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/1997_AIME_Problems/Problem_2 | null | 125 | The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles , of which $s$ are squares . The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | [
"To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\\choose 2} = 36$ . Similarly, there are ${9\\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles.\nFor $s$ , there are $8^2$ unit squares $7^2$ of the $2\\times2$ sq... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_14 | C | 4 | The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers $1,2,3$ . Then $A+B=$
\[\begin{tabular}{|c|c|c|}\hline 1 & &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad ... | [
"The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.\n\\[\\begin{tabular}{|c|c|c|}\\hline 1 & 3 &\\\\ \\hline & 2 & A\\\\ \\hline & & B\\\\ \\hline\\end{tabular}\\]\nThe last square in the top row cannot be either 1 or 3, so it must be 2.\n\\[\\begin{tabular}{|c|c|c|}\\hline 1 &... |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_1 | null | 7 | The number
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | [
"$\\frac 2{\\log_4{2000^6}} + \\frac 3{\\log_5{2000^6}}$\n$=\\frac{\\log_4{16}}{\\log_4{2000^6}}+\\frac{\\log_5{125}}{\\log_5{2000^6}}$\n$=\\frac{\\log{16}}{\\log{2000^6}}+\\frac{\\log{125}}{\\log{2000^6}}$\n$=\\frac{\\log{2000}}{\\log{2000^6}}$\n$=\\frac{\\log{2000}}{6\\log{2000}}$\n$=\\frac{1}{6}$\nTherefore, $m+... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_18 | D | 46 | The number $2013$ has the property that its units digit is the sum of its other digits, that is $2+0+1=3$ . How many integers less than $2013$ but greater than $1000$ have this property?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58$ | [
"We take cases on the thousands digit, which must be either $1$ or $2$ :\nIf the number is of the form $\\overline{1bcd},$ where $b, c, d$ are digits, then we must have $d = 1 + b + c.$ Since $d \\le 9,$ we must have $b + c \\le 9 - 1 = 8.$ By casework on the value of $b$ , we find that there are $1 + 2 + \\dots + ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_20 | B | 2 | The number $2013$ is expressed in the form
where $a_1 \ge a_2 \ge \cdots \ge a_m$ and $b_1 \ge b_2 \ge \cdots \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | [
"The prime factorization of $2013$ is $61\\cdot11\\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_15 | B | 2 | The number $2013$ is expressed in the form
where $a_1 \ge a_2 \ge \cdots \ge a_m$ and $b_1 \ge b_2 \ge \cdots \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | [
"The prime factorization of $2013$ is $61\\cdot11\\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and ... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_23 | C | 1,024 | The number $2017$ is prime. Let $S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}$ . What is the remainder when $S$ is divided by $2017?$
$\textbf{(A) }32\qquad \textbf{(B) }684\qquad \textbf{(C) }1024\qquad \textbf{(D) }1576\qquad \textbf{(E) }2016\qquad$ | [
"Note that $2014\\equiv -3 \\mod2017$ . We have for $k\\ge1$ \\[\\dbinom{2014}{k}\\equiv \\frac{(-3)(-4)(-5)....(-2-k)}{k!}\\mod 2017\\] \\[\\equiv (-1)^k\\dbinom{k+2}{k} \\mod 2017\\] Therefore \\[\\sum \\limits_{k=0}^{62} \\dbinom{2014}{k}\\equiv \\sum \\limits_{k=0}^{62}(-1)^k\\dbinom{k+2}{2} \\mod 2017\\] This... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_14 | B | 14 | The number $25^{64}\cdot 64^{25}$ is the square of a positive integer $N$ . In decimal representation, the sum of the digits of $N$ is
$\mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 35$ | [
"Taking the root, we get $N=\\sqrt{25^{64}\\cdot 64^{25}}=5^{64}\\cdot 8^{25}$\nNow, we have $N=5^{64}\\cdot 8^{25}=5^{64}\\cdot (2^{3})^{25}=5^{64}\\cdot 2^{75}$\nCombining the $2$ 's and $5$ 's gives us $(2\\cdot 5)^{64}\\cdot 2^{(75-64)}=(2\\cdot 5)^{64}\\cdot 2^{11}=10^{64}\\cdot 2^{11}$\nThis is the number $20... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_11 | C | 17 | The number $64$ has the property that it is divisible by its unit digit. How many whole numbers between 10 and 50 have this property?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$ | [
"Casework by the units digit $u$ will help organize the answer.\n$u=0$ gives no solutions, since no real numbers are divisible by $0$\n$u=1$ has $4$ solutions, since all numbers are divisible by $1$\n$u=2$ has $4$ solutions, since every number ending in $2$ is even (ie divisible by $2$ ).\n$u=3$ has $1$ solution: $... |
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_22 | A | 162 | The number $6545$ can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
$\text{(A)}\ 162 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 173 \qquad \text{(D)}\ 174 \qquad \text{(E)}\ 222$ | [
"The prime factorization of $6545$ is $5\\cdot7\\cdot11\\cdot17 =385\\cdot17$ , which contains a three digit number, but we want 6545 to be expressed as ab x cd. Now we do trial and error: \\[5\\cdot7=35 \\text{, } 11\\cdot17=187 \\text{ X}\\] \\[5\\cdot11=55 \\text{, } 7\\cdot17=119 \\text{ X}\\] \\[5\\cdot... |
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_35 | D | 3 | The number $695$ is to be written with a factorial base of numeration, that is, $695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!$ where $a_1, a_2, a_3 ... a_n$ are integers such that $0 \le a_k \le k,$ and $n!$ means $n(n-1)(n-2)...2 \times 1$ . Find $a_4$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}... | [
"This problem can be approached similarly to other base number problems.\nSince $120 < 695 < 720$ , divide $695$ by $120$ . The quotient is $5$ and the remainder is $95$ , so rewrite the number as \\[695 = 5 \\cdot 120 + 95\\] Similarly, dividing $95$ by $24$ results in a quotient of $3$ and a remainder of $23$ , ... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_5 | E | 7 | The number $N$ is a two-digit number.
• When $N$ is divided by $9$ , the remainder is $1$
• When $N$ is divided by $10$ , the remainder is $3$
What is the remainder when $N$ is divided by $11$
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$ | [
"From the second bullet point, we know that the second digit must be $3$ , for a number divisible by $10$ ends in zero. Since there is a remainder of $1$ when $N$ is divided by $9$ , the multiple of $9$ must end in a $2$ for it to have the desired remainder $\\pmod {10}.$ We now look for this one:\n$9(1)=9\\\\ 9(2)... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_5 | null | 936 | The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers . Find $abc$ | [
"We begin by equating the two expressions:\n\\[a\\sqrt{2}+b\\sqrt{3}+c\\sqrt{5} = \\sqrt{104\\sqrt{6}+468\\sqrt{10}+144\\sqrt{15}+2006}\\]\nSquaring both sides yields:\n\\[2ab\\sqrt{6} + 2ac\\sqrt{10} + 2bc\\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\\sqrt{6}+468\\sqrt{10}+144\\sqrt{15}+2006\\]\nSince $a$ $b$ , and $c$ ar... |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_17 | B | 10 | The number $\text{N}$ is between $9$ and $17$ . The average of $6$ $10$ , and $\text{N}$ could be
$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$ | [
"We know that $9<N<17$ and we wish to bound $\\frac{6+10+N}{3}=\\frac{16+N}{3}$\nFrom what we know, we can deduce that $25<N+16<33$ , and thus \\[8.\\overline{3}<\\frac{N+16}{3}<11\\]\nThe only answer choice that falls in this range is choice $\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_25 | C | 929 | The number $a=\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying \[\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\] is $420$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfl... | [
"Let $w=\\lfloor x \\rfloor$ and $f=\\{x\\}$ denote the whole part and the fractional part of $x,$ respectively, for which $0\\leq f<1$ and $x=w+f.$\nWe rewrite the given equation as \\[w\\cdot f=a\\cdot(w+f)^2. \\hspace{38.75mm}(1)\\] Since $a\\cdot(w+f)^2\\geq0,$ it follows that $w\\cdot f\\geq0,$ from which $w\\... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_I_Problems/Problem_2 | null | 925 | The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$ , can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$ , and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$ , ... | [
"We have these equations: $196a+14b+c=225a+15c+b=222a+37c$ .\nTaking the last two we get $3a+b=22c$ . Because $c \\neq 0$ otherwise $a \\ngtr 0$ , and $a \\leq 5$ $c=1$\nThen we know $3a+b=22$ .\nTaking the first two equations we see that $29a+14c=13b$ . Combining the two gives $a=4, b=10, c=1$ . Then we see that $... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_5 | null | 417 | The number $r$ can be expressed as a four-place decimal $0.abcd,$ where $a, b, c,$ and $d$ represent digits , any of which could be zero. It is desired to approximate $r$ by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to $r$ is $\frac 27.$ What is the number of p... | [
"The nearest fractions to $\\frac 27$ with numerator $1$ are $\\frac 13, \\frac 14$ ; and with numerator $2$ are $\\frac 26, \\frac 28 = \\frac 13, \\frac 14$ anyway. For $\\frac 27$ to be the best approximation for $r$ , the decimal must be closer to $\\frac 27 \\approx .28571$ than to $\\frac 13 \\approx .33333$ ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_6 | C | 4 | The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5$ . What is the missing number in the top row?
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(... | [
"Let the value in the empty box in the middle row be $x$ , and the value in the empty box in the top row be $y$ $y$ is the answer we're looking for.\n\nFrom the diagram, $600 = 30x$ , making $x = 20$\n\nIt follows that $20 = 5y$ , so $y = \\boxed{4}$",
"Another way to do this problem is to realize what makes up t... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_23 | A | 2,900 | The number of $x$ -intercepts on the graph of $y=\sin(1/x)$ in the interval $(0.0001,0.001)$ is closest to
$\mathrm{(A)}\ 2900 \qquad\mathrm{(B)}\ 3000 \qquad\mathrm{(C)}\ 3100 \qquad\mathrm{(D)}\ 3200 \qquad\mathrm{(E)}\ 3300$ | [
"The function $f(x) = \\sin x$ has roots in the form of $\\pi n$ for all integers $n$ . Therefore, we want $\\frac{1}{x} = \\pi n$ on $\\frac{1}{10000} \\le x \\le \\frac{1}{1000}$ , so $1000 \\le \\frac 1x = \\pi n \\le 10000$ . There are $\\frac{10000-1000}{\\pi} \\approx \\boxed{2900}$ solutions for $n$ on this ... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_33 | D | 36 | The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:
$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$ | [
"It must be assumed that the pipes have an equal height.\nWe can represent the amount of water carried per unit time by cross sectional area.\nCross sectional of Pipe with diameter $6 in$ \\[\\pi r^2 = \\pi \\cdot 3^2 = 9\\pi\\]\nCross sectional area of pipe with diameter $1 in$\n\\[\\pi r^2 = \\pi \\cdot 0.5 \\cdo... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45 | A | 4,850 | The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$ | [
"Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be $\\binom{100}{2}=4950$ . However this also counts the 100 sides of the polygon, so the actual answer is $4950-100=\\boxed{4850}$",
"The formula for the number of d... |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_9 | D | 28 | The number of digits in $4^{16}5^{25}$ (when written in the usual base $10$ form) is
$\mathrm{(A) \ }31 \qquad \mathrm{(B) \ }30 \qquad \mathrm{(C) \ } 29 \qquad \mathrm{(D) \ }28 \qquad \mathrm{(E) \ } 27$ | [
"We can rewrite this as $2^{32}5^{25}$ . We can also combine some of the factors to introduce factors of $10$ , whose digit count is simple to evaluate because it simply adds $0$ s. Thus, we have $2^{32}5^{25}=2^72^{25}5^{25}=2^710^{25}$ . We can see that this final number is $2^7$ with $25$ $0$ s annexed onto it. ... |
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_1 | B | 7 | The number of distinct lines representing the altitudes, medians, and interior angle bisectors of a triangle that is isosceles, but not equilateral, is:
$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 3$ | [
"\nAs shown in the diagram above, all nine altitudes, medians, and interior angle bisectors are distinct, except for the three coinciding lines from the vertex opposite to the base. Thusly, there are $7$ distinct lines, so our answer is $\\boxed{7}$ , and we are done."
] |
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_6 | E | 4 | The number of distinct pairs $(x,y)$ of real numbers satisfying both of the following equations:
\[x=x^2+y^2 \ \ y=2xy\] is
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad \textbf{(E) }4$ | [
"If $x=x^2+y^2$ and $y=2xy$ , then we can break this into two cases.\nCase 1: $y = 0$\nIf $y = 0$ , then $x = x^2$ and $0 = 0$\nTherefore, $x = 0$ or $x = 1$\nThis yields 2 solutions\nCase 2: $x = \\frac{1}{2}$\nIf $x = \\frac{1}{2}$ , this means that $y = y$ , and $\\frac{1}{2} = \\frac{1}{4} + y^2$\nBecause y can... |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_28 | C | 3 | The number of distinct pairs of integers $(x, y)$ such that $0<x<y$ and $\sqrt{1984}=\sqrt{x}+\sqrt{y}$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ }4 \qquad \mathrm{(E) \ } 7$ | [
"We can simplify $\\sqrt{1984}$ to $8\\sqrt{31}$ . Therefore, the only solutions are $a\\sqrt{31}+b\\sqrt{31}$ such that $a+b=8$ and $0<a<b$ . The only solutions to this are $a=1, b=7; a=2, b=6; a=3, b=5$ . Each of these gives distinct pairs of $(x, y)$ , so there are $3$ pairs, $\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_14 | B | 84 | The number of geese in a flock increases so that the difference between the populations in year $n+2$ and year $n$ is directly proportional to the population in year $n+1$ . If the populations in the years $1994$ $1995$ , and $1997$ were $39$ $60$ , and $123$ , respectively, then the population in $1996$ was
$\textbf{(... | [
"Let $x$ be the population in $1996$ , and let $k$ be the constant of proportionality.\nIf $n=1994$ , then the difference in population between $1996$ and $1994$ is directly proportional to the population in $1995$\nTranslating this sentence, $(x - 39) = k(60)$\nSimilarly, letting $n=1995$ gives the sentence $(123 ... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_9 | D | 6 | The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ % alcohol to a lotion containing $30$ % alcohol is:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$ | [
"Say we add $N$ ounces of water to the shaving lotion. Since half of a $9$ ounce bottle of shaving lotion is alcohol, we know that we have $\\frac{9}{2}$ ounces of alcohol. We want $\\frac{9}{2}=0.3(9+N)$ (because we want the amount of alcohol, $\\frac{9}{2}$ , to be $30\\%$ , or $0.3$ , of the total amount of shav... |
https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_18 | B | 4 | The number of points common to the graphs of $(x-y+2)(3x+y-4)=0 \text{ and } (x+y-2)(2x-5y+7)=0$ is:
$\text{(A) } 2\quad \text{(B) } 4\quad \text{(C) } 6\quad \text{(D) } 16\quad \text{(E) } \infty$ | [
"By the Zero Product Property $x-y+2=0$ or $3x+y-4=0$ in the first equation and $x+y-2=0$ or $2x-5y+7=0$ in the second equation. Thus, from the first equation, $y = x+2$ or $y =-3x+4$ , and from the second equation, $y=-x+2$ or $y = \\frac{2}{5}x + \\frac{7}{5}$\nIf a point is common to the two graphs, then the po... |
https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_10 | C | 3 | The number of points equidistant from a circle and two parallel tangents to the circle is:
$\text{(A) } 0\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } \infty$ | [
" The distance between the two parallel tangents is the length of the circle's diameter, so the distance from a point that satisfies the conditions and the two tangents is the length of the circle's radius. From the diagram, there are $\\boxed{3}$ points that satisfies the conditions."
] |
https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_30 | B | 2 | The number of real solutions $(x,y,z,w)$ of the simultaneous equations $2y = x + \frac{17}{x}, 2z = y + \frac{17}{y}, 2w = z + \frac{17}{z}, 2x = w + \frac{17}{w}$ is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$ | [
"Consider the cases $x>0$ and $x<0$ , and also note that by AM-GM, for any positive number $a$ , we have $a+\\frac{17}{a} \\geq 2\\sqrt{17}$ , with equality only if $a = \\sqrt{17}$ . Thus, if $x>0$ , considering each equation in turn, we get that $y \\geq \\sqrt{17}, z \\geq \\sqrt{17}, w \\geq \\sqrt{17}$ , and f... |
https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_1 | B | 1 | The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:
$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 3 \qquad \textbf{(E) }\ \text{more than 4}$ | [
"Notice that $a^0=1, a>0$ . So $2^0=1$ . So $2x^2-7x+5=0$ . Evaluating the discriminant, we see that it is equal to $7^2-4*2*5=9$ . So this means that the equation has two real solutions. Therefore, select $\\boxed{1}$"
] |
https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_1 | C | 2 | The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:
$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 3 \qquad \textbf{(E) }\ \text{more than 4}$ | [
"Solution by e_power_pi_times_i\nTake the logarithm with a base of $2$ to both sides, resulting in the equation $2x^2-7x+5 = 0$ . Factoring results in $(2x-5)(x-1) = 0$ , so there are $\\boxed{2}$ real solutions."
] |
https://artofproblemsolving.com/wiki/index.php/1966_AHSME_Problems/Problem_12 | E | 3 | The number of real values of $x$ that satisfy the equation \[(2^{6x+3})(4^{3x+6})=8^{4x+5}\] is:
$\text{(A) zero} \qquad \text{(B) one} \qquad \text{(C) two} \qquad \text{(D) three} \qquad \text{(E) greater than 3}$ | [
"We know that $2^{6x+3}\\cdot4^{3x+6}=2^{6x+3}\\cdot(2^2)^{3x+6}=2^{6x+3}\\cdot2^{6x+12}=2^{12x+15}$ .\nWe also know that $8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}$ .\nThere are infinite solutions to the equation $2^{12x+15}=2^{12x+15}$ , so the answer is $\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_10 | C | 880 | The number of revolutions of a wheel, with fixed center and with an outside diameter of $6$ feet, required to cause a point on the rim to go one mile is:
$\textbf{(A)}\ 880 \qquad\textbf{(B)}\ \frac{440}{\pi} \qquad\textbf{(C)}\ \frac{880}{\pi} \qquad\textbf{(D)}\ 440\pi\qquad\textbf{(E)}\ \text{none of these}$ | [
"We know that the radius of the wheel is $3$ feet, so the total circumference of the wheel is $6\\pi$ feet. We also know that one mile is equivalent to $5280$ feet. It takes $\\frac{5280}{6\\pi}$ revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is $\\boxed{880}$"
] |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_43 | C | 3 | The number of scalene triangles having all sides of integral lengths, and perimeter less than $13$ is:
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 18$ | [
"We can write all possible triangles adding up to 12 or less \\[(2, 4, 5)=11\\] \\[(3, 4, 5)=12\\] \\[(2, 3, 4)=9\\]\nThis leaves $\\boxed{3}$ scalene triangles."
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_21 | B | 2 | The number of sets of two or more consecutive positive integers whose sum is 100 is
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | [
"If the first number of a group of $n$ consecutive numbers is $a$ , the $n^\\text{th}$ number is $a+n-1$ . We know that the sum of the group of numbers is $100$ , so \\[\\frac{n(2a+n-1)}{2} = 100\\] \\[2a+n-1=\\frac{200}{n}\\] \\[2a = 1-n + \\frac{200}{n}\\] We know that $n$ and $a$ are positive integers, so we ch... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_29 | D | 5 | The number of significant digits in the measurement of the side of a square whose computed area is $1.1025$ square inches to
the nearest ten-thousandth of a square inch is:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 1$ | [
"There are 5 significant digits, $1$ $1$ $0$ $2$ , and $5$ . The answer is $\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_31 | D | 127 | The number of solutions in positive integers of $2x+3y=763$ is:
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 254\qquad \textbf{(C)}\ 128 \qquad \textbf{(D)}\ 127 \qquad \textbf{(E)}\ 0$ | [
"Solving for $x$ in the equation yields $x =rfthe meaning of theta 0$ . Solving the inequality results in $y \\le 254 \\frac{1}{3}$ . From the two conditions, $y$ can be an odd number from $1$ to $253$ , so there are $127$ solutions where $x$ and $y$ are integers. The answer is $\\boxed{127}$",
"We can solve f... |
https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_4 | D | 8 | The number of solutions to $\{1,~2\}\subseteq~X~\subseteq~\{1,~2,~3,~4,~5\}$ , where $X$ is a subset of $\{1,~2,~3,~4,~5\}$ is
$\textbf{(A) }2\qquad \textbf{(B) }4\qquad \textbf{(C) }6\qquad \textbf{(D) }8\qquad \textbf{(E) }\text{None of these}$ | [
"$X$ has to contain $\\{1,~2\\}$ , so only $\\{3,~4,~5\\}$ matters. There are two choices for the elements; the element is either in $X$ or outside of $X$ . With this combinatorics in mind, the answer is simply $2^3=\\boxed{8}.$ ~lopkiloinm"
] |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_26 | E | 8 | The number of terms in an A.P. (Arithmetic Progression) is even. The sum of the odd and even-numbered terms are 24 and 30, respectively. If the last term exceeds the first by 10.5, the number of terms in the A.P. is
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf... | [
"Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference. That means the last term is $a+r(n-1)$\nWe can write an equation on the difference between the last and first term based on the conditions. \\[a+r(n-1)-a =10.5\\] \\[rn-r=10.5\\] Also, half of the terms add up to $24$ while t... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_16 | B | 5 | The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | [
"Use properties of exponents to move the squares outside the brackets use difference of squares.\n\\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\\]\nUsing the binomial theorem, we can see that the number of terms is $\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_20 | C | 2 | The number of the distinct solutions to the equation
$|x-|2x+1||=3$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$ | [
"We can create a tree of possibilities, progressively eliminating the absolute value signs by creating different cases.\n\nSo we have $4$ possible solutions: $-2, \\frac{2}{3}, 2,$ and $-\\frac{4}{3}$ . Checking for extraneous solutions, we find that the only ones that work are $2$ and $-\\frac{4}{3}$ , so there ar... |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_21 | C | 2 | The number of triples $(a, b, c)$ of positive integers which satisfy the simultaneous equations
$ab+bc=44$
$ac+bc=23$
is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$ | [
"We can factor the second equation to get $c(a+b)=23$ , so we see that $c$ must be a factor of $23$ , and since this is prime $c=1$ or $c=23$ . However, if $c=23$ , then $a+b=1$ , which is impossible for the field of positive integers. Therefore, $c=1$ for all possible solutions. Substituting this into the original... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_11 | C | 6.5 | The numbers $-2, 4, 6, 9$ and $12$ are rearranged according to these rules:
What is the average of the first and last numbers?
$\textbf{(A)}\ 3.5 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6.5 \qquad \textbf{(D)}\ 7.5 \qquad \textbf{(E)}\ 8$ | [
"From rule 1, the largest number, $12$ , can be second or third. From rule 2, because there are five places, the smallest number $-2$ can either be third or fourth. The median, $6$ can be second, third, or fourth. Because we know the middle three numbers, the first and last numbers are $4$ and $9$ , disregarding th... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_15 | null | 85 | The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ ar... | [
"It is helpful to consider the cube $ABCDEFGH$ , where the vertices of the cube represent the faces of the octahedron, and the edges of the cube represent adjacent octahedral faces. Each assignment of the numbers $1,2,3,4,5,6,7$ , and $8$ to the faces of the octahedron corresponds to a permutation of $ABCDEFGH$ , a... |
https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10 | null | 432 | The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there? | [
"Suppose that the two identical digits are both $1$ . Since the thousands digit must be $1$ , only one of the other three digits can be $1$ . This means the possible forms for the number are\nBecause the number must have exactly two identical digits, $x\\neq y$ $x\\neq1$ , and $y\\neq1$ . Hence, there are $3\\cdot9... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_9 | B | 1,011 | The numbers $16$ and $25$ are a pair of consecutive positive squares whose difference is $9$ . How many pairs of consecutive positive perfect squares have a difference of less than or equal to $2023$
$\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017$ | [
"Let x be the square root of the smaller of the two perfect squares. Then, $(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \\le 2023$ . Thus, $x \\le 1011$ . So there are $\\boxed{1011}$ numbers that satisfy the equation.",
"The smallest number that can be expressed as the difference of a pair of consecutive positive squares... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_2 | C | 30 | The numbers $3, 5, 7, a,$ and $b$ have an average (arithmetic mean) of $15$ . What is the average of $a$ and $b$
$\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60$ | [
"The arithmetic mean of the numbers $3, 5, 7, a,$ and $b$ is equal to $\\frac{3+5+7+a+b}{5}=\\frac{15+a+b}{5}=15$ . Solving for $a+b$ , we get $a+b=60$ . Dividing by $2$ to find the average of the two numbers $a$ and $b$ gives $\\frac{60}{2}=\\boxed{30}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_16 | D | 112 | The numbers $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$ | [
"Let $A = \\log(a)$ and $B = \\log(b)$\nThe first three terms of the arithmetic sequence are $3A + 7B$ $5A + 12B$ , and $8A + 15B$ , and the $12^\\text{th}$ term is $nB$\nThus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \\Rightarrow A = 2B$\nSince the first three terms in the sequence are $13B$ $22B$ , and $31B$ , the $... |
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_8 | A | 2 | The numbers $x,\,y,\,z$ are proportional to $2,\,3,\,5$ . The sum of $x, y$ , and $z$ is $100$ . The number y is given by the equation $y = ax - 10$ . Then a is:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac{3}{2}\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ \frac{5}{2}\qquad \textbf{(E)}\ 4$ | [
"In order to solve the problem, we first need to find each of the three variables. We can use the proportions the problem gives us to find the value of one part, and, by extension, the values of the variables (as $x$ would have $2$ parts, $y$ would have $3$ , and $z$ would have $5$ ). One part, after some algebra, ... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_4 | D | 3 | The numbers from $1$ to $49$ are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number $7.$ How many of these four numbers are prime? [asy] ... | [
"We fill out the grid, as shown below: From the four numbers that appear in the shaded squares, $\\boxed{3}$ of them are prime: $19,23,$ and $47.$",
"Note that given time constraint, it's better to only count from perfect squares (in pink), as shown below: From the four numbers that appear in the shaded squares... |
https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_13 | null | 401 | The numbers in the sequence $101$ $104$ $109$ $116$ $\ldots$ are of the form $a_n=100+n^2$ , where $n=1,2,3,\ldots$ For each $n$ , let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$ . Find the maximum value of $d_n$ as $n$ ranges through the positive integers | [
"If $(x,y)$ denotes the greatest common divisor of $x$ and $y$ , then we have $d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)$ . Now assuming that $d_n$ divides $100+n^2$ , it must divide $2n+1$ if it is going to divide the entire expression $100+n^2+2n+1$\nThus the equation turns into $d_n=(100+n^2,2n+1)$ . Now note tha... |
https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_1 | null | 220 | The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples ... | [
"In the arithmetic sequence, let $a$ be the first term and $d$ be the common difference, where $d>0.$ The sum of the first six terms is \\[a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.\\] We are given that \\begin{align*} 6a+15d &= 990, \\\\ 2a &= a+5d. \\end{align*} The second equation implies that $a=5d.$ Substit... |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_11 | E | 81 | The numbers on the faces of this cube are consecutive whole numbers. The sum of the two numbers on each of the three pairs of opposite faces are equal. The sum of the six numbers on this cube is
[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); label("$15$",(1... | [
"The only possibilities for the numbers are $11,12,13,14,15,16$ and $10,11,12,13,14,15$\nIn the second case, the common sum would be $(10+11+12+13+14+15)/3=25$ , so $11$ must be paired with $14$ , which\nit isn't.\nThus, the only possibility is the first case and the sum of the six numbers is $81\\rightarrow \\boxe... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_14 | A | 28 | The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
2024-AMC8-q14.png
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \t... | [
"We can execute Dijkstra's algorithm by hand to find the shortest path from $A$ to every other town, including $Z$ . This effectively proves that, assuming we execute the algorithm correctly, that we will have found the shortest distance. The distance estimates for each step of the algorithm (from $A$ to each node)... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_14 | null | 28 | The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
2024-AMC8-q14.png
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \t... | [
"We can simply see that path $A \\rightarrow X \\rightarrow M \\rightarrow Y \\rightarrow C \\rightarrow Z$ will give us the smallest value. Adding, $5+2+6+5+10 = \\boxed{28}$ . This is nice as it’s also the smallest value, solidifying our answer."
] |
https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_6 | null | 33 | The pages of a book are numbered $1_{}^{}$ through $n_{}^{}$ . When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of $1986_{}^{}$ . What was the number of the page that was added twice? | [
"Denote the page number as $x$ , with $x < n$ . The sum formula shows that $\\frac{n(n + 1)}{2} + x = 1986$ . Since $x$ cannot be very large, disregard it for now and solve $\\frac{n(n+1)}{2} = 1986$ . The positive root for $n \\approx \\sqrt{3972} \\approx 63$ . Quickly testing, we find that $63$ is too large, but... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_25 | B | 40 | The parabola $P$ has focus $(0,0)$ and goes through the points $(4,3)$ and $(-4,-3)$ . For how many points $(x,y)\in P$ with integer coordinates is it true that $|4x+3y|\leq 1000$
$\textbf{(A) }38\qquad \textbf{(B) }40\qquad \textbf{(C) }42\qquad \textbf{(D) }44\qquad \textbf{(E) }46\qquad$ | [
"The axis of $P$ is inclined at an angle $\\theta$ relative to the coordinate axis, where $\\tan\\theta = \\tfrac 34$ . We rotate the coordinate axis by angle $\\theta$ anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let $(\\widetilde{x}, \\widetilde{y})$ b... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_12 | null | 4 | The parabola $y=ax^2+bx+c$ has vertex $(p,p)$ and $y$ -intercept $(0,-p)$ , where $p\ne 0$ . What is $b$
$\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p$ | [
"A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p)^2+p$ . Because the $y$ -intercept is $(0,-p)$ and $p\\ne 0$ , it follows that $a=-2/p$ . Thus \\[y=-\\frac{2}{p}(x^2-2px+p^2)+p=-\\frac{2}{p}x^2+4x-p,\\] so $\\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_12 | B | 1.5 | The parabolas $y=ax^2 - 2$ and $y=4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12$ . What is $a+b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1.5\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2.5\qquad\textbf{(E)}\ 3$ | [
"Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by $4 - (-2)$ (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are $(2, 0), (-2, 0).... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_20 | D | 16 | The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qqu... | [
"Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\\left(\\frac{d-c}{a-b},\\frac{ad-bc}{a-b}\\right),\\left(\\frac{c-d}{a-b},\\frac{bc-ad}{a-b}\\right)$ . Because $72= 4\\cdot 18$ , we have that $4(c-d)\\left(\\frac{c-d}{a-b}\\right) = (c+d)\\left(\\frac{c+d}{a-b}\\right)$ or th... |
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_1 | B | 100 | The percent that $M$ is greater than $N$ is:
$(\mathrm{A})\ \frac{100(M-N)}{M} \qquad (\mathrm{B})\ \frac{100(M-N)}{N} \qquad (\mathrm{C})\ \frac{M-N}{N} \qquad (\mathrm{D})\ \frac{M-N}{N} \qquad (\mathrm{E})\ \frac{100(M+N)}{N}$ | [
"$M-N$ is the amount by which $M$ is greater than $N$ . We divide this by $N$ to get the percent by which $N$ is increased in the form of a decimal, and then multiply by $100$ to make it a percentage. Therefore, the answer is $\\boxed{100}$"
] |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_16 | E | 9 | The perimeter of one square is $3$ times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9$ | [
"Let $a$ be the sidelength of one square, and $b$ be the sidelength of the other, where $a>b$ . If the perimeter of one is $3$ times the other's, then $a=3b$ . The area of the larger square over the area of the smaller square is\n\\[\\frac{a^2}{b^2} = \\frac{(3b)^2}{b^2} = \\frac{9b^2}{b^2} = \\boxed{9}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_13 | C | 28 | The perimeter of the polygon shown is
$\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 48$
$\text{(E)}\ \text{cannot be determined from the information given}$ | [
"You might have not seen this coming but there is a very simple way to do this. If we try to make a rectangle out of this, we have to take out both of the lines that are taking out part of the rectangle we want to make. but now we see that to finish the rectangle, we have to use those same irregular lines!\n\nSo al... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_14 | null | 98 | The perimeter of triangle $APM$ is $152$ , and the angle $PAM$ is a right angle . A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$ . Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$ | [
"Let the circle intersect $\\overline{PM}$ at $B$ . Then note $\\triangle OPB$ and $\\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point . Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have \\[\\frac{19}{AM} = \\frac{152-... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_19 | C | 32 | The pie charts below indicate the percent of students who prefer golf, bowling, or tennis at East Junior High School and West Middle School. The total number of students at East is 2000 and at West, 2500. In the two schools combined, the percent of students who prefer tennis is
[asy] unitsize(18); draw(circle((0,0),4... | [
"In the first school, $2000 \\cdot 22\\% = 2000 \\cdot 0.22 = 440$ students prefer tennis.\nIn the second school, $2500 \\cdot 40\\% = 2500 \\cdot 0.40 = 1000$ students prefer tennis.\nIn total, $440 + 1000 = 1440$ students prefer tennis out of a total of $2000 + 2500 = 4500$ students\nThis means $\\frac{1440}{4500... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_12 | A | 13 | The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was $... | [
"For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a $3:2$ ratio. Therefore, assume they made $3x$ and $2x$ two- and three- point shots, respectively, and thus $3x+1$ free throws. The total number of points is \\[2 \\times (3x) + 3 \\times (2x) + 1 \\t... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_6 | A | 13 | The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was $... | [
"For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a $3:2$ ratio. Therefore, assume they made $3x$ and $2x$ two- and three- point shots, respectively, and thus $3x+1$ free throws. The total number of points is \\[2 \\times (3x) + 3 \\times (2x) + 1 \\t... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_5 | B | 5 | The point $(-1, -2)$ is rotated $270^{\circ}$ counterclockwise about the point $(3, 1)$ . What are the coordinates of its new position?
$\textbf{(A)}\ (-3, -4) \qquad \textbf{(B)}\ (0,5) \qquad \textbf{(C)}\ (2,-1) \qquad \textbf{(D)}\ (4,3) \qquad \textbf{(E)}\ (6,-3)$ | [
"$(-1,-2)$ is $4$ units west and $3$ units south of $(3,1)$ . Performing a counterclockwise rotation of $270^{\\circ}$ , which is equivalent to a clockwise rotation of $90^{\\circ}$ , the answer is $3$ units west and $4$ units north of $(3,1)$ , or $\\boxed{0,5}$",
"We write $(-1, -2)$ as $-1-2i.$ We'd like to ro... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_9 | E | 32 | The point $(-3,2)$ is rotated $90^\circ$ clockwise around the origin to point $B$ . Point $B$ is then reflected over the line $x=y$ to point $C$ . What are the coordinates of $C$
$\mathrm{(A)}\ (-3,-2) \qquad \mathrm{(B)}\ (-2,-3) \qquad \mathrm{(C)}\ (2,-3) \qquad \mathrm{(D)}\ (2,3) \qquad \mathrm{(E)}\ (3,2)$ | [
"The entire situation is in the picture below. The correct answer is $\\boxed{3,2}$"
] |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_4 | D | 50 | The point $O$ is the center of the circle circumscribed about $\triangle ABC,$ with $\angle BOC=120^\circ$ and $\angle AOB=140^\circ,$ as shown. What is the degree measure of $\angle ABC?$
$\textbf{(A) } 35 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 50 \qquad\textbf{(E) } 60$ | [
"Because all the central angles of a circle add up to $360^\\circ,$\n\\begin{align*} \\angle BOC + \\angle AOB + \\angle AOC &= 360\\\\ 120 + 140 + \\angle AOC &= 360\\\\ \\angle AOC &= 100. \\end{align*}\nTherefore, the measure of $\\text{arc}AC$ is also $100^\\circ.$ Since the measure of an inscribed angle is equ... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_9 | D | 7 | The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ... | [
"The final image of $P$ is $(-6,3)$ . We know the reflection rule for reflecting over $y=-x$ is $(x,y) \\rightarrow (-y, -x)$ . So before the reflection and after rotation the point is $(-3,6)$\nBy definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and ... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_5 | D | 7 | The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ... | [
"The final image of $P$ is $(-6,3)$ . We know the reflection rule for reflecting over $y=-x$ is $(x,y) \\rightarrow (-y, -x)$ . So before the reflection and after rotation the point is $(-3,6)$\nBy definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and ... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_3 | B | 10,001,988 | The point in the $xy$ -plane with coordinates $(1000, 2012)$ is reflected across the line $y=2000$ . What are the coordinates of the reflected point?
$\textbf{(A)}\ (998,2012)\qquad\textbf{(B)}\ (1000,1988)\qquad\textbf{(C)}\ (1000,2024)\qquad\textbf{(D)}\ (1000,4012)\qquad\textbf{(E)}\ (1012,2012)$ | [
"The line $y = 2000$ is a horizontal line located $12$ units beneath the point $(1000, 2012)$ . When a point is reflected about a horizontal line, only the $y$ - coordinate will change. The $x$ - coordinate remains the same. Since the $y$ -coordinate of the point is $12$ units above the line of reflection, the new ... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8 | null | 315 | The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ | [
"Consider the points on the complex plane . The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:\n\\[(a+11i)\\left(\\mathrm{cis}\\,60^{\\circ}\\right) = (a+11i)\\left(\\frac 12+\\frac{\\sqrt{3}i}2\\right)=b+37i.\\]\nEquating the real and imaginary parts, we have:\n\\begin{align*}b&=... |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_5 | A | 33 | The points $(6,12)$ and $(0,-6)$ are connected by a straight line. Another point on this line is:
$\textbf{(A) \ }(3,3) \qquad \textbf{(B) \ }(2,1) \qquad \textbf{(C) \ }(7,16) \qquad \textbf{(D) \ }(-1,-4) \qquad \textbf{(E) \ }(-3,-8)$ | [
"The slope of this line is $\\frac{y_2-y_1}{x_2-x_1}=\\frac{12+6}{6-0}=3$ . Hence, its equation is $y=3x-6$ . The only given point which satisfies these conditions is $\\boxed{3,3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_12 | null | 118 | The points $A$ $B$ and $C$ lie on the surface of a sphere with center $O$ and radius $20$ . It is given that $AB=13$ $BC=14$ $CA=15$ , and that the distance from $O$ to $\triangle ABC$ is $\frac{m\sqrt{n}}k$ , where $m$ $n$ , and $k$ are positive integers, $m$ and $k$ are relatively prime, and $n$ is not divisible by t... | [
"Let $D$ be the foot of the perpendicular from $O$ to the plane of $ABC$ . By the Pythagorean Theorem on triangles $\\triangle OAD$ $\\triangle OBD$ and $\\triangle OCD$ we get:\n\\[DA^2=DB^2=DC^2=20^2-OD^2\\]\nIt follows that $DA=DB=DC$ , so $D$ is the circumcenter of $\\triangle ABC$\nBy Heron's Formula the area ... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_10 | E | 6 | The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge -to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
2003amc10a10.gif
$\mathrm{(A) \... | [
"2003amc10a10solution.gif\nLet the squares be labeled $A$ $B$ $C$ , and $D$\nWhen the polygon is folded, the \"right\" edge of square $A$ becomes adjacent to the \"bottom edge\" of square $C$ , and the \"bottom\" edge of square $A$ becomes adjacent to the \"bottom\" edge of square $D$\nSo, any \"new\" square that i... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_13 | E | 6 | The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge -to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
2003amc10a10.gif
$\mathrm{(A) \... | [
"2003amc10a10solution.gif\nLet the squares be labeled $A$ $B$ $C$ , and $D$\nWhen the polygon is folded, the \"right\" edge of square $A$ becomes adjacent to the \"bottom edge\" of square $C$ , and the \"bottom\" edge of square $A$ becomes adjacent to the \"bottom\" edge of square $D$\nSo, any \"new\" square that i... |
https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | null | 816 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | [
"Using the geometric series formula, $1 - x + x^2 + \\cdots - x^{17} = \\frac {1 - x^{18}}{1 + x} = \\frac {1-x^{18}}{y}$ . Since $x = y - 1$ , this becomes $\\frac {1-(y - 1)^{18}}{y}$ . We want $a_2$ , which is the coefficient of the $y^3$ term in $-(y - 1)^{18}$ (because the $y$ in the denominator reduces the de... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_8 | null | 30 | The polynomial $P(x)$ is cubic . What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$ | [
"We can see that $Q_1$ and $Q_2$ must have a root in common for them to both be factors of the same cubic.\nLet this root be $a$\nWe then know that $a$ is a root of $Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$ , so $x = \\frac{-k}{5}$\nWe then know that $\\frac{-k}{5}$ is a root of $Q_{1}$... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_13 | null | 482 | The polynomial $P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17}$ has $34$ complex roots of the form $z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34,$ with $0 < a_1 \le a_2 \le a_3 \le \cdots \le a_{34} < 1$ and $r_k>0.$ Given that $a_1 + a_2 + a_3 + a_4 + a_5 = m/n,$ where $m$ and $n$ are relatively prime positi... | [
"We see that the expression for the polynomial $P$ is very difficult to work with directly, but there is one obvious transformation to make: sum the geometric series\n\\begin{align*} P(x) &= \\left(\\frac{x^{18} - 1}{x - 1}\\right)^2 - x^{17} = \\frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17}\\\\ &= \\frac{x^{36... |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_8 | null | 53 | The polynomial $f(z)=az^{2018}+bz^{2017}+cz^{2016}$ has real coefficients not exceeding $2019$ , and $f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i$ . Find the remainder when $f(1)$ is divided by $1000$ | [
"We have $\\frac{1+\\sqrt{3}i}{2} = \\omega$ where $\\omega = e^{\\frac{i\\pi}{3}}$ is a primitive 6th root of unity. Then we have\n\\begin{align*} f(\\omega) &= a\\omega^{2018} + b\\omega^{2017} + c\\omega^{2016}\\\\ &= a\\omega^2 + b\\omega + c \\end{align*}\nWe wish to find $f(1) = a+b+c$ . We first look at the ... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_23 | C | 250,500 | The polynomial $x^3 - 2004 x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $n$ are possible?
$\mathrm{(A)}\ 250,\!000 \qquad\mathrm{(B)}\ 250,\!250 \qquad\mathrm{(C)}\ 250,\!500 \qquad\mathrm{(D)}\ 250,... | [
"Let the roots be $r,s,r + s$ , and let $t = rs$ . Then\nand by matching coefficients, $2(r + s) = 2004 \\Longrightarrow r + s = 1002$ . Then our polynomial looks like \\[x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0\\] and we need the number of possible products $t = rs = r(1002 - r)$ . Because $m=t+1002^2$ is an inte... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_21 | A | 78 | The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$
$\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118$ | [
"By Vieta's Formulas , we know that $a$ is the sum of the three roots of the polynomial $x^3-ax^2+bx-2010$ . Again Vieta's Formulas tell us that $2010$ is the product of the three integer roots. Also, $2010$ factors into $2\\cdot3\\cdot5\\cdot67$ . But, since the polynomial has only three roots, two of the four pri... |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_38 | B | 7 | The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$ , the population was one more than a perfect square. Now, with an additional increase of $100$ , the population is again a perfect square.
The original population is a multiple of:
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 7\... | [
"Let $a^2$ $=$ original population count, $b^2+1$ $=$ the second population count, and $c^2$ $=$ the third population count \nWe first see that $a^2 + 100 = b^2 + 1$ or $99$ $=$ $b^2-a^2$ .\nWe then factor the right side getting $99$ $=$ $(b-a)(b+a)$ . \nSince we can only have an nonnegative integral population, cl... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_9 | B | 160 | The population of a small town is $480$ . The graph indicates the number of females and males in the town, but the vertical scale-values are omitted. How many males live in the town?
[asy] draw((0,13)--(0,0)--(20,0)); draw((3,0)--(3,10)--(8,10)--(8,0)); draw((3,5)--(8,5)); draw((11,0)--(11,5)--(16,5)--(16,0)); labe... | [
"The graph show that the ratio of men to total population is $\\frac{1}{3}$ , so the total number of men is $\\frac{1}{3} \\times 480= \\boxed{160}$"
] |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_9 | null | 160 | The population of a small town is $480$ . The graph indicates the number of females and males in the town, but the vertical scale-values are omitted. How many males live in the town?
[asy] draw((0,13)--(0,0)--(20,0)); draw((3,0)--(3,10)--(8,10)--(8,0)); draw((3,5)--(8,5)); draw((11,0)--(11,5)--(16,5)--(16,0)); labe... | [
"The graph shows $3$ equal squares, each with value $x$ . So $3x = 480$ , so $x = \\boxed{160}$"
] |
https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_8 | E | 500,000 | The population of the United States in $1980$ was $226,504,825$ . The area of the country is $3,615,122$ square miles. There are $(5280)^{2}$ square feet in one square mile. Which number below best approximates the average number of square feet per person?
$\textbf{(A)}\ 5,000\qquad \textbf{(B)}\ 10,000\qquad \textbf{(... | [
"With about $230$ million people and under $4$ million square miles, there are about $60$ people per square mile. Since a square mile is about $(5000 \\ \\text{ft})^{2} = 25$ million square feet, that gives approximately $\\frac{25}{60}$ of a million square feet per person. $\\frac{25}{60}$ is approximately half, s... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_13 | E | 16 | The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \q... | [
"Let the two primes be $a$ and $b$ . We would have $a-b=2$ and $a^{3}-b^{3}=31106$ . Using difference of cubes, we would have $(a-b)(a^{2}+ab+b^{2})=31106$ . Since we know $a-b$ is equal to $2$ $(a-b)(a^{2}+ab+b^{2})$ would become $2(a^{2}+ab+b^{2})=31106$ . Simplifying more, we would get $a^{2}+ab+b^{2}=15553$\nNo... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_13 | null | 16 | The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \q... | [
"Let the two primes be $x + 1$ and $x - 1$ . Then, plugging it into the second condition, we get $(x + 1)^3 - (x - 1)^3 = 31106.$ Expanding the left side, \\[6x^2 + 2 = 31106 \\implies x^2 = 5184.\\] Taking the square root of both sides, we get that $x = 72$ and the larger prime is $73$ . The smallest prime larger ... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8 | null | 937 | The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$ | [
"We have that $N^2 - N = N(N - 1)\\equiv 0\\mod{10000}$\nThus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$ . Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$ , which is impossible for a number that is divisible by either $2$ or ... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_10 | null | 503 | The probability that a set of three distinct vertices chosen at random from among the vertices of a regular n-gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$ | [
"Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater).\nBreak up the problem into two ca... |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_4 | C | 74 | The product $(1.8)(40.3+.07)$ is closest to
$\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737$ | [
"Approximating $40.37$ instead of $1.8$ is more effective because larger numbers are less affected by absolute changes (e.g $1001$ is much closer relatively to $1000$ than $2$ is to $1$ ). $74$ is the closest to $72$ , so the answer is $\\boxed{74}$"
] |
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_2 | C | 12 | The product $8\times .25\times 2\times .125 =$
$\text{(A)}\ \frac18 \qquad \text{(B)}\ \frac14 \qquad \text{(C)}\ \frac12 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | [
"Converting the decimals to fractions , this is \\begin{align*} 8\\times \\frac{1}{4} \\times 2\\times \\frac{1}{8} &= \\frac{8\\times 2}{4\\times 8} \\\\ &= \\frac{16}{32} \\\\ &= \\frac{1}{2} \\rightarrow \\boxed{12}"
] |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_1 | null | 336 | The product $N$ of three positive integers is $6$ times their sum , and one of the integers is the sum of the other two. Find the sum of all possible values of $N$ | [
"Let the three integers be $a, b, c$ $N = abc = 6(a + b + c)$ and $c = a + b$ . Then $N = ab(a + b) = 6(a + b + a + b) = 12(a + b)$ . Since $a$ and $b$ are positive, $ab = 12$ so $\\{a, b\\}$ is one of $\\{1, 12\\}, \\{2, 6\\}, \\{3, 4\\}$ so $a + b$ is one of $13, 8, 7$ so the sum of all possible values of $N$ i... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_8 | D | 150 | The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?
$\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\te... | [
"Let the lengths of the two congruent sides of the triangle be $x$ , then the product desired is $x^2$\nNotice that the product of the base and twice the height is $4$ times the area of the triangle.\nSet the vertex angle to be $a$ , we derive the equation:\n$x^2=4\\left(\\frac{1}{2}x^2\\sin(a)\\right)$\n$\\sin(a)=... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_18 | D | 8 | The product of the two $99$ -digit numbers
$303,030,303,...,030,303$ and $505,050,505,...,050,505$
has thousands digit $A$ and units digit $B$ . What is the sum of $A$ and $B$
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10$ | [
"We can first make a small example to find out $A$ and $B$ . So,\n$303\\times505=153015$\nThe ones digit plus thousands digit is $5+3=8$\nNote that the ones and thousands digits are, added together, $8$ . (and so on...) So the answer is $\\boxed{8}$ This is a direct multiplication way."
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_11 | B | 77 | The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares
$\mathrm{(A)}\ 50 \qquad\mathrm{(B)}\ 77 \qquad\mathrm{(C)}\ 110 \qquad\mathrm{(D)}\ 149 \qquad\mathrm{(E)}\ 194$ | [
"Let the three consecutive positive integers be $a-1$ $a$ , and $a+1$ . Since the mean is $a$ , the sum of the integers is $3a$ . So $8$ times the sum is just $24a$ . With this, we now know that $a(a-1)(a+1)=24a\\Rightarrow(a-1)(a+1)=24$ $24=4\\times6$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=\\bo... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_7 | B | 77 | The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares
$\mathrm{(A)}\ 50 \qquad\mathrm{(B)}\ 77 \qquad\mathrm{(C)}\ 110 \qquad\mathrm{(D)}\ 149 \qquad\mathrm{(E)}\ 194$ | [
"Let the three consecutive positive integers be $a-1$ $a$ , and $a+1$ . Since the mean is $a$ , the sum of the integers is $3a$ . So $8$ times the sum is just $24a$ . With this, we now know that $a(a-1)(a+1)=24a\\Rightarrow(a-1)(a+1)=24$ $24=4\\times6$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=\\bo... |
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