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/1953_AHSME_Problems/Problem_39 | A | 1 | The product, $\log_a b \cdot \log_b a$ is equal to:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$ | [
"\\[a^x=b\\] \\[b^y=a\\] \\[{a^x}^y=a\\] \\[xy=1\\] \\[\\log_a b\\log_b a=1\\] As a result, the answer should be $\\boxed{1}$",
"Apply the change of base formula to $\\log_a b$ and $\\log_b a$ . For simplicity, let us convert to base-10 log.\nBy change of base, the expression becomes $\\frac{\\log b}{\\log a} * \... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_16 | D | 8 | The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$ , and none of $m,n,$ and $p$ is zero. What is the value of $n/p$
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}$ | [
"Let $x^2 + px + m = 0$ have roots $a$ and $b$ . Then\n\\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\\]\nso $p = -(a+b)$ and $m = ab$ . Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so\n\\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\\]\nand $m = -2(a+b)$ and $n = 4ab$ . Thus $\\frac{n}{p} = \\frac{4... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_12 | D | 8 | The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$ , and none of $m,n,$ and $p$ is zero. What is the value of $n/p$
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}$ | [
"Let $x^2 + px + m = 0$ have roots $a$ and $b$ . Then\n\\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\\]\nso $p = -(a+b)$ and $m = ab$ . Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so\n\\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\\]\nand $m = -2(a+b)$ and $n = 4ab$ . Thus $\\frac{n}{p} = \\frac{4... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_31 | A | 20 | The rails on a railroad are $30$ feet long. As the train passes over the point where the rails are joined, there is an audible click.
The speed of the train in miles per hour is approximately the number of clicks heard in:
$\textbf{(A)}\ 20\text{ seconds} \qquad \textbf{(B)}\ 2\text{ minutes} \qquad \textbf{(C)}\ 1\fr... | [
"We assume that the clicks are heard at the head of the train. Then if the train's speed in miles per hour is $x$ , we can convert it to clicks per minute: \\[\\frac{x\\text{ mile}}{\\text{hr}}\\cdot\\left(\\frac{1\\text{ hr}}{60\\text{ min}}\\right)\\cdot\\left(\\frac{5280\\text{ ft}}{1\\text{ mile}}\\right)\\cdot... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_1 | D | 5 | The ratio $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ is closest to which of the following numbers?
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$ | [
"We factor $\\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ as $\\frac{10^{2000}(1+100)}{10^{2001}(1+1)}=\\frac{101}{20}$ . As $\\frac{101}{20}=5.05$ , our answer is $\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_3 | B | 18 | The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 50$ | [
"Let $m$ be Mary's age. Then $\\frac{m}{30}=\\frac{3}{5}$ . Solving for $m$ , we obtain $m=\\boxed{18}.$",
"We can see this is a combined ratio of $8$ $(5+3)$ . We can equalize by doing $30\\div5=6$ , and $6\\cdot3=\\boxed{18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\\cdot8=30+18$ .... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_3 | B | 18 | The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 50$ | [
"Let $m$ be Mary's age. Then $\\frac{m}{30}=\\frac{3}{5}$ . Solving for $m$ , we obtain $m=\\boxed{18}.$",
"We can see this is a combined ratio of $8$ $(5+3)$ . We can equalize by doing $30\\div5=6$ , and $6\\cdot3=\\boxed{18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\\cdot8=30+18$ .... |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_16 | D | 6 | The ratio of boys to girls in Mr. Brown's math class is $2:3$ . If there are $30$ students in the class, how many more girls than boys are in the class?
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 10$ | [
"Let the number of boys be $2x$ . It follows that the number of girls is $3x$ . These two values add up to $30$ students, so \\[2x+3x=5x=30\\Rightarrow x=6\\]\nThe difference between the number of girls and the number of boys is $3x-2x=x$ , which is $6$ , so the answer is $\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_31 | C | 3 | The ratio of the interior angles of two regular polygons with sides of unit length is $3: 2$ . How many such pairs are there?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{infinitely many}$ | [
"The formula for the measure of the interior angle of a regular polygon with $n$ -sides is $180 - \\frac{360}{n}$ . Letting our two polygons have side length $r$ and $k$ , we have that the ratio of the interior angles is $\\frac{180 - \\frac{360}{r}}{180 - \\frac{360}{k}} = \\frac{(r-2) \\cdot k}{(k-2) \\cdot r} = ... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_7 | C | 135 | The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | [
"We can set up a system of equations where $x$ and $y$ are the two acute angles. WLOG, assume that $x$ $<$ $y$ in order for the complement of $x$ to be greater than the complement of $y$ . Therefore, $5x$ $=$ $4y$ and $90$ $-$ $x$ $=$ $2$ $(90$ $-$ $y)$ . Solving for $y$ in the first equation and substituting into ... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_4 | C | 135 | The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | [
"By: dragonfly\nWe set up equations to find each angle. The larger angle will be represented as $x$ and the smaller angle will be represented as $y$ , in degrees. This implies that\n$4x=5y$\nand\n$2\\times(90-x)=90-y$\nsince the larger the original angle, the smaller the complement.\nWe then find that $x=75$ and $y... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_12 | B | 27 | The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$ . To the nearest whole percent, what percent of its games did the team lose?
$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$ | [
"The ratio means that for every $11$ games won, $4$ are lost, so the team has won $11x$ games, lost $4x$ games, and played $15x$ games for some positive integer $x$ . The percentage of games lost is just $\\dfrac{4x}{15x}\\times100=\\dfrac{4}{15}\\times 100=26.\\overline{6}\\%\\approx\\boxed{27}$"
] |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_12 | null | 27 | The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$ . To the nearest whole percent, what percent of its games did the team lose?
$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$ | [
"The Won/Lost ratio is 11/4 so, for some number $N$ , the team won $11N$ games and lost $4N$ games. Thus, the team played $15N$ games and the fraction of games lost is $\\dfrac{4N}{15N}=\\dfrac{4}{15}\\approx 0.27=\\boxed{27}.$ -Clara Garza"
] |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_27 | E | 23 | The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:
$\textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 1:3\qquad\textbf{(C)}\ 1:\sqrt3\qquad\textbf{(D)}\ \sqrt3:2 \qquad\textbf{(E)}\ 2:3$ | [
"If the radius of the circle is $r$ , then the perimeter of the first triangle is $3\\left(\\frac{2r}{\\sqrt3}\\right)=2r\\sqrt3$ , and the perimeter of the second is $3r\\sqrt3$ . So the ratio is $\\boxed{23}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | B | 0 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | [
"We square $x+\\frac{1}{x}=\\sqrt5$ to get $x^2+2+\\frac{1}{x^2}=5$ . We subtract 2 on both sides for $x^2+\\frac{1}{x^2}=3$ and square again, and see that $x^4+2+\\frac{1}{x^4}=9$ so $x^4+\\frac{1}{x^4}=7$ . We can factor out $x^7$ from our original expression of $x^{11}-7x^7+x^3$ to get that it is equal to $x^7(x... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5 | null | 98 | The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$ | [
"We note that $8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3$ . Therefore, we have that $9x^3 = (x+1)^3$ , so it follows that $x\\sqrt[3]{9} = x+1$ . Solving for $x$ yields $\\frac{1}{\\sqrt[3]{9}-1} = \\frac{\\sqrt[3]{81}+\\sqrt[3]{9}+1}{8}$ , so the answer is $\\boxed{98}$",
"Let $r$ be t... |
https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_14 | null | 594 | The rectangle $ABCD^{}_{}$ below has dimensions $AB^{}_{} = 12 \sqrt{3}$ and $BC^{}_{} = 13 \sqrt{3}$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P^{}_{}$ . If triangle $ABP^{}_{}$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segment... | [
"\nOur triangular pyramid has base $12\\sqrt{3} - 13\\sqrt{3} - 13\\sqrt{3} \\triangle$ . The area of this isosceles triangle is easy to find by $[ACD] = \\frac{1}{2}bh$ , where we can find $h_{ACD}$ to be $\\sqrt{399}$ by the Pythagorean Theorem . Thus $A = \\frac 12(12\\sqrt{3})\\sqrt{399} = 18\\sqrt{133}$\n\nTo ... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_18 | A | 320 | The rectangle shown has length $AC=32$ , width $AE=20$ , and $B$ and $F$ are midpoints of $\overline{AC}$ and $\overline{AE}$ , respectively. The area of quadrilateral $ABDF$ is
[asy] pair A,B,C,D,EE,F; A = (0,20); B = (16,20); C = (32,20); D = (32,0); EE = (0,0); F = (0,10); draw(A--C--D--EE--cycle); draw(B--D--F); d... | [
"The area of the quadrilateral $ABDF$ is equal to the areas of the two right triangles $\\triangle BCD$ and $\\triangle EFD$ subtracted from the area of the rectangle $ABCD$ . Because $B$ and $F$ are midpoints, we know the dimensions of the two right triangles.\n\n\\[(20)(32)-\\frac{(16)(20)}{2}-\\frac{(10)(32)}{2}... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_11 | D | 20 | The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$ . What is the length $\textit{AB}$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$ | [
"In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$ . However, we need to find the region containing all points within $3$ units of a segmen... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_22 | C | 1,152 | The regular octagon $ABCDEFGH$ has its center at $J$ . Each of the vertices and the center are to be associated with one of the digits $1$ through $9$ , with each digit used once, in such a way that the sums of the numbers on the lines $AJE$ $BJF$ $CJG$ , and $DJH$ are all equal. In how many ways can this be done?
$\... | [
"First of all, note that $J$ must be $1$ $5$ , or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:\n\nWe also notice that $A+E = B+F = C+G = D+H$\nWLOG, assume that $J = 1$ . Thus the pairs of vertices must ... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_15 | E | 4 | The remainder when the product $1492\cdot 1776\cdot 1812\cdot 1996$ is divided by 5 is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | [
"To determine a remainder when a number is divided by $5$ , you only need to look at the last digit. If the last digit is $0$ or $5$ , the remainder is $0$ . If the last digit is $1$ or $6$ , the remainder is $1$ , and so on.\nTo determine the last digit of $1492\\cdot 1776\\cdot 1812\\cdot 1996$ , you only need ... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4 | null | 447 | The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ | [
"Notice repeating decimals can be written as the following:\n$0.\\overline{ab}=\\frac{10a+b}{99}$\n$0.\\overline{abc}=\\frac{100a+10b+c}{999}$\nwhere a,b,c are the digits. Now we plug this back into the original fraction:\n$\\frac{10a+b}{99}+\\frac{100a+10b+c}{999}=\\frac{33}{37}$\nMultiply both sides by $999*99.$ ... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_16 | D | 30 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}... | [
"Let $a$ $b$ $c$ be the three roots of the polynomial. The lengthened prism's volume is \\[V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+c) + 8.\\] By Vieta's formulas, we know that a cubic polynomial $Ax^3+Bx^2+Cx+D$ with roots $a$ $b$ $c$ satisfies: \\begin{alignat*}{8} a+b+c &= -\\... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_15 | D | 30 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}... | [
"Let $a$ $b$ $c$ be the three roots of the polynomial. The lengthened prism's volume is \\[V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+c) + 8.\\] By Vieta's formulas, we know that a cubic polynomial $Ax^3+Bx^2+Cx+D$ with roots $a$ $b$ $c$ satisfies: \\begin{alignat*}{8} a+b+c &= -\\... |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_15 | D | 79 | The sale ad read: "Buy three tires at the regular price and get the fourth tire for 3 dollars." Sam paid 240 dollars for a set of four tires at the sale. What was the regular price of one tire?
$\text{(A)}\ 59.25\text{ dollars} \qquad \text{(B)}\ 60\text{ dollars} \qquad \text{(C)}\ 70\text{ dollars} \qquad \text{(D)}... | [
"Let the regular price of one tire be $x$ . We have \\begin{align*} 3x+3=240 &\\Rightarrow 3x=237 \\\\ &\\Rightarrow x=79 \\end{align*}\n$\\boxed{79}$ Good Job!"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_11 | C | 0 | The sales tax rate in Rubenenkoville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtract... | [
"The price Jack rings up is $\\textdollar{(90.00)(1.06)(0.80)}$ . The price Jill rings up is $\\textdollar{(90.00)(0.80)(1.06)}$ . By the commutative property of multiplication, these quantities are the same, and the difference is $\\boxed{0}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_25 | A | 17 | The sequence $(a_n)$ is defined recursively by $a_0=1$ $a_1=\sqrt[19]{2}$ , and $a_n=a_{n-1}a_{n-2}^2$ for $n\geq 2$ . What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ ... | [
"Let $b_i=19\\text{log}_2a_i$ . Then $b_0=0, b_1=1,$ and $b_n=b_{n-1}+2b_{n-2}$ for all $n\\geq 2$ . The characteristic polynomial of this linear recurrence is $x^2-x-2=0$ , which has roots $2$ and $-1$\nTherefore, $b_n=k_12^{n}+k_2(-1)^n$ for constants to be determined $k_1, k_2$ . Using the fact that $b_0=0, b_1... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_14 | null | 983 | The sequence $(a_n)$ satisfies $a_0=0$ and $a_{n + 1} = \frac85a_n + \frac65\sqrt {4^n - a_n^2}$ for $n\geq 0$ . Find the greatest integer less than or equal to $a_{10}$ | [
"We can now simply start to compute the values $b_i$ by hand:\n\\begin{align*} b_1 & = \\frac 35 \\\\ b_2 & = \\frac 45\\cdot \\frac 35 + \\frac 35 \\sqrt{1 - \\left(\\frac 35\\right)^2} = \\frac{24}{25} \\\\ b_3 & = \\frac 45\\cdot \\frac {24}{25} + \\frac 35 \\sqrt{1 - \\left(\\frac {24}{25}\\right)^2} = \\frac{9... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_7 | null | 41 | The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$ . Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$ | [
"The best way to solve this problem is to get the iterated part out of the exponent: \\[5^{(a_{n + 1} - a_n)} = \\frac {1}{n + \\frac {2}{3}} + 1\\] \\[5^{(a_{n + 1} - a_n)} = \\frac {n + \\frac {5}{3}}{n + \\frac {2}{3}}\\] \\[5^{(a_{n + 1} - a_n)} = \\frac {3n + 5}{3n + 2}\\] \\[a_{n + 1} - a_n = \\log_5{\\left(\... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_24 | B | 2,419 | The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$ ’s separated by blocks of $2$ ’s with $n$ $2$ ’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\t... | [
"The sum of the first $1$ numbers is $1$\nThe sum of the next $2$ numbers is $2 + 1$\nThe sum of the next $3$ numbers is $2 + 2 + 1$\nIn general, we can write \"the sum of the next $n$ numbers is $1 + 2(n-1)$ \", where the word \"next\" follows the pattern established above.\nThus, we first want to find what triang... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_24 | null | 2,419 | The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$ ’s separated by blocks of $2$ ’s with $n$ $2$ ’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\t... | [
"The $k$ th appearance of 1 is at position $1 + 2 + \\dots + k = \\frac{k(k + 1)}{2}$ . Then there are $k$ 1's and $\\frac{k(k + 1)}{2} - k = \\frac{k(k - 1)}{2}$ 2's among the first $\\frac{k(k + 1)}{2}$ numbers, so the sum of these $\\frac{k(k + 1)}{2}$ terms is $k + k(k - 1) = k^2$\nWhen $k = 49$ $\\frac{k(k + 1... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_7 | C | 10 | The sequence $S_1, S_2, S_3, \cdots, S_{10}$ has the property that every term beginning with the third is the sum of the previous two. That is, \[S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.\] Suppose that $S_9 = 110$ and $S_7 = 42$ . What is $S_4$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\te... | [
"$S_9 = 110$ $S_7 = 42$\n$S_8 = S_9 - S_ 7 = 110 - 42 = 68$\n$S_6 = S_8 - S_7 = 68 - 42 = 26$\n$S_5 = S_7 - S_6 = 42 - 26 = 16$\n$S_4 = S_6 - S_5 = 26 - 16 = 10$\nTherefore, the answer is $\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_6 | null | 561 | The sequence $\{a_n\}$ is defined by \[a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \frac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2.\] The sequence $\{b_n\}$ is defined by \[b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \frac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2.\] Find $\frac {b_{32}}{a_{32}}$ | [
"Rearranging the definitions, we have \\[\\frac{a_n}{a_{n-1}} = \\frac{a_{n-1}}{a_{n-2}} + 1,\\quad \\frac{b_n}{b_{n-1}} = \\frac{b_{n-1}}{b_{n-2}} + 1\\] from which it follows that $\\frac{a_n}{a_{n-1}} = 1+ \\frac{a_{n-1}}{a_{n-2}} = \\cdots = (n-1) + \\frac{a_{1}}{a_0} = n$ and $\\frac{b_n}{b_{n-1}} = (n-1) + \\... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_9 | B | 12 | The sequence $a_0,a_1,a_2,\cdots$ is a strictly increasing arithmetic sequence of positive integers such that \[2^{a_7}=2^{27} \cdot a_7.\] What is the minimum possible value of $a_2$
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 22$ | [
"We can rewrite the given equation as $2^{a_7-27}=a_7$ . Hence, $a_7$ must be a power of $2$ and larger than $27$ . The first power of 2 that is larger than $27$ , namely $32$ , does satisfy the equation: $2^{32 - 27} = 2^5 = 32$ . In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is lin... |
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_9 | null | 46 | The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$ | [
"\\[\\log_8 a_1+\\log_8 a_2+\\ldots+\\log_8 a_{12}= \\log_8 a+\\log_8 (ar)+\\ldots+\\log_8 (ar^{11}) \\\\ = \\log_8(a\\cdot ar\\cdot ar^2\\cdot \\cdots \\cdot ar^{11}) = \\log_8 (a^{12}r^{66})\\]\nSo our question is equivalent to solving $\\log_8 (a^{12}r^{66})=2006$ for $a, r$ positive integers $a^{12}r^{66}=8^{2... |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_20 | E | 179 | The sequence $a_{1},a_{2},a_{3},\ldots$ satisfies $a_{1} = 19,a_{9} = 99$ , and, for all $n\geq 3$ $a_{n}$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2$
$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$ | [
"Let $m$ be the arithmetic mean of $a_1$ and $a_2$ . We can then write $a_1=m-x$ and $a_2=m+x$ for some $x$\nBy definition, $a_3=m$\nNext, $a_4$ is the mean of $m-x$ $m+x$ and $m$ , which is again $m$\nRealizing this, one can easily prove by induction that $\\forall n\\geq 3;~ a_n=m$\nIt follows that $m=a_9=99$ . F... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_9 | null | 262 | The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$ | [
"Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for $b_2$ . When we get to $b_2=9$ and $a_2=91$ , we have $a_4=271$ and $b_4=729$ , which works, therefore, the answer is $b_3+a_3=81+181=\\boxed{262}$",
"Using the same reasoning ( $100$ isn't ver... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_25 | E | 225 | The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$ $3\le|y|\le7$ . How many squares of side at least $6$ have their four vertices in $G$
[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8... | [
"Consider any square that meets the requirements described in the problem. Then, take the vertices of the square and translate them to the first quadrant (This is the \"mapping\" described in Solution 2). For example, consider a square with vertices $(7, 7), (-7, 7), (-7, -7),$ and $(7, -7)$\n\nAfter following the ... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_25 | null | 225 | The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$ $3\le|y|\le7$ . How many squares of side at least $6$ have their four vertices in $G$
[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8... | [
"We need to find a reasonably easy way to count the squares.\nFirst, obviously the maximum distance between two points in the same quadrant is $4\\sqrt 2 < 6$ , hence each square has exactly one vertex in each quadrant.\nGiven any square, we can circumscribe another axes-parallel square around it. In the picture be... |
https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_10 | null | 144 | The sets $A = \{z : z^{18} = 1\}$ and $B = \{w : w^{48} = 1\}$ are both sets of complex roots of unity . The set $C = \{zw : z \in A ~ \mbox{and} ~ w \in B\}$ is also a set of complex roots of unity. How many distinct elements are in $C_{}^{}$ | [
"The least common multiple of $18$ and $48$ is $144$ , so define $n = e^{2\\pi i/144}$ . We can write the numbers of set $A$ as $\\{n^8, n^{16}, \\ldots n^{144}\\}$ and of set $B$ as $\\{n^3, n^6, \\ldots n^{144}\\}$ $n^x$ can yield at most $144$ different values. All solutions for $zw$ will be in the form of $n^{8... |
https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_14 | null | 750 | The shortest distances between an interior diagonal of a rectangular parallelepiped $P$ , and the edges it does not meet are $2\sqrt{5}$ $\frac{30}{\sqrt{13}}$ , and $\frac{15}{\sqrt{10}}$ . Determine the volume of $P$ | [
"In the above diagram, we focus on the line that appears closest and is parallel to $BC$ . All the blue lines are perpendicular lines to $BC$ and their other points are on $AB$ , the main diagonal. The green lines are projections of the blue lines onto the bottom face; all of the green lines originate in the corner... |
https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_38 | D | 9 | The sides $PQ$ and $PR$ of triangle $PQR$ are respectively of lengths $4$ inches, and $7$ inches. The median $PM$ is $3\frac{1}{2}$ inches. Then $QR$ , in inches, is:
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$ | [
"By the Median Formula $PM = \\frac12\\sqrt{2PQ^2+2PR^2-QR^2}$\nPlugging in the numbers given in the problem, we get \\[\\frac72=\\frac12\\sqrt{2\\cdot4^2+2\\cdot7^2-QR^2}\\]\nSolving, \\[7=\\sqrt{2(16)+2(49)-QR^2}\\] \\[49=32+98-QR^2\\] \\[QR^2=81\\] \\[QR=9=\\boxed{9}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1958_AHSME_Problems/Problem_36 | D | 65 | The sides of a triangle are $30$ $70$ , and $80$ units. If an altitude is dropped upon the side of length $80$ , the larger segment cut off on this side is:
$\textbf{(A)}\ 62\qquad \textbf{(B)}\ 63\qquad \textbf{(C)}\ 64\qquad \textbf{(D)}\ 65\qquad \textbf{(E)}\ 66$ | [
"Let the shorter segment be $x$ and the altitude be $y$ . The larger segment is then $80-x$ . By the Pythagorean Theorem \\[30^2-y^2=x^2 \\qquad(1)\\] and \\[(80-x)^2=70^2-y^2 \\qquad(2)\\] Adding $(1)$ and $(2)$ and simplifying gives $x=15$ . Therefore, the answer is $80-15=\\boxed{65}$"
] |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_17 | B | 4 | The sides of a triangle have lengths $6.5$ $10$ , and $s$ , where $s$ is a whole number. What is the smallest possible value of $s$
[asy] pair A,B,C; A=origin; B=(10,0); C=6.5*dir(15); dot(A); dot(B); dot(C); draw(B--A--C); draw(B--C,dashed); label("$6.5$",3.25*dir(15),NNW); label("$10$",(5,0),S); label("$s$",(8,1),NE... | [
"By Triangle Inequality $6.5 + s >10$ and therefore $s>3.5$ . The smallest whole number that satisfies this is $\\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_10 | D | 8 | The sides of a triangle with positive area have lengths $4$ $6$ , and $x$ . The sides of a second triangle with positive area have lengths $4$ $6$ , and $y$ . What is the smallest positive number that is not a possible value of $|x-y|$
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \... | [
"Since $6$ and $4$ are fixed sides, the smallest possible side has to be larger than $6-4=2$ and the largest possible side has to be smaller than $6+4=10$ . This gives us the triangle inequality $2<x<10$ and $2<y<10$ $7$ can be attained by letting $x=9.1$ and $y=2.1$ . However, $8=10-2$ cannot be attained. Thus, th... |
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_15 | null | 554 | The sides of rectangle $ABCD$ have lengths $10$ and $11$ . An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$ . The maximum possible area of such a triangle can be written in the form $p\sqrt{q}-r$ , where $p$ $q$ , and $r$ are positive integers, and $q$ is not divisible by the square... | [
"Consider points on the complex plane $A (0,0),\\ B (11,0),\\ C (11,10),\\ D (0,10)$ . Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at $A$ , and the ... |
https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_23 | B | 13 | The six edges of a tetrahedron $ABCD$ measure $7, 13, 18, 27, 36$ and $41$ units. If the length of edge $AB$ is $41$ , then the length of edge $CD$ is
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$ | [
"By the triangle inequality in $\\triangle ABC$ , we find that $BC$ and $CA$ must sum to greater than $41$ , so they must be (in some order) $7$ and $36$ $13$ and $36$ $18$ and $27$ $18$ and $36$ , or $27$ and $36$ . We try $7$ and $36$ , and now by the triangle inequality in $\\triangle ABD$ , we must use the rema... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_5 | E | 9 | The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$ | [
"First, modulo $2$ or $5$ $\\underline{20210A} \\equiv A$ .\nHence, $A \\neq 0, 2, 4, 5, 6, 8$\nSecond modulo $3$ $\\underline{20210A} \\equiv 2 + 0 + 2 + 1 + 0 + A \\equiv 5 + A$ .\nHence, $A \\neq 1, 4, 7$\nThird, modulo $11$ $\\underline{20210A} \\equiv A + 1 + 0 - 0 - 2 - 2 \\equiv A - 3$ .\nHence, $A \\neq 3$\... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_4 | E | 9 | The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$ | [
"First, modulo $2$ or $5$ $\\underline{20210A} \\equiv A$ .\nHence, $A \\neq 0, 2, 4, 5, 6, 8$\nSecond modulo $3$ $\\underline{20210A} \\equiv 2 + 0 + 2 + 1 + 0 + A \\equiv 5 + A$ .\nHence, $A \\neq 1, 4, 7$\nThird, modulo $11$ $\\underline{20210A} \\equiv A + 1 + 0 - 0 - 2 - 2 \\equiv A - 3$ .\nHence, $A \\neq 3$\... |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_6 | B | 21 | The smallest product one could obtain by multiplying two numbers in the set $\{ -7,-5,-1,1,3 \}$ is
$\text{(A)}\ -35 \qquad \text{(B)}\ -21 \qquad \text{(C)}\ -15 \qquad \text{(D)}\ -1 \qquad \text{(E)}\ 3$ | [
"To get the smallest possible product, we want to multiply the smallest negative number by the largest positive number. These are $-7$ and $3$ , respectively, and their product is $-21$ , which is $\\boxed{21}$"
] |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_3 | C | 3 | The smallest sum one could get by adding three different numbers from the set $\{ 7,25,-1,12,-3 \}$ is
$\text{(A)}\ -3 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 21$ | [
"To find the smallest sum, we just have to find the smallest 3 numbers and add them together.\nObviously, the numbers are $-3, -1, 7$ , and adding them gets us $3$\n$\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11 | null | 288 | The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid? | [
"First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$ . The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$ , and one of the legs is $3\\sqrt{2}$ . We apply the Pythagorean Theorem to deduce that the height is $6$\nNext, we complete t he ... |
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_12 | D | 1 | The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:
$\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0$ | [
"First, square both sides. This gives us\n\\[\\sqrt{5x-1}^2+2\\cdot\\sqrt{5x-1}\\cdot\\sqrt{x-1}+\\sqrt{x-1}^2=4 \\Longrightarrow 5x-1+2\\cdot\\sqrt{(5x-1)\\cdot(x-1)}+x-1=4 \\Longrightarrow 2\\cdot\\sqrt{5x^2-6x+1}+6x-2=4\\] Then, adding $-6x$ to both sides gives us\n\\[2\\cdot\\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \\Lon... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_14 | D | 31 | The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$ , where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$ , can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$
$\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C... | [
"We apply the Change of Base Formula, then rearrange: \\begin{align*} \\frac{\\log_2{4}}{\\log_2{(3x)}}&=\\frac{\\log_2{8}}{\\log_2{(2x)}} \\\\ \\frac{2}{\\log_2{(3x)}}&=\\frac{3}{\\log_2{(2x)}} \\\\ 3\\log_2{(3x)}&=2\\log_2{(2x)}. \\\\ \\end{align*} By the logarithmic identity $n\\log_b{a}=\\log_b{\\left(a^n\\righ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_22 | A | 20 | The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible... | [
"We solve each equation separately:\nNote that the problem is equivalent to finding the area of a parallelogram with consecutive vertices $(x_1,y_1)=\\left(\\sqrt{10}, \\sqrt{6}\\right),(x_2,y_2)=\\left(\\sqrt{3},1\\right),(x_3,y_3)=\\left(-\\sqrt{10},-\\sqrt{6}\\right),$ and $(x_4,y_4)=\\left(-\\sqrt{3}, -1\\right... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_6 | null | 12 | The solutions to the system of equations
are $(x_1,y_1)$ and $(x_2,y_2)$ . Find $\log_{30}\left(x_1y_1x_2y_2\right)$ | [
"Let $A=\\log_{225}x$ and let $B=\\log_{64}y$\nFrom the first equation: $A+B=4 \\Rightarrow B = 4-A$\nPlugging this into the second equation yields $\\frac{1}{A}-\\frac{1}{B}=\\frac{1}{A}-\\frac{1}{4-A}=1 \\Rightarrow A = 3\\pm\\sqrt{5}$ and thus, $B=1\\pm\\sqrt{5}$\nSo, $\\log_{225}(x_1x_2)=\\log_{225}(x_1)+\\log_... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_22 | C | 35 | The square
is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $g$
$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 62 \qquad \textbf... | [
"All the unknown entries can be expressed in terms of $b$ .\nSince $100e = beh = ceg = def$ , it follows that $h = \\frac{100}{b}, g = \\frac{100}{c}$ ,\nand $f = \\frac{100}{d}$ . Comparing rows $1$ and $3$ then gives $50bc = 2 \\cdot \\frac{100}{b} \\cdot \\frac{100}{c}$ ,\nfrom which $c = \\frac{20}{b}$ .\nCompa... |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_3 | B | 100 | The stronger Goldbach conjecture states that any even integer greater than 7 can be written as the sum of two different prime numbers. For such representations of the even number 126, the largest possible difference between the two primes is
$\textbf{(A)}\ 112\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 92\qquad\textbf{(... | [
"We can guess and check small primes, subtract it from $126$ , and see if the result is a prime because the further away the two numbers are, the greater the difference will be. Since $126 = 2 \\cdot 3^2 \\cdot 7$ , we can eliminate $2$ $3$ , and $7$ as an option because subtracting these would result in a composi... |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_11 | A | 2,001 | The student locker numbers at Olympic High are numbered consecutively beginning with locker number $1$ . The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number $9$ and four centers to label locker number $10$ . If it costs $137.94 to label all the lockers, ... | [
"Since all answers are over $2000$ , work backwards and find the cost of the first $1999$ lockers. The first $9$ lockers cost $0.18$ dollars, while the next $90$ lockers cost $0.04\\cdot 90 = 3.60$ . Lockers $100$ through $999$ cost $0.06\\cdot 900 = 54.00$ , and lockers $1000$ through $1999$ inclusive cost $0.08... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_20 | B | 17 | The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 17\qquad \textbf{(C)}\ 24\qquad \t... | [
"Let $b$ be the number of boys and $g$ be the number of girls.\n\\[\\frac23 b = \\frac34 g \\Rightarrow b = \\frac98 g\\]\nFor $g$ and $b$ to be integers, $g$ must cancel out with the denominator, and the smallest possible value is $8$ . This yields $9$ boys. The minimum number of students is $8+9=\\boxed{17}$",
... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_7 | E | 20 | The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E?
[asy] real[] r={6, 8, 4, 2, 5}; int i; for(i=0; i<5; i=i+1) { filldraw((4i,0)--(4i+3,0)--(4i+3,2r[i])--(4i,2... | [
"From the bar graph, we can see that $5$ students chose candy E. There are $6+8+4+2+5=25$ total students in Mrs. Sawyers class. The percent that chose E is $\\frac{5}{25} \\cdot 100 = \\boxed{20}$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9 | D | 2,023 | The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$ | [
"Note that $\\frac{n}{(n+1)!} = \\frac{1}{n!} - \\frac{1}{(n+1)!}$ , and therefore this sum is a telescoping sum, which is equivalent to $1 - \\frac{1}{2022!}$ . Our answer is $1 + 2022 = \\boxed{2023}$",
"We add $\\frac{1}{2022!}$ to the original expression \\[\\left(\\frac{1}{2!}+\\frac{2}{3!}+\\frac{3}{4!}+\\c... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19 | E | 424 | The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?
$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$ | [
"Let $n$ be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to $25n$ since $(n-2k)+\\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \\dots +(n+2k)=25n$ . Now, $25n=10000 \\rightarrow n=400$ . Remembering that this is the 13th integer, we wish t... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_3 | B | 8 | The sum of $5$ consecutive even integers is $4$ less than the sum of the first $8$ consecutive odd counting numbers. What is the smallest of the even integers?
$\textbf{(A) } 6 \qquad\textbf{(B) } 8 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 14$ | [
"It is a well-known fact that the sum of the first $n$ odd numbers is $n^2$ . This makes the sum of the first $8$ odd numbers equal to $64$\nLet $n$ be equal to the smallest of the $5$ even integers. Then $(n+2)$ is the next highest, $(n+4)$ even higher, and so on.\nThis sets up the equation $n+(n+2)+(n+4)+(n+6)+... |
https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_28 | D | 5 | The sum of $n$ terms of an arithmetic progression is $153$ , and the common difference is $2$ .
If the first term is an integer, and $n>1$ , then the number of possible values for $n$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 6$ | [
"Let the progression start at $a$ , have common difference $2$ , and end at $a + 2(n-1)$\nThe average term is $\\frac{a + (a + 2(n-1))}{2}$ , or $a + n - 1$ . Since the number of terms is $n$ , and the sum of the terms is $153$ , we have:\n$n(a+n-1) = 153$\nSince $n$ is a positive integer, it must be a factor of $... |
https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_6 | D | 130 | The sum of all but one of the interior angles of a convex polygon equals $2570^\circ$ . The remaining angle is
$\text{(A)} \ 90^\circ \qquad \text{(B)} \ 105^\circ \qquad \text{(C)} \ 120^\circ \qquad \text{(D)}\ 130^\circ\qquad \text{(E)}\ 144^\circ$ | [
"Note that the sum of the interior angles of a convex polygon of $n$ sides is $180(n-2)^\\circ$ , and each interior angle belongs to $[0, 180^\\circ)$ . Therefore, we must have $n - 2 = \\lfloor \\frac{2570^\\circ}{180^\\circ} \\rfloor = 15$ . Then the missing angle must be $180*15^\\circ - 2570^\\circ = 130^\\circ... |
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_7 | A | 5,880 | The sum of all integers between 50 and 350 which end in 1 is
$\textbf{(A)}\ 5880\qquad\textbf{(B)}\ 5539\qquad\textbf{(C)}\ 5208\qquad\textbf{(D)}\ 4877\qquad\textbf{(E)}\ 4566$ | [
"The numbers that we are adding are $51,61,71 \\cdots 341$ . The numbers are part of an arithmetic series with first term $51$ , last term $341$ , common difference $10$ , and $30$ terms. Using the arithmetic series formula, the sum of the terms is $\\tfrac{30 \\cdot 392}{2} = \\boxed{5880}$"
] |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_25 | C | 2 | The sum of all numbers of the form $2k + 1$ , where $k$ takes on integral values from $1$ to $n$ is:
$\textbf{(A)}\ n^2\qquad\textbf{(B)}\ n(n+1)\qquad\textbf{(C)}\ n(n+2)\qquad\textbf{(D)}\ (n+1)^2\qquad\textbf{(E)}\ (n+1)(n+2)$ | [
"The sum of the odd integers $2k-1$ from $1$ to $n$ is $n^2$ . However, in this problem, the sum is instead $2k+1$ , starting at $3$ rather than $1$ . To rewrite this, we note that $2k-1$ is $2$ less than $2k+1$ for every $k$ we add, so for $n$ $k$ 's, we subtract $2n$ , giving us $n^2+2n$ ,which factors as $n(n+2)... |
https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_20 | E | 72 | The sum of all real $x$ such that $(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3$ is
$\textbf{(A) } \frac32 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } \frac52 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \frac72$ | [
"Note that $(2^x-4)+(4^x-2)=4^x+2^x-6,$ so we let $a=2^x-4$ and $b=4^x-2.$ The original equation becomes \\[a^3+b^3=(a+b)^3.\\] We expand the right side, then rearrange: \\begin{align*} a^3+b^3 &= a^3+3a^2b+3ab^2+b^3 \\\\ 0 &= 3a^2b+3ab^2 \\\\ 0 &= 3ab(a+b). \\end{align*}\nTogether, the answer is $2+\\frac12+1=\\bo... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | E | 4 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | [
"The sum of an infinite geometric series is of the form: \\[\\begin{split} S & = \\frac{a_1}{1-r} \\end{split}\\] where $a_1$ is the first term and $r$ is the ratio whose absolute value is less than 1.\nWe know that the second term is the first term multiplied by the ratio. \nIn other words: \\[\\begin{split} a_1 ... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14 | E | 4 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | [
"The sum of the geometric sequence is $\\frac{a}{1 - r}$ where $a$ is the first term and $r$ is the common ratio. We know the second term, $ar,$ is equal to $1.$ Thus $ar = 1 \\Rightarrow a = \\frac{1}{r}.$ This means, \\[S = \\frac{a}{1 - r} = \\frac{1/r}{1 - r} = \\frac{1}{r(1 - r)}.\\] In order to minimize $S,$ ... |
https://artofproblemsolving.com/wiki/index.php/1970_AHSME_Problems/Problem_19 | C | 5 | The sum of an infinite geometric series with common ratio $r$ such that $|r|<1$ is $15$ , and the sum of the squares of the terms of this series is $45$ . The first term of the series is
$\textbf{(A) } 12\quad \textbf{(B) } 10\quad \textbf{(C) } 5\quad \textbf{(D) } 3\quad \textbf{(E) 2}$ | [
"We know that the formula for the sum of an infinite geometric series is $S = \\frac{a}{1-r}$\nSo we can apply this to the conditions given by the problem.\nWe have two equations:\n\\begin{align*} 15 &= \\frac{a}{1-r} \\\\ 45 &= \\frac{a^{2}}{1-r^{2}} \\end{align*}\nWe get\n\\begin{align*} a &= 15 - 15r \\\\ a^{2} ... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_15 | D | 4 | The sum of four two-digit numbers is $221$ . None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | [
"$221$ can be written as the sum of four two-digit numbers, let's say $\\overline{ae}$ $\\overline{bf}$ $\\overline{cg}$ , and $\\overline{dh}$ . Then $221= 10(a+b+c+d)+(e+f+g+h)$ . The last digit of $221$ is $1$ , and $10(a+b+c+d)$ won't affect the units digits, so $(e+f+g+h)$ must end with $1$ . The smallest valu... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_15 | null | 4 | The sum of four two-digit numbers is $221$ . None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | [
"Alternatively, we know that a number is congruent to the sum of its digits mod 9, so $221 \\equiv 5 \\equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \\equiv -d$ , where $d$ is some digit. Clearly, $\\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_7 | D | 6 | The sum of seven integers is $-1$ . What is the maximum number of the seven integers that can be larger than $13$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | [
"If the first six integers are $14$ , the last number can be $(-14\\cdot 6) - 1 = -85$ . The sum of all seven integers will be $-1$\nHowever, if all seven integers are over $13$ , the smallest possible sum is $14\\cdot 7 = 98$\nThus, the answer is $6$ , which is option $\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_17 | B | 338 | The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$ | [
"The arithmetic mean of these numbers is $\\frac{\\frac{2013}{3}}{2}=\\frac{671}{2}=335.5$ . Therefore the numbers are $333$ $334$ $335$ $336$ $337$ $338$ , so the answer is $\\boxed{338}$",
"Let the $4^{\\text{th}}$ number be $x$ . Then our desired number is $x+2$\nOur integers are $x-3,x-2,x-1,x,x+1,x+2$ , so w... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_17 | null | 338 | The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$ | [
"Since there are $6$ numbers, we divide $2013$ by $6$ to find the mean of the numbers. $\\frac{2013}{6} = 335 \\frac{1}{2}$ .\nThen, $335 \\frac{1}{2} + \\frac{1}{2} = 336$ (the fourth number). Fifth: $337$ ; Sixth: $\\boxed{338}$"
] |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_6 | null | 348 | The sum of the areas of all triangles whose vertices are also vertices of a $1$ by $1$ by $1$ cube is $m + \sqrt{n} + \sqrt{p},$ where $m, n,$ and $p$ are integers . Find $m + n + p.$ | [
"Since there are $8$ vertices of a cube , there are ${8 \\choose 3} = 56$ total triangles to consider. They fall into three categories: there are those which are entirely contained within a single face of the cube (whose sides are two edges and one face diagonal ), those which lie in a plane perpendicular to one f... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_23 | null | 11 | The sum of the base- $10$ logarithms of the divisors of $10^n$ is $792$ . What is $n$
$\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15$ | [
"Every factor of $10^n$ will be of the form $2^a \\times 5^b , a\\leq n , b\\leq n$ . Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property $\\log(a \\times b) = \\log(a)+\\log(b)$ . For any factor $2^a \\times 5^b$ ... |
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_10 | A | 4 | The sum of the digits in base ten of $(10^{4n^2+8}+1)^2$ , where $n$ is a positive integer, is
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4n \qquad \textbf{(C)}\ 2+2n \qquad \textbf{(D)}\ 4n^2 \qquad \textbf{(E)}\ n^2+n+2$ | [
"We see that the result of this expression will always be in the form $(100\\text{ some number of zeros }001)^2.$ Multiplying these together yields: \\[110\\text{ some number of zeros }011.\\] This works because of the way they are multiplied. Therefore, the answer is $\\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_16 | D | 10 | The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is $6$ . How many two-digit numbers have this property?
$\textbf{(A) } 5\qquad \textbf{(B) } 7\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 19$ | [
"Let the number be $10a+b$ where $a$ and $b$ are the tens and units digits of the number.\nSo $(10a+b)-(a+b)=9a$ must have a units digit of $6$\nThis is only possible if $9a=36$ , so $a=4$ is the only way this can be true.\nSo the numbers that have this property are $40, 41, 42, 43, 44, 45, 46, 47, 48, 49$\nTherefo... |
https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_5 | null | 542 | The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms. | [
"Since the sum of the first $2011$ terms is $200$ , and the sum of the first $4022$ terms is $380$ , the sum of the second $2011$ terms is $180$ .\nThis is decreasing from the first 2011, so the common ratio is less than one.\nBecause it is a geometric sequence and the sum of the first 2011 terms is $200$ , second ... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_22 | A | 255 | The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$
$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$ | [
"The sum of the first $m$ odd integers is given by $m^2$ . The sum of the first $n$ even integers is given by $n(n+1)$\nThus, $m^2 = n^2 + n + 212$ . Since we want to solve for n, rearrange as a quadratic equation: $n^2 + n + (212 - m^2) = 0$\nUse the quadratic formula: $n = \\frac{-1 + \\sqrt{1 - 4(212 - m^2)}}{2}... |
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_6 | E | 80 | The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80$ | [
"Solution by e_power_pi_times_i\nWhen the $n$ th odd positive integer is subtracted from the $n$ th even positive integer, the result is $1$ . Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is $80\\cdot1 = \\boxed{80}$"
] |
https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_8 | null | 61 | The sum of the following seven numbers is exactly 19: $a_1 = 2.56$ $a_2 = 2.61$ $a_3 = 2.65$ $a_4 = 2.71$ $a_5 = 2.79$ $a_6 = 2.82$ $a_7 = 2.86$ . It is desired to replace each $a_i$ by an integer approximation $A_i$ $1\le i \le 7$ , so that the sum of the $A_i$ 's is also 19 and so that $M$ , the maximum of the "error... | [
"If any of the approximations $A_i$ is less than 2 or more than 3, the error associated with that term will be larger than 1, so the largest error will be larger than 1. However, if all of the $A_i$ are 2 or 3, the largest error will be less than 1. So in the best case, we write 19 as a sum of 7 numbers, each of ... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_23 | B | 784 | The sum of the lengths of the twelve edges of a rectangular box is $140$ , and
the distance from one corner of the box to the farthest corner is $21$ . The total surface area of the box is
$\text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812$ | [
"Let $x, y$ , and $z$ be the unique lengths of the edges of the box. Each box has $4$ edges of each length, so: \\[4x + 4y + 4z = 140 \\ \\Longrightarrow \\ x + y + z = 35.\\] The spacial diagonal (longest distance) is given by $\\sqrt{x^2 + y^2 + z^2}$ . Thus, we have $\\sqrt{x^2 + y^2 + z^2} = 21$ , so $x^2 + y... |
https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_20 | B | 1 | The sum of the numerical coefficients of all the terms in the expansion of $(x-2y)^{18}$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ -1\qquad \textbf{(E)}\ -19$ | [
"For any polynomial, even a polynomial with more than one variable, the sum of all the coefficients (including the constant, which is the coefficient of $x^0y^0$ ) is found by setting all variables equal to $1$ . Note that we don't have to worry about whether a constant is a coefficient of an \"invisible\" $x^0y^0... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_11 | E | 14 | The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$
$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$ | [
"$AMC10+AMC12=123422$\n$AMC00+AMC00=123400$\n$AMC+AMC=1234$\n$2\\cdot AMC=1234$\n$AMC=\\frac{1234}{2}=617$\nSince $A$ $M$ , and $C$ are digits, $A=6$ $M=1$ $C=7$\nTherefore, $A+M+C = 6+1+7 = \\boxed{14}$",
"We know that $AMC12$ is $2$ more than $AMC10$ . We set up $AMC10=x$ and $AMC12=x+2$ . We have $x+x+2=123422... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12A_Problems/Problem_5 | E | 14 | The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$
$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$ | [
"$AMC10+AMC12=123422$\n$AMC00+AMC00=123400$\n$AMC+AMC=1234$\n$2\\cdot AMC=1234$\n$AMC=\\frac{1234}{2}=617$\nSince $A$ $M$ , and $C$ are digits, $A=6$ $M=1$ $C=7$\nTherefore, $A+M+C = 6+1+7 = \\boxed{14}$",
"We know that $AMC12$ is $2$ more than $AMC10$ . We set up $AMC10=x$ and $AMC12=x+2$ . We have $x+x+2=123422... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_13 | A | 28 | The sum of three numbers is $20$ . The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?
$\mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800$ | [
"Let the numbers be $x$ $y$ , and $z$ in that order. The given tells us that\n\\begin{eqnarray*}y&=&7z\\\\ x&=&4(y+z)=4(7z+z)=4(8z)=32z\\\\ x+y+z&=&32z+7z+z=40z=20\\\\ z&=&\\frac{20}{40}=\\frac{1}{2}\\\\ y&=&7z=7\\cdot\\frac{1}{2}=\\frac{7}{2}\\\\ x&=&32z=32\\cdot\\frac{1}{2}=16 \\end{eqnarray*}\nTherefore, the pro... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_3 | E | 5 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \tex... | [
"Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$\nWe have \\[6x+(x+40)+x=8x+40=96,\\] from which $x=7.$\nTherefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\\boxed{5}.$",
"Solve this using a... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_2 | E | 5 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \tex... | [
"Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$\nWe have \\[6x+(x+40)+x=8x+40=96,\\] from which $x=7.$\nTherefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\\boxed{5}.$",
"Solve this using a... |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_16 | C | 30 | The sum of three numbers is $98$ . The ratio of the first to the second is $\frac {2}{3}$ ,
and the ratio of the second to the third is $\frac {5}{8}$ . The second number is:
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33$ | [
"Let the $3$ numbers be $a,$ $b,$ and $c$ . We see that \\[a+b+c = 98\\] and \\[\\frac{a}{b} = \\frac{2}{3} \\Rrightarrow 3a = 2b\\] \\[\\frac{b}{c} = \\frac{5}{8} \\Rrightarrow 8b = 5c\\] Writing $a$ and $c$ in terms of $b$ we have $a = \\frac{2}{3} b$ and $c = \\frac{8}{5} b$ . \nSubstituting in the sum, we have ... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_7 | B | 72 | The sum of two angles of a triangle is $\frac{6}{5}$ of a right angle, and one of these two angles is $30^{\circ}$ larger than the other. What is the degree measure of the largest angle in the triangle?
$\textbf{(A)}\ 69 \qquad\textbf{(B)}\ 72 \qquad\textbf{(C)}\ 90 \qquad\textbf{(D)}\ 102 \qquad\textbf{(E)}\ 108$ | [
"The sum of two angles in a triangle is $\\frac{6}{5}$ of a right angle $\\longrightarrow \\frac{6}{5} \\times 90 = 108$\nIf $x$ is the measure of the first angle, then the measure of the second angle is $x+30$ \\[x + x + 30 = 108 \\longrightarrow 2x = 78 \\longrightarrow x = 39\\]\nNow we know the measure of two a... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3 | D | 14,238 | The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(... | [
"The units digit of a multiple of $10$ will always be $0$ . We add a $0$ whenever we multiply by $10$ . So, removing the units digit is equal to dividing by $10$\nLet the smaller number (the one we get after removing the units digit) be $a$ . This means the bigger number would be $10a$\nWe know the sum is $10a+a = ... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_3 | D | 14,238 | The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(... | [
"The units digit of a multiple of $10$ will always be $0$ . We add a $0$ whenever we multiply by $10$ . So, removing the units digit is equal to dividing by $10$\nLet the smaller number (the one we get after removing the units digit) be $a$ . This means the bigger number would be $10a$\nWe know the sum is $10a+a = ... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_5 | C | 4 | The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$ | [
"Let the two real numbers be $x,y$ . We are given that $x+y=4xy,$ and dividing both sides by $xy$ $\\frac{x}{xy}+\\frac{y}{xy}=4.$\n\\[\\frac{1}{y}+\\frac{1}{x}=\\boxed{4}.\\]",
"Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a frac... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_2 | C | 4 | The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$ | [
"Let $x, y$ be our two numbers. Then $x+y = 4xy$ . Thus,\n$\\frac{1}{x} + \\frac{1}{y} = \\frac{x+y}{xy} = 4$\n$\\boxed{4}$",
"We can let $x=y.$ Then, we have that $2x=4x^2$ making $x=\\tfrac{1}{2}.$ The answer is $\\dfrac{2}{x}=4=\\boxed{4}.$ Solasky talk"
] |
https://artofproblemsolving.com/wiki/index.php/1958_AHSME_Problems/Problem_13 | B | 12 | The sum of two numbers is $10$ ; their product is $20$ . The sum of their reciprocals is:
$\textbf{(A)}\ \frac{1}{10}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 4$ | [
"$x+y=10$\n$xy=20$\n$\\frac1x+\\frac1y=\\frac{y}{xy}+\\frac{x}{xy}=\\frac{x+y}{xy}=\\frac{10}{20}=\\boxed{12}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_6 | 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 ... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.