for which \[\underbrace{\log_2\log_2\log_2\ldots\log_2B}_{k\text{ times}}\] is defined?
$\textbf{(A)}\ 2009\qquad \textbf... | [
"(Note: This for the people who need clear, concise reasoning in the form of mostly words, instead of a compact proof written in set theory symbols and the like. I figured if I had trouble understanding the above solutions, others would too.)\nWe begin by contemplating what $B$ actually is. Calculating the first fe... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_15 | B | 47 | The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
$\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66... | [
"Let the amount of people be $p$ , horses be $h$ , sheep be $s$ , cows be $c$ , and ducks be $d$ . \nWe know \\[3h=p\\] \\[4c=s\\] \\[3p=d\\] Then the total amount of people, horses, sheep, cows, and ducks may be written as $p+h+s+c+d = 3h+h+4c+c+(3\\times3h)$ . This is equivalent to $13h+5c$ . Looking through the ... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_7 | C | 4.5 | The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?
[asy] size(2... | [
"The area of a triangle is $\\frac12 bh$ . If we let $CD$ be the base of the triangle, then the height is $AB$ , and the area is $\\frac12 \\cdot 3 \\cdot 3 = \\boxed{4.5}$",
"We can see that there is a big triangle encasing $ACD$ . The area of that triangle is $\\frac12 \\cdot 3 \\cdot 6 = 9$ . We can easily see... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_4 | C | 13 | The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$
[asy] unitsize(8mm); for (int i=0; i<7; ++i) { draw((i,0)--(i,7),gray); draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4... | [
"We count $3 \\cdot 3=9$ unit squares in the middle, and $8$ small triangles, which gives 4 rectangles each with an area of $1$ . Thus, the answer is $9+4=\\boxed{13}$",
"We can see here that there are $9$ total squares in the middle. We also see that the triangles that make the corners of the shape have an area ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_19 | C | 64 | The two circles pictured have the same center $C$ . Chord $\overline{AD}$ is tangent to the inner circle at $B$ $AC$ is $10$ , and chord $\overline{AD}$ has length $16$ . What is the area between the two circles?
[asy] unitsize(45); import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaul... | [
"Since $\\triangle ACD$ is isosceles, $CB$ bisects $AD$ . Thus $AB=BD=8$ . From the Pythagorean Theorem, $CB=6$ . Thus the area between the two circles is $100\\pi - 36\\pi=64\\pi$ $\\boxed{64}$"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_17 | B | 18 | The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
$\mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } ... | [
"If Jack's current age is $\\overline{ab}=10a+b$ , then Bill's current age is $\\overline{ba}=10b+a$\nIn five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$\nWe are given that $10a+b+5=2(10b+a+5)$\nThus $8a=19b+5 \\Rightarrow a=\\dfrac{19b+5}{8}$\nFor $b=1$ we get $a=3$ . For $b=2$ and $b=3$ t... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_15 | B | 18 | The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
$\mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } ... | [
"If Jack's current age is $\\overline{ab}=10a+b$ , then Bill's current age is $\\overline{ba}=10b+a$\nIn five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$\nWe are given that $10a+b+5=2(10b+a+5)$\nThus $8a=19b+5 \\Rightarrow a=\\dfrac{19b+5}{8}$\nFor $b=1$ we get $a=3$ . For $b=2$ and $b=3$ t... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_9 | C | 3 | The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | [
"We find that the area of the triangle is $\\frac{6\\times 2\\sqrt{3}}{2} =6\\sqrt{3}$ . By the Pythagorean Theorem , we have that the length of the hypotenuse is $\\sqrt{(2\\sqrt{3})^2+6^2}=4\\sqrt{3}$ . Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.\nLet $h$... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_12 | D | 79 | The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$1$",(0,.5)); l... | [
"The possible sums are \\[\\begin{tabular}{c|ccc} & 1 & 3 & 5 \\\\ \\hline 2 & 3 & 5 & 7 \\\\ 4 & 5 & 7 & 9 \\\\ 6 & 7 & 9 & 11 \\end{tabular}\\]\nOnly $9$ is not prime, so there are $7$ prime numbers and $9$ total numbers for a probability of $\\boxed{79}$"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_4 | null | 185 | The two squares shown share the same center $O_{}$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n.$
[asy] //code taken from thread for problem real alpha = 25; pair W=dir(225), ... | [
"Triangles $AOB$ $BOC$ $COD$ , etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length $1$ in the circumcircle of the squares pass through $B$ , etc.), and each area is $\\frac{\\frac{43}{99}\\cdot\\frac{1}{2}}{2}$ . Since the area of a tr... |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_6 | A | 0 | The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | [
"Within six consecutive integers, there must be a number with a factor of $5$ and an even integer with a factor of $2$ . Multiplied together, these would produce a number that is a multiple of $10$ and has a units digit of $\\boxed{0}$",
"We can easily compute the product of the first 6 positive integers: $(1*2*3... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_14 | E | 9 | The units digit of $3^{1001} 7^{1002} 13^{1003}$ is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$ | [
"First, we notice that $3^0$ is congruent to $1 \\ \\text{(mod 10)}$ $3^1$ is $3 \\ \\text{(mod 10)}$ $3^2$ is $9 \\ \\text{(mod 10)}$ $3^3$ is $7 \\ \\text{(mod 10)}$ $3^4$ is $1 \\ \\text{(mod 10)}$ , and so on. This turns out to be a cycle repeating every $4$ terms, so $3^{1001}$ is congruent to $3 \\ \\text{(mo... |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_7 | D | 10,000,000,000 | The value of $\frac{(487,000)(12,027,300)+(9,621,001)(487,000)}{(19,367)(.05)}$ is closest to
$\text{(A)}\ 10,000,000 \qquad \text{(B)}\ 100,000,000 \qquad \text{(C)}\ 1,000,000,000 \qquad \text{(D)}\ 10,000,000,000 \qquad \text{(E)}\ 100,000,000,000$ | [
"We can make the approximations \\begin{align*} 487,000 &\\approx 500,000 \\\\ 12,027,300 &\\approx 12,000,000 \\\\ 9,621,001 &\\approx 10,000,000 \\\\ 19,367 &\\approx 20,000. \\end{align*}\nUsing these instead of the original numbers for an estimate, we have \\begin{align*} \\frac{(500,000)(12,000,000)+(10,000,0... |
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_5 | A | 4 | The value of $\left(256\right)^{.16}\left(256\right)^{.09}$ is:
$\textbf{(A)}\ 4 \qquad \\ \textbf{(B)}\ 16\qquad \\ \textbf{(C)}\ 64\qquad \\ \textbf{(D)}\ 256.25\qquad \\ \textbf{(E)}\ -16$ | [
"When we multiply numbers with exponents, we add the exponents together and leave the bases unchanged. We can apply this concept to computate $256^{0.16} \\cdot 256^{0.09}$ \\[256^{0.16} \\cdot 256^{0.09} = 256^{0.16+0.09}=256^{0.25}.\\] Now we can convert the decimal exponent to a fraction: \\[256^{0.25} = 256^{\\... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_25 | D | 5 | The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$ | [
"$\\log_{5}\\frac{(125)(625)}{25}$ can be simplified to $\\log_{5}\\ (125)(25)$ since $25^2 = 625$ $125 = 5^3$ and $5^2 = 25$ so $\\log_{5}\\ 5^5$ would be the simplest form. In $\\log_{5}\\ 5^5$ $5^x = 5^5$ . Therefore, $x = 5$ and the answer is $\\boxed{5}$",
"$\\log_{5}\\frac{(125)(625)}{25}$ can be also repre... |
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_9 | B | 14 | The value of $x - y^{x - y}$ when $x = 2$ and $y = -2$ is:
$\textbf{(A)}\ -18 \qquad \textbf{(B)}\ -14\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 256$ | [
"Just plug in the numbers and follow the order of operations: \\[2-(-2)^{2-(-2)}\\] \\[2-(-2)^4\\] \\[2-16\\] \\[\\boxed{14}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_3 | null | 103 | The value of $x$ that satisfies $\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [
"Let $\\log _{2^x}3^{20}=\\log _{2^{x+3}}3^{2020}=n$ . Based on the equation, we get $(2^x)^n=3^{20}$ and $(2^{x+3})^n=3^{2020}$ . Expanding the second equation, we get $8^n\\cdot2^{xn}=3^{2020}$ . Substituting the first equation in, we get $8^n\\cdot3^{20}=3^{2020}$ , so $8^n=3^{2000}$ . Taking the 100th root, we ... |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_20 | D | 3 | The value of the expression $\frac{(304)^5}{(29.7)(399)^4}$ is closest to
$\text{(A)}\ .003 \qquad \text{(B)}\ .03 \qquad \text{(C)}\ .3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 30$ | [
"\\[\\frac{(304)^5}{(29.7)(399)^4} \\approx \\frac{300^5}{30\\cdot400^4} = \\frac{3^5 \\cdot 10^{10}}{3\\cdot 4^4 \\cdot 10^9} = \\frac{3^4\\cdot 10}{4^4} = \\frac{810}{256}\\] Which is closest to $3\\rightarrow\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_9 | E | 16 | The vertex of the parabola $y = x^2 - 8x + c$ will be a point on the $x$ -axis if the value of $c$ is:
$\textbf{(A)}\ - 16 \qquad \textbf{(B) }\ - 4 \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 8 \qquad \textbf{(E) }\ 16$ | [
"Notice that if the vertex of a parabola is on the x-axis, then the x-coordinate of the vertex must be a solution to the quadratic. Since the quadratic is strictly increasing on either side of the vertex, the solution must have double multiplicity, or the quadratic is a perfect square trinomial. This means that for... |
https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_12 | null | 344 | The vertices of $\triangle ABC$ are $A = (0,0)\,$ $B = (0,420)\,$ , and $C = (560,0)\,$ . The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's. Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ $P_3\,$ $P_4, \dots$ are generated by rolling the die ... | [
"If we have points $(p,q)$ and $(r,s)$ and we want to find $(u,v)$ so $(r,s)$ is the midpoint of $(u,v)$ and $(p,q)$ , then $u=2r-p$ and $v=2s-q$ . So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have \\[P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\\bmo... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_17 | null | 12 | The vertices of a quadrilateral lie on the graph of $y=\ln{x}$ , and the $x$ -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$ . What is the $x$ -coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qqua... | [
"Let the coordinates of the quadrilateral be $(n,\\ln(n)),(n+1,\\ln(n+1)),(n+2,\\ln(n+2)),(n+3,\\ln(n+3))$ . We have by shoelace's theorem, that the area is \\begin{align*} &\\frac{\\ln(n)(n+1) + \\ln(n+1)(n+2) + \\ln(n+2)(n+3)+n\\ln(n+3)}{2} - \\frac{\\ln(n+1)(n) + \\ln(n+2)(n+1) + \\ln(n+3)(n+2)+\\ln(n)(n+3)}{2} ... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_5 | null | 144 | The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the no... | [
"First, we determine which possible combinations of digits $1$ through $9$ will yield sums that are multiples of $3$ . It is simplest to do this by looking at each of the digits $\\bmod{3}$\nWe see that the numbers $1, 4,$ and $7$ are congruent to $1 \\pmod{3}$ , that the numbers $2, 5,$ and $8$ are congruent to $2... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_24 | C | 108 | The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$ | [
"WLOG, let the centroid of $\\triangle ABC$ be $I = (-1,-1)$ . The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, $A = (1,1)$ , so $AI = BI = CI = 2\\sqrt{2}$ , so since $\\triangle AIB$ is isosceles and ... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_24 | null | 108 | The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$ | [
"Without loss of generality, let the centroid of $\\triangle ABC$ be $G = (1, 1)$ . Assuming we don't know one vertex is $(-1, -1)$ we let the vertices be $A\\left(x_1, \\frac{1}{x_1}\\right), B\\left(x_2, \\frac{1}{x_2}\\right), C\\left(x_3, \\frac{1}{x_3}\\right).$\nSince the centroid coordinates are the average ... |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_25 | B | 32 | The volume of a certain rectangular solid is $8$ cm , its total surface area is $32$ cm , and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is
$\mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 40 \qquad \mathr... | [
"As the dimensions are in geometric progression, let them be $\\frac{b}{r}$ $b$ , and $br$ cm, so the volume is $\\left(\\frac{b}{r}\\right)(b)(br) = b^3$ , giving $b^3 = 8$ and thus $b = 2$ . The surface area condition now yields \\begin{align*}2\\left(\\frac{2}{r}\\right)(2)+2(2)(2r)+2(2r)\\left(\\frac{2}{r}\\rig... |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_32 | A | 9 | The volume of a pyramid whose base is an equilateral triangle of side length 6 and whose other edges are each of length $\sqrt{15}$ is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 9/2 \qquad \textbf{(C)}\ 27/2 \qquad \textbf{(D)}\ \frac{9\sqrt3}{2} \qquad \textbf{(E)}\ \text{none of these}$ | [
"\nDraw an altitude towards the equilateral triangle base. By symmetry (this can also be proved by HL), the base of the altitude is equidistant from the three points of the equilateral triangle. This means that the distance from the base of the altitude to one of the points of the equilateral triangle is $2\\sqrt... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_21 | B | 24 | The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^{2}$ $8\text{ in}^{2}$ , and $6\text{ in}^{2}$ respectively is:
$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ ... | [
"If the sidelengths of the cubes are expressed as $a, b,$ and $c,$ then we can write three equations:\n\\[ab=12, bc=8, ac=6.\\]\nThe volume is $abc.$ Notice symmetry in the equations. We can find $abc$ my multiplying all the equations and taking the positive square root.\n\\begin{align*} (ab)(bc)(ac) &= (12)(8)(6)\... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_10 | null | 4 | The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$ . At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a cou... | [
"We divide this up into casework. The \"directions\" the bug can go are $\\text{Clockwise}$ $\\text{Counter-Clockwise}$ , and $\\text{Switching}$ . Let an $I$ signal going clockwise (because it has to be in the inner circle), an $O$ signal going counter-clockwise, and an $S$ switching between inner and outer circ... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_13 | D | 36 | The word " HELP " in block letters is painted in black with strokes $1$ unit wide on a $5$ by $15$ rectangular white sign with dimensions as shown. The area of the white portion of the sign, in square units, is
[asy] unitsize(12); fill((0,0)--(0,5)--(1,5)--(1,3)--(2,3)--(2,5)--(3,5)--(3,0)--(2,0)--(2,2)--(1,2)--(1,0)-... | [
"Count the number of black squares in each letter. H has 11, E has 11, L has 7, and P has 10, giving the number of black squares to be $11+11+7+10=39$ . The total number of squares is $(15)(5)=75$ and the number of white squares is $75-39=\\boxed{36}$"
] |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_4 | null | 450 | The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $100$ workers can produce $300$ widgets and $200$ whoosits. In two hours, $60$ workers can produce $240$ widgets and $300$ whoo... | [
"Suppose that it takes $x$ hours for one worker to create one widget, and $y$ hours for one worker to create one whoosit.\nTherefore, we can write that (note that two hours is similar to having twice the number of workers, and so on):\nSolve the system of equations with the first two equations to find that $(x,y) =... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_4 | B | 4 | The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?
$\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$ | [
"The palindrome right after 2002 is 2112. The product of the digits of 2112 is $\\boxed{4}$",
"The palindrome formula is to add 110 to the number in order to get the next palindrome, a palindrome needs to be in the form as ABBA . We can use this in this case to get 2112. 2*1*1*2=4. Therefore, the answer is $\\box... |
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_18 | A | 12 | The yearly changes in the population census of a town for four consecutive years are,
respectively, 25% increase, 25% increase, 25% decrease, 25% decrease.
The net change over the four years, to the nearest percent, is:
$\textbf{(A)}\ -12 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1\qquad \te... | [
"A 25% increase means the new population is $\\frac{5}{4}$ of the original population. A 25% decrease means the new population is $\\frac{3}{4}$ of the original population.\nThus, after four years, the population is $1 \\cdot \\frac{5}{4} \\cdot \\frac{5}{4} \\cdot \\frac{3}{4} \\cdot \\frac{3}{4} = \\frac{225}{25... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_23 | C | 16 | The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$ | [
"By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.\nBecause the zeros are integral, the discriminant of the function, $a^2 - 8a$ , is a perfect square, say $k^2$ . Then adding 16 to both sides and completing the square yields \\[(a - 4)^2 = k^2 + 16.\\] Therefore $(a-... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | C | 16 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$ | [
"The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.\nThe quadratic formula gives the roots of the quadratic equation: $x=\\frac{a\\pm\\sqrt{a^2-8a}}{2}$\nAs long as the numerator is an even integer, the roots are both integers. But first o... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_16 | B | 3 | There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start runn... | [
"If we have horses, $a_1, a_2, \\ldots, a_n$ , then any number that is a multiple of all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that $\\... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_12 | B | 3 | There are $10$ horses, named Horse 1, Horse 2, $\ldots$ , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in t... | [
"We know that Horse $k$ will be at the starting point after $n$ minutes if $k|n$ . Thus, we are looking for the smallest $n$ such that at least $5$ of the numbers $\\{1,2,\\cdots,10\\}$ divide $n$ . Thus, $n$ has at least $5$ positive integer divisors.\nWe quickly see that $12$ is the smallest number with at least ... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_17 | C | 13 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know... | [
"Consider the $10$ people to be standing in a circle, where two people opposite each other form a diameter of the circle.\nLet us use casework on the number of pairs that form a diameter of the circle.\nCase 1: $0$ diameters\nThere are $2$ ways: either $1$ pairs with $2$ $3$ pairs with $4$ , and so on or $10$ pairs... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_15 | C | 13 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know... | [
"Consider the $10$ people to be standing in a circle, where two people opposite each other form a diameter of the circle.\nLet us use casework on the number of pairs that form a diameter of the circle.\nCase 1: $0$ diameters\nThere are $2$ ways: either $1$ pairs with $2$ $3$ pairs with $4$ , and so on or $10$ pairs... |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_19 | B | 40 | There are $120$ seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?
$\text{(A)}\ 30 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 41 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 119$ | [
"Let $p$ be a person seated and $o$ is an empty seat\nThe pattern of seating that results in the fewest occupied seats is $\\text{opoopoopoo...po}$ .\nWe can group the seats in 3s like this: $\\text{opo opo opo ... opo}.$\nThere are a total of $40=\\boxed{40}$ groups"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_14 | D | 95,000 | There are $20$ cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all $20$ cities?
[asy] // made by SirCalcsALot size(300); pen shortdashed... | [
"We can see that the dotted line is exactly halfway between $4500$ and $5000$ , so it is at $4750$ . As this is the average population of all $20$ cities, the total population is simply $4750 \\cdot 20 = \\boxed{95000}$"
] |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_13 | C | 3 | There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at l... | [
"By PIE (Property of Inclusion/Exclusion), we have\n$|A_1 \\cup A_2 \\cup A_3| = \\sum |A_i| - \\sum |A_i \\cap A_j| + |A_1 \\cap A_2 \\cap A_3|.$ Number of people in at least two sets is $\\sum |A_i \\cap A_j| - 2|A_1 \\cap A_2 \\cap A_3| = 9.$ So, $20 = (10 + 13 + 9) - (9 + 2x) + x,$ which gives $x = \\boxed{3}.$... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_17 | D | 12 | There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | [
"Note that these $z$ such that $z^{24}=1$ are $e^{\\frac{ni\\pi}{12}}$ for integer $0\\leq n<24$ . So\n$z^6=e^{\\frac{ni\\pi}{2}}$\nThis is real if $\\frac{n}{2}\\in \\mathbb{Z} \\Leftrightarrow (n$ is even $)$ . Thus, the answer is the number of even $0\\leq n<24$ which is $\\boxed{12}$",
"$z = \\sqrt[24]{1} = 1... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12 | null | 548 | There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | [
"Let $a_{n}$ be the number of ways to form $n$ -letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical.\nNote that, at the end of each $n$ -letter string, there are $3$ possibilities for the last letter chain: it must be either $1$ $2$ , or $3$ letters long. Removing this last... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_14 | null | 840 | There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $\mid z \mid = 1$ . These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$ , where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \l... | [
"$z$ can be written in the form $\\text{cis\\,}\\theta$ . Rearranging, we find that $\\text{cis\\,}{28}\\theta = \\text{cis\\,}{8}\\theta+1$\nSince the real part of $\\text{cis\\,}{28}\\theta$ is one more than the real part of $\\text{cis\\,} {8}\\theta$ and their imaginary parts are equal, it is clear that either ... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_11 | D | 5 | There are $52$ people in a room. what is the largest value of $n$ such that the statement "At least $n$ people in this room have birthdays falling in the same month" is always true?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 12$ | [
"Pretend you have $52$ people you want to place in $12$ boxes, because there are $12$ months in a year. By the Pigeonhole Principle , one box must have at least $\\left\\lceil \\frac{52}{12} \\right\\rceil$ people $\\longrightarrow \\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_14 | null | 440 | There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$ | [
"Be wary of \"position\" versus \"number\" in the solution!\nEach POSITION in the 30-position permutation is uniquely defined by an ordered triple $(i, j, k)$ . The $n$ th position is defined by this ordered triple where $i$ is $n \\mod 2$ $j$ is $n \\mod 3$ , and $k$ is $n \\mod 5$ . There are 2 choices for $i$ , ... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_15 | null | 332 | There are $n$ mathematicians seated around a circular table with $n$ seats numbered $1,$ $2,$ $3,$ $...,$ $n$ in clockwise order. After a break they again sit around the table. The mathematicians note that there is a positive integer $a$ such that
Find the number of possible values of $n$ with $1 < n < 1000.$ | [
"It is a well-known fact that the set $0, a, 2a, ... (n-1)a$ forms a complete set of residues if and only if $a$ is relatively prime to $n$\nThus, we have $a$ is relatively prime to $n$ . In addition, for any seats $p$ and $q$ , we must have $ap - aq$ not be equivalent to either $p - q$ or $q - p$ modulo $n$ to sat... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_25 | D | 7,425 | There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
$\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$ | [
"We begin by narrowing down the possibilities. If the larger number were twice the smaller number, then the smallest possibility for the larger number is $2457\\times2=4914$ , since $2457$ is the smallest number in the set. The largest possibility would have to be twice the largest number in the set such that when ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_9 | B | 3 | There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$ , and $1001B + 3003A = 5005$ . What is the average of A, B, and C?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E) }\text{Not uniquely determined}$ | [
"Notice that we don't need to find what $A, B,$ and $C$ actually are, just their average. In other words, if we can find $A+B+C$ , we will be done.\nAdding up the equations gives $1001(A+B+C)=9009=1001(9)$ so $A+B+C=9$ and the average is $\\frac{9}{3}=3$ . Our answer is $\\boxed{3}$",
"Start by isolating $B$ and ... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_24 | D | 27,720 | There are exactly $77,000$ ordered quadruplets $(a, b, c, d)$ such that $\gcd(a, b, c, d) = 77$ and $\operatorname{lcm}(a, b, c, d) = n$ . What is the smallest possible value for $n$
$\textbf{(A)}\ 13,860\qquad\textbf{(B)}\ 20,790\qquad\textbf{(C)}\ 21,560 \qquad\textbf{(D)}\ 27,720 \qquad\textbf{(E)}\ 41,580$ | [
"Let $A=\\frac{a}{77},\\ B=\\frac{b}{77}$ , etc., so that $\\gcd(A,B,C,D)=1$ . Then for each prime power $p^k$ in the prime factorization of $N=\\frac{n}{77}$ , at least one of the prime factorizations of $(A,B,C,D)$ has $p^k$ , at least one has $p^0$ , and all must have $p^m$ with $0\\le m\\le k$\nLet $f(k)$ be th... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_19 | E | 78 | There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and \[5x^2+kx+12=0\] has at least one integer solution for $x$ . What is $N$
$\textbf{(A) }6\qquad \textbf{(B) }12\qquad \textbf{(C) }24\qquad \textbf{(D) }48\qquad \textbf{(E) }78\qquad$ | [
"Factor the quadratic into \\[\\left(5x + \\frac{12}{n}\\right)\\left(x + n\\right) = 0\\] where $-n$ is our integer solution. Then, \\[k = \\frac{12}{n} + 5n,\\] which takes rational values between $-200$ and $200$ when $|n| \\leq 39$ , excluding $n = 0$ . This leads to an answer of $2 \\cdot 39 = \\boxed{78}$",
... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_13 | B | 3 | There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$ . What is $b$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$ | [
"$\\sqrt[a]{N\\sqrt[b]{N\\sqrt[c]{N}}}$ can be simplified to $N^{\\frac{1}{a}+\\frac{1}{ab}+\\frac{1}{abc}}.$\nThe equation is then $N^{\\frac{1}{a}+\\frac{1}{ab}+\\frac{1}{abc}}=N^{\\frac{25}{36}}$ which implies that $\\frac{1}{a}+\\frac{1}{ab}+\\frac{1}{abc}=\\frac{25}{36}.$\n$a$ has to be $2$ since $\\frac{25}{3... |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_5 | A | 119 | There are many two-digit multiples of 7, but only two of the multiples have a digit sum of 10. The sum of these two multiples of 7 is
$\text{(A)}\ 119 \qquad \text{(B)}\ 126 \qquad \text{(C)}\ 140 \qquad \text{(D)}\ 175 \qquad \text{(E)}\ 189$ | [
"Writing out all two digit numbers that have a digital sum of $10$ , you get $19, 28, 37, 46, 55, 64, 73, 82,$ and $91$ . The two numbers on that list that are divisible by $7$ are $28$ and $91$ . Their sum is $28+91=119$ , choice $\\boxed{119}$",
"Writing out all the two digit multiples of $7$ , you get $14, 2... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_10 | null | 80 | There are nonzero integers $a$ $b$ $r$ , and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$ . For each possible combination of $a$ and $b$ , let ${p}_{a,b}$ be the sum of the zeros of $P(x)$ . Find the sum of the ${p}_{a,b}$ 's for all possible combinations of $a$ and ... | [
"Since $r+si$ is a root, by the Complex Conjugate Root Theorem, $r-si$ must be the other imaginary root. Using $q$ to represent the real root, we have\n$(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65$\nApplying difference of squares, and regrouping, we have\n$(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65$\nSo mat... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | null | 880 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $... | [
"Add the two equations to get that $\\log x+\\log y+2(\\log(\\gcd(x,y))+2(\\log(\\text{lcm}(x,y)))=630$ .\nThen, we use the theorem $\\log a+\\log b=\\log ab$ to get the equation, $\\log (xy)+2(\\log(\\gcd(x,y))+\\log(\\text{lcm}(x,y)))=630$ .\nUsing the theorem that $\\gcd(x,y) \\cdot \\text{lcm}(x,y)=x\\cdot y$ ,... |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_23 | C | 36 | There are positive integers that have these properties:
The product of the digits of the largest integer with both properties is
$\text{(A)}\ 7 \qquad \text{(B)}\ 25 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 60$ | [
"Five-digit numbers will have a minimum of $1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$ as the sum of their squares if the five digits are distinct and non-zero. If there is a zero, it will be forced to the left by rule #2.\nNo digit will be greater than $7$ , as $8^2 = 64$\nTrying four digit numbers $WXYZ$ , we have $w^2 +... |
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | null | 330 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | [
"By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are complex conjugates. Let $z=m+\\sqrt{n}\\cdot i$ and $\\overline{z}=m-\\sqrt{n}\\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\\overline{z}.$... |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_11 | B | 4 | There are several sets of three different numbers whose sum is $15$ which can be chosen from $\{ 1,2,3,4,5,6,7,8,9 \}$ . How many of these sets contain a $5$
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$ | [
"Let the three-element set be $\\{ a,b,c \\}$ and suppose that $a=5$\nWe need $b+c=10$ and $b\\neq c$ . This gives us four solutions, so there are $4$ sets with a $5$ also with the desired properties $\\rightarrow \\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_12 | B | 5,724 | There are twenty-four $4$ -digit numbers that use each of the four digits $2$ $4$ $5$ , and $7$ exactly once. Listed in numerical order from smallest to largest, the number in the $17\text{th}$ position in the list is
$\text{(A)}\ 4527 \qquad \text{(B)}\ 5724 \qquad \text{(C)}\ 5742 \qquad \text{(D)}\ 7245 \qquad \tex... | [
"For each choice of the thousands digit, there are $6$ numbers with that as the thousands digit. Thus, the six smallest are in the two thousands, the next six are in the four thousands, and then we need $5$ more numbers.\nWe can just list from here: $5247,5274,5427,5472,5724 \\rightarrow \\boxed{5724}$"
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_23 | D | 23 | There are two cards; one is red on both sides and the other is red on one side and blue on the other. The cards have the same probability (1/2) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is red, then the probability that the under-side is also red is
$\textbf{... | [
"There are three red faces, and two are on the card that is completely red, so our answer is $\\frac{2}{3}$ , which is $\\boxed{23}$"
] |
https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_21 | B | 392 | There are two natural ways to inscribe a square in a given isosceles right triangle.
If it is done as in Figure 1 below, then one finds that the area of the square is $441 \text{cm}^2$ .
What is the area (in $\text{cm}^2$ ) of the square inscribed in the same $\triangle ABC$ as shown in Figure 2 below?
[asy] draw((0,... | [
"We are given that the area of the inscribed square is $441$ , so the side length of that square is $21$ . Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$ , then the legs of the larger isosceles right triangle ( $BC$ and $AB$ ) are equal to $42$ \nWe now have that $3S=42\... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_10 | A | 16 | There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$ . What is the sum of those values of $a$
$\textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20$ | [
"quadratic equation has exactly one root if and only if it is a perfect square . So set\n$4x^2 + ax + 8x + 9 = (mx + n)^2$\n$4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2$\nTwo polynomials are equal only if their coefficients are equal, so we must have\n$m^2 = 4, n^2 = 9$\n$m = \\pm 2, n = \\pm 3$\n$a + 8= 2mn = \\pm 2... |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_25 | B | 9 | There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ . Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\text... | [
"Multiply out the $7!$ to get\n\\[5 \\cdot 6! = (3 \\cdot 4 \\cdots 7)a_2 + (4 \\cdots 7)a_3 + (5 \\cdot 6 \\cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .\\]\nBy Wilson's Theorem (or by straightforward division), $a_7 + 7(a_6 + 6a_5 + \\cdots) \\equiv 5 \\cdot 6! \\equiv -5 \\equiv 2 \\pmod{7}$ , so $a_7 = 2$ . Then we move $a... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_4 | null | 21 | There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ unique integers $a_k$ $1\le k\le r$ ) with each $a_k$ either $1$ or $- 1$ such that \[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\] Find $n_1 + n_2 + \cdots + n_r$ | [
"In base $3$ , we find that $\\overline{2008}_{10} = \\overline{2202101}_{3}$ . In other words,\nIn order to rewrite as a sum of perfect powers of $3$ , we can use the fact that $2 \\cdot 3^k = 3^{k+1} - 3^k$\nThe answer is $7+5+4+3+2+0 = \\boxed{021}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_I_Problems/Problem_4 | null | 80 | There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$ | [
"Completing the square $y^2 = x^2 + 84x + 2008 = (x+42)^2 + 244$ . Thus $244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)$ by difference of squares\nSince $244$ is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since $244 = 2^2 \\cdot 61$ , the factors mus... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10 | null | 944 | There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .) | [
"Define $\\left\\{ x \\right\\} = x - \\left\\lfloor x \\right\\rfloor$\nFirst, we bound $U$\nWe establish an upper bound of $U$ . We have \\begin{align*} U & \\leq \\sum_{n=1}^{2023} \\frac{n^2 - na}{5} \\\\ & = \\frac{1}{5} \\sum_{n=1}^{2023} n^2 - \\frac{a}{5} \\sum_{n=1}^{2023} n \\\\ & = \\frac{1012 \\cdot 202... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_21 | C | 137 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | [
"First, substitute $2^{17}$ with $x$ . \nThen, the given equation becomes $\\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+x^{14}...-x^1+x^0$ by sum of powers factorization.\nNow consider only $x^{16}-x^{15}$ . This equals $x^{15}(x-1)=x^{15} \\cdot (2^{17}-1)$ .\nNote that $2^{17}-1$ equals $2^{16}+2^{15}+...+1$ , by differen... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_21 | null | 137 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | [
"Notice that the only answer choices that are spaced one apart are $136$ and $137$ . It's likely that people will forget to include the final term so the answer is $\\boxed{137}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_19 | C | 137 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | [
"First, substitute $2^{17}$ with $x$ . \nThen, the given equation becomes $\\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+x^{14}...-x^1+x^0$ by sum of powers factorization.\nNow consider only $x^{16}-x^{15}$ . This equals $x^{15}(x-1)=x^{15} \\cdot (2^{17}-1)$ .\nNote that $2^{17}-1$ equals $2^{16}+2^{15}+...+1$ , by differen... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_19 | null | 137 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | [
"Notice that the only answer choices that are spaced one apart are $136$ and $137$ . It's likely that people will forget to include the final term so the answer is $\\boxed{137}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_2 | null | 107 | There is a $40\%$ chance of rain on Saturday and a $30\%$ chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positiv... | [
"Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\\dfrac{4}{5}x+\\dfrac{2}{5}2x = \\dfrac{3}{10} \\implies \\dfrac{7}{5}x=\\dfrac{3}{10}$ $\\implies x=\\dfrac{3}{14}$ . Therefore, the probability that it doesn't rain on either day is $\\left(1-\\dfrac{3}{14}\... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_9 | null | 902 | There is a collection of $25$ indistinguishable white chips and $25$ indistinguishable black chips. Find the number of ways to place some of these chips in the $25$ unit cells of a $5\times5$ grid such that: | [
"The problem says \"some\", so not all cells must be occupied.\nWe start by doing casework on the column on the left. There can be 5,4,3,2, or 1 black chip. The same goes for white chips, so we will multiply by 2 at the end. There is $1$ way to select $5$ cells with black chips. Because of the 2nd condition, there ... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_2 | null | 697 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$ | [
"Let $z = a + 164i$\nThen \\[\\frac {a + 164i}{a + 164i + n} = 4i\\] and \\[a + 164i = \\left (4i \\right ) \\left (a + n + 164i \\right ) = 4i \\left (a + n \\right ) - 656.\\]\nBy comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,\nwe conclude that \\[a = -656.\\]\... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_23 | B | 6 | There is a list of seven numbers. The average of the first four numbers is $5$ , and the average of the last four numbers is $8$ . If the average of all seven numbers is $6\frac{4}{7}$ , then the number common to both sets of four numbers is
$\text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qqu... | [
"Remember that if a list of $n$ numbers has an average of $k$ , then the sum $S$ of all the numbers on the list is $S = nk$\nSo if the average of the first $4$ numbers is $5$ , then the first four numbers total $4 \\cdot 5 = 20$\nIf the average of the last $4$ numbers is $8$ , then the last four numbers total $4 \\... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_13 | null | 220 | There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ | [
"Because $0 < x < 1$ , we have \\begin{align*} P \\left( x \\right) & = \\sum_{a=0}^6 \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\binom{6}{a} x^{2310a} \\left( - 1 \\right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\\\ & = \\sum_{a=0}^6 \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_6 | C | 10 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | [
"\\[\\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \\cdot n! \\\\ \\Rightarrow \\ &n![n+1 + (n+2)(n+1)] = 440 \\cdot n! \\\\ \\Rightarrow \\ &n + 1 + n^2 + 3n + 2 = 440 \\\\ \\Rightarrow \\ &n^2 + 4n - 437 = 0\\end{split}\\]\nSolving by the quadratic formula, $n = \\frac{-4\\pm \\sqrt{16+437\\cdot4}}{2} = \\frac{-4\\... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | C | 10 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | [
"\\[\\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \\cdot n! \\\\ \\Rightarrow \\ &n![n+1 + (n+2)(n+1)] = 440 \\cdot n! \\\\ \\Rightarrow \\ &n + 1 + n^2 + 3n + 2 = 440 \\\\ \\Rightarrow \\ &n^2 + 4n - 437 = 0\\end{split}\\]\nSolving by the quadratic formula, $n = \\frac{-4\\pm \\sqrt{16+437\\cdot4}}{2} = \\frac{-4\\... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 | null | 112 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [
"Define $a$ to be $\\log_{20x} (22x) = \\log_{2x} (202x)$ , what we are looking for. Then, by the definition of the logarithm, \\[\\begin{cases} (20x)^{a} &= 22x \\\\ (2x)^{a} &= 202x. \\end{cases}\\] Dividing the first equation by the second equation gives us $10^a = \\frac{11}{101}$ , so by the definition of lo... |
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_7 | null | 650 | There is a set of 1000 switches, each of which has four positions, called $A, B, C$ , and $D$ . When the position of any switch changes, it is only from $A$ to $B$ , from $B$ to $C$ , from $C$ to $D$ , or from $D$ to $A$ . Initially each switch is in position $A$ . The switches are labeled with the 1000 different in... | [
"For each $i$ th switch (designated by $x_{i},y_{i},z_{i}$ ), it advances itself only one time at the $i$ th step; thereafter, only a switch with larger $x_{j},y_{j},z_{j}$ values will advance the $i$ th switch by one step provided $d_{i}= 2^{x_{i}}3^{y_{i}}5^{z_{i}}$ divides $d_{j}= 2^{x_{j}}3^{y_{j}}5^{z_{j}}$ . ... |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_14 | D | 7 | There is a set of five positive integers whose average (mean) is 5, whose median is 5, and whose only mode is 8. What is the difference between the largest and smallest integers in the set?
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | [
"When these numbers are ordered in ascending order, 5, the median, falls right in the middle, which is the third integer from the left. Since there is a unique mode of 8, both integers to the right of 5 must be 8s. Since the mean is 5, the sum of the integers is 25, which means the 2 leftmost integers have to sum t... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24 | B | 9 | There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf... | [
"The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value.\nDerive $f(x)$ so that the acceleration $f''(x)=0$ . Using the power rule, \\begin{align*} f(x) &= x^3-ax^2+bx-a \\\\ f’(x) &= 3x^2-2ax+b \\\\ f’’(x) &= 6x-2a \\end{align*} So $x=\\frac{a}{3}... |
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