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/1991_AIME_Problems/Problem_4 | null | 159 | How many real numbers $x^{}_{}$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$ | [
"The range of the sine function is $-1 \\le y \\le 1$ . It is periodic (in this problem) with a period of $\\frac{2}{5}$\nThus, $-1 \\le \\frac{1}{5} \\log_2 x \\le 1$ , and $-5 \\le \\log_2 x \\le 5$ . The solutions for $x$ occur in the domain of $\\frac{1}{32} \\le x \\le 32$ . When $x > 1$ the logarithm function... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_10 | C | 2 | How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | [
"The first thing one would want to do is place a possible letter that works and then stem off of it. For example, if we start with an $a$ , we can only place a $c$ or $d$ next to it. Unfortunately, after that step, we can't do too much, since:\n$acbd$ is not allowed because of the $cb$ , and $acdb$ is not allowed... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_19 | D | 11 | How many rectangles are in this figure?
[asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(20,0); C=(20,20); D=(0,20); draw(A--B--C--D--cycle); E=(-10,-5); F=(13,-5); G=(13,5); H=(-10,5); draw(E--F--G--H--cycle); I=(10,-20); J=(18,-20); K=(18,13); L=(10,13); draw(I--J--K--L--cycle);[/asy]
$\textbf{(A)}\ 8\qquad\textbf{(B)... | [
"The figure can be divided into $7$ sections. The number of rectangles with just one section is $3.$ The number of rectangles with two sections is $5.$ There are none with only three sections. The number of rectangles with four sections is $3.$\n$3+5+3=\\boxed{11}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_15 | A | 6 | How many right triangles have integer leg lengths $a$ and $b$ and a hypotenuse of length $b+1$ , where $b<100$
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$ | [
"By the Pythagorean theorem, $a^2+b^2=b^2+2b+1$\nThis means that $a^2=2b+1$\nWe know that $a,b>0$ and that $b<100$\nWe also know that $a^2$ is odd and thus $a$ is odd, since the right side of the equation is odd. $2b$ is even. $2b+1$ is odd.\nSo $a=1,3,5,7,9,11,13$ , but if $a=1$ , then $b=0$ . Thus $a\\neq1.$\n$a=... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25 | C | 65 | How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | [
"Let $f(n)$ be the number of valid sequences of length $n$ (satisfying the conditions given in the problem).\nWe know our valid sequence must end in a $0$ . Then, since we cannot have two consecutive $0$ s, it must end in a $10$ . Now, we only have two cases: it ends with $010$ , or it ends with $110$ which is equi... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_23 | C | 65 | How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | [
"Let $f(n)$ be the number of valid sequences of length $n$ (satisfying the conditions given in the problem).\nWe know our valid sequence must end in a $0$ . Then, since we cannot have two consecutive $0$ s, it must end in a $10$ . Now, we only have two cases: it ends with $010$ , or it ends with $110$ which is equi... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_12 | D | 380 | How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
$\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382$ | [
"There are $\\binom{20}{2}$ selections; however, we count these twice, therefore\n$2\\cdot\\binom{20}{2} = \\boxed{380}$ . The wording of the question implies D, not E."
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_12 | E | 382 | How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
$\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382$ | [
"Consider the 20 term sequence of $0$ 's and $1$ 's. Keeping all other terms 1, a sequence of $k>0$ consecutive 0's can be placed in $21-k$ locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence ... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_9 | C | 3 | How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | [
"Notice that if the consecutive positive integers have a sum of $15$ , then their average (which could be a fraction) must be a divisor of $15$ . If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$ , and $1$ is clearly not possible. The other two possibilities both work:\nIf... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_8 | C | 3 | How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | [
"Notice that if the consecutive positive integers have a sum of $15$ , then their average (which could be a fraction) must be a divisor of $15$ . If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$ , and $1$ is clearly not possible. The other two possibilities both work:\nIf... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_19 | C | 2 | How many solutions does the equation $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$
$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$ | [
"The ranges of $\\frac{\\pi}2 \\sin x$ and $\\frac{\\pi}2 \\cos x$ are both $\\left[-\\frac{\\pi}2, \\frac{\\pi}2 \\right],$ which is included in the range of $\\arcsin,$ so we can use it with no issues. \\begin{align*} \\frac{\\pi}2 \\cos x &= \\arcsin \\left( \\cos \\left( \\frac{\\pi}2 \\sin x\\right)\\right) \\... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_9 | E | 5 | How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | [
"We count the intersections of the graphs of $y=\\tan(2x)$ and $y=\\cos\\left(\\frac x2\\right):$\nThe graphs of $y=\\tan(2x)$ and $y=\\cos\\left(\\frac x2\\right)$ intersect once on each of the five branches of $y=\\tan(2x),$ as shown below: Therefore, the answer is $\\boxed{5}.$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_14 | D | 50 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}... | [
"The region is a right triangle which contains the following lattice points: $(0,0); (1,0)\\rightarrow(1,3); (2,0)\\rightarrow(2,6); (3,0)\\rightarrow(3,9); (4,0)\\rightarrow(4,12); (5,0)\\rightarrow(5,15)$\n\nSquares $1\\times 1$ :\nSuppose that the top-right corner is $(x,y)$ , with $2\\le x\\le 5$ . Then to incl... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_11 | D | 50 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}... | [
"Solution by e_power_pi_times_i\nRevised by Kinglogic and RJ5303707\n (red shows lattice points within the triangle)\nIf we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is $16$ squares $(y=5.1*\\pi)$ , and the limit for the... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24 | E | 1,296 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $... | [
"For some $n$ , let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \\ldots c_n$ each between $0$ and $n$ , and let $n$ cars come into this circle so that the $i$ th car tries to park at spot $c_i$ , but if it is already taken then it instead keeps going counter... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24 | E | 1,296 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $... | [
"For some $n$ , let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \\ldots c_n$ each between $0$ and $n$ , and let $n$ cars come into this circle so that the $i$ th car tries to park at spot $c_i$ , but if it is already taken then it instead keeps going counter... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_21 | D | 12 | How many subsets containing three different numbers can be selected from the set \[\{ 89,95,99,132, 166,173 \}\] so that the sum of the three numbers is even?
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$ | [
"To have an even sum with three numbers, we must add either $E+O+O$ , or $E + E + E$ , where $O$ represents an odd number, and $E$ represents an even number.\nSince there are not three even numbers in the given set, $E+E+E$ is impossible. Thus, we must choose two odd numbers, and one even number.\nThere are $2$ ch... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_5 | D | 240 | How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A) } 128 \qquad \textbf{(B) } 192 \qquad \textbf{(C) } 224 \qquad \textbf{(D) } 240 \qquad \textbf{(E) } 256$ | [
"Since an element of a subset is either in or out, the total number of subsets of the $8$ -element set is $2^8 = 256$ . However, since we are only concerned about the subsets with at least $1$ prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Becau... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_5 | D | 240 | How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}$ | [
"Well, there are 4 composite numbers, and you can list them in a 1 number format, a 2 number, 3 number, and a 4 number format. Now, we can use combinations.\n$\\binom{4}{1} + \\binom{4}{2} + \\binom{4}{3} + \\binom{4}{4} = 15$ . Using the answer choices, the only multiple of 15 is $\\boxed{240}$",
"Subsets of $\\... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_13 | D | 5 | How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the remaining numbers is 6?
$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$ | [
"Since there will be $9$ elements after removal, and their mean is $6$ , we know their sum is $54$ . We also know that the sum of the set pre-removal is $66$ . Thus, the sum of the $2$ elements removed is $66-54=12$ . There are only $\\boxed{5}$",
"We can simply remove $5$ subsets of $2$ numbers while leaving onl... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_7 | B | 21 | How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$ | [
"$73-13 = 60$ , so the amount of terms in the sequence $13$ $16$ $19$ $\\dotsc$ $70$ $73$ is the same as in the sequence $0$ $3$ $6$ $\\dotsc$ $57$ $60$\nIn this sequence, the terms are the multiples of $3$ going up to $60$ , and there are $20$ multiples of $3$ in $60$\nHowever, the number 0 must also be included, ... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_18 | C | 69 | How many three-digit numbers are divisible by 13?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77$ | [
"Let $k$ be any positive integer so that $13k$ is a multiple of $13$ . For the smallest three-digit number, $13k>100$ and $k>\\frac{100}{13} \\approx 7.7$ . For the greatest three-digit number, $13k<999$ and $k<\\frac{999}{13} \\approx 76.8$ . The number $k$ can range from $8$ to $76$ so there are $\\boxed{69}$ thr... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_13 | B | 60 | How many three-digit numbers are not divisible by $5$ , have digits that sum to less than $20$ , and have the first digit equal to the third digit?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$ | [
"We use a casework approach to solve the problem. These three digit numbers are of the form $\\overline{xyx}$ .( $\\overline{abc}$ denotes the number $100a+10b+c$ ). We see that $x\\neq 0$ and $x\\neq 5$ , as $x=0$ does not yield a three-digit integer and $x=5$ yields a number divisible by 5.\nThe second condition... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_7 | A | 52 | How many three-digit numbers have at least one $2$ and at least one $3$
$\text{(A) }52 \qquad \text{(B) }54 \qquad \text{(C) }56 \qquad \text{(D) }58 \qquad \text{(E) }60$ | [
"We can do this problem with some simple case work.\nCase 1: The hundreds place is not $2$ or $3.$ This means that the tens place and ones place must be $2$ and $3$ respectively or $3$ and $2$ respectively. This case covers $1, 4, 5, 6, 7, 8,$ and $9,$ so it gives us $2 \\cdot 7 = 14$ cases.\nCase 2: The hundreds p... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_14 | E | 45 | How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
$\textbf{(A) } 41\qquad \textbf{(B) } 42\qquad \textbf{(C) } 43\qquad \textbf{(D) } 44\qquad \textbf{(E) } 45$ | [
"If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.\nDoing some casework\nIf the middle digit is $1$ , possible numbers range from $111$ to $210$ . So there are $2$ numbers in this case.\nIf the middle digit is $2$ , possibl... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_12 | B | 14 | How many three-digit positive integers $N$ satisfy the following properties?
$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$ | [
"Multiples of $5$ will always end in $0$ or $5$ , and since the numbers have to be a three-digit numbers (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with $5$ . Since the numbers must be divisible by 7, all possibilities have to be in the rang... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_17 | D | 13 | How many three-digit positive integers $\underline{a} \ \underline{b} \ \underline{c}$ are there whose nonzero digits $a,b,$ and $c$ satisfy \[0.\overline{\underline{a}~\underline{b}~\underline{c}} = \frac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?\] (The bar indicates repetition, thus $0.\overline{\unde... | [
"We rewrite the given equation, then rearrange: \\begin{align*} \\frac{100a+10b+c}{999} &= \\frac13\\left(\\frac a9 + \\frac b9 + \\frac c9\\right) \\\\ 100a+10b+c &= 37a + 37b + 37c \\\\ 63a &= 27b+36c \\\\ 7a &= 3b+4c. \\end{align*} Now, this problem is equivalent to counting the ordered triples $(a,b,c)$ that sa... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_3 | D | 450 | How many three-digit positive integers have an odd number of even digits?
$\textbf{(A) }150\qquad\textbf{(B) }250\qquad\textbf{(C) }350\qquad\textbf{(D) }450\qquad\textbf{(E) }550$ | [
"We use simple case work to solve this problem.\nCase 1: even, even, even = $4 \\cdot 5 \\cdot 5 = 100$\nCase 2: even, odd, odd = $4 \\cdot 5 \\cdot 5 = 100$\nCase 3: odd, even, odd = $5 \\cdot 5 \\cdot 5 = 125$\nCase 4: odd, odd, even = $5 \\cdot 5 \\cdot 5 = 125$\nSimply sum up the cases to get your answer. $100 ... |
https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_4 | E | 5 | How many triangles are in this figure? (Some triangles may overlap other triangles.)
[asy] draw((0,0)--(42,0)--(14,21)--cycle); draw((14,21)--(18,0)--(30,9)); [/asy]
$\text{(A)}\ 9 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 5$ | [
"By inspection, we have that there $5$ triangles: Each of the $3$ small triangles, $1$ medium triangle made of the rightmost two small triangles, and the $1$ large triangle.\n$\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_19 | null | 4 | How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$ | [
"The triangle can be seen as having the base on the $x$ axis and height $|5\\sin\\theta|$ . The length of the base is $10$ , thus the height must be $2$ . The equation $|\\sin\\theta| = \\frac 25$ has $\\boxed{4}$ solutions, one in each quadrant.",
"Alternatively, we use shoelace to get: \\[\\frac {1}{2}|(-5*0+5*... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_23 | B | 2,148 | How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$ , inclusive?
$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$ | [
"We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are $25$ points in all, from $(1,1)$ to $(5,5)$ , so $\\dbinom{25}3$ is $\\frac{25\\cdot 24\\cdot 23}{3\\cdot 2 \\cdot 1}$ , which simplifies to $2300$ .\nNow we count the ones that are on the same line. W... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_11 | C | 17 | How many two-digit numbers have digits whose sum is a perfect square?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19$ | [
"There is $1$ integer whose digits sum to $1$ $10$\nThere are $4$ integers whose digits sum to $4$ $13, 22, 31, \\text{and } 40$\nThere are $9$ integers whose digits sum to $9$ $18, 27, 36, 45, 54, 63, 72, 81, \\text{and } 90$\nThere are $3$ integers whose digits sum to $16$ $79, 88, \\text{and } 97$\nTwo digits ca... |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_13 | E | 8 | How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of is a perfect square?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | [
"Let $N = 10t + u$ , where $t$ is the tens digit and $u$ is the units digit.\nThe condition of the problem is that $10t + u + 10u + t$ is a perfect square.\nSimplifying and factoring, we want $11(t+u)$ to be a perfect square.\nThus, $t+u$ must at least be a multiple of $11$ , and since $t$ and $u$ are digits, the o... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_2 | B | 18 | How many two-digit positive integers have at least one $7$ as a digit?
$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 30$ | [
"Ten numbers $(70,71,\\dots,79)$ have $7$ as the tens digit. Nine numbers $(17,27,\\dots,97)$ have it as the ones digit. Number $77$ is in both sets.\nThus the result is $10+9-1=18 \\Rightarrow$ $\\boxed{18}$",
"We use complementary counting. The complement of having at least one $7$ as a digit is having no $7$ s... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_11 | D | 42 | How many unordered pairs of edges of a given cube determine a plane?
$\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66$ | [
"Without loss of generality, choose one of the $12$ edges of the cube to be among the two selected. We now calculate the probability that a randomly-selected second edge makes the pair satisfy the condition in the problem statement.\nFor two lines in space to determine a common plane, they must either intersect or ... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_13 | D | 6 | How many values of $\theta$ in the interval $0<\theta\le 2\pi$ satisfy \[1-3\sin\theta+5\cos3\theta = 0?\] $\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8$ | [
"We rearrange to get \\[5\\cos3\\theta = 3\\sin\\theta-1.\\] We can graph two functions in this case: $y=5\\cos{3x}$ and $y=3\\sin{x} -1$ .\nUsing transformation of functions, we know that $5\\cos{3x}$ is just a cosine function with amplitude $5$ and period $\\frac{2\\pi}{3}$ . Similarly, $3\\sin{x} -1$ is just a s... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_13 | E | 432 | How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$ | [
"The $5$ and $7$ can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of $5$ or $7$ . There are 3 ways to paint each, giving us $\\underline{9}$ ways to paint both. The $2$ is the most restrictive number. There are $\\underline{3}$ ways to paint $2$ , but with... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | E | 36 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B... | [
"Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors to these three letters, so $6$ must be multiplied after the letters are permuted in the grid. \nWLOG assume that A is in the center. \\[\\begin{tabular}{ c c c } ? & ? & ? \\\\ ? & A & ? \\\\ ? & ? & ? \\end{tabular}\\] In t... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_14 | E | 144 | How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$ | [
"Clearly, the integers from $8$ through $14$ must be in different pairs, and $7$ must pair with $14.$\nNote that $6$ can pair with either $12$ or $13.$ From here, we consider casework:\nTogether, the answer is $72+72=\\boxed{144}.$",
"As said in Solution 1, clearly, the integers from $8$ through $14$ must be in d... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_10 | E | 144 | How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$ | [
"Clearly, the integers from $8$ through $14$ must be in different pairs, and $7$ must pair with $14.$\nNote that $6$ can pair with either $12$ or $13.$ From here, we consider casework:\nTogether, the answer is $72+72=\\boxed{144}.$",
"As said in Solution 1, clearly, the integers from $8$ through $14$ must be in d... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_14 | C | 337 | How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$ | [
"The amount of twos in our sum ranges from $0$ to $1008$ , with differences of $3$ because $2 \\cdot 3 = \\operatorname{lcm}(2, 3)$\nThe possible amount of twos is $\\frac{1008 - 0}{3} + 1 \\Rightarrow \\boxed{337}$",
"You can also see that you can rewrite the word problem into an equation $2x + 3y$ $2016$ . Ther... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_4 | E | 24 | How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\t... | [
"We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.\nIgnoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:\nPeriods $1, 3, 5$\nPeriods $1, 3, 6$\nPeriods $1, 4... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_3 | E | 24 | How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\t... | [
"We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.\nIgnoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:\nPeriods $1, 3, 5$\nPeriods $1, 3, 6$\nPeriods $1, 4... |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_7 | B | 6 | How many whole numbers are between $\sqrt{8}$ and $\sqrt{80}$
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$ | [
"No... of course you're not supposed to know what the square root of 8 is, or the square root of 80. There aren't any formulas, either. Approximation seems like the best strategy.\nClearly it must be true that for any positive integers $a$ $b$ , and $c$ with $a>b>c$ \\[\\sqrt{a}>\\sqrt{b}>\\sqrt{c}\\]\nIf we let $a... |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_15 | C | 138 | How many whole numbers between $100$ and $400$ contain the digit $2$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148$ | [
"This is a very common type of counting problem that you'll see quite often. Doing this the simple way would take too long, and might even make lots of mistakes.\nIf you ever learned about complementary counting , this would be the best time to utilize it. Instead of counting how many DO have 2's, why don't we coun... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_22 | D | 728 | How many whole numbers between 1 and 1000 do not contain the digit 1?
$\textbf{(A)}\ 512 \qquad \textbf{(B)}\ 648 \qquad \textbf{(C)}\ 720 \qquad \textbf{(D)}\ 728 \qquad \textbf{(E)}\ 800$ | [
"Note that this is the same as finding how many numbers with up to three digits do not contain 1.\nSince there are 10 total possible digits, and only one of them is not allowed (1), each place value has its choice of 9 digits, for a total of $9*9*9=729$ such numbers. However, we overcounted by one; 0 is not between... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_19 | D | 162 | How many whole numbers between 99 and 999 contain exactly one 0?
$\text{(A)}\ 72\qquad\text{(B)}\ 90\qquad\text{(C)}\ 144\qquad\text{(D)}\ 162\qquad\text{(E)}\ 180$ | [
"Numbers with exactly one zero have the form $\\overline{a0b}$ or $\\overline{ab0}$ , where the $a,b \\neq 0$ . There are $(9\\cdot1\\cdot9)+(9\\cdot9\\cdot1) = 81+81 = \\boxed{162}$"
] |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_9 | B | 21 | How many whole numbers from $1$ through $46$ are divisible by either $3$ or $5$ or both?
$\text{(A)}\ 18 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 25 \qquad \text{(E)}\ 27$ | [
"There are $\\left\\lfloor \\frac{46}{3}\\right\\rfloor =15$ numbers divisible by $3$ $\\left\\lfloor\\frac{46}{5}\\right\\rfloor =9$ numbers divisible by $5$ , so at first we have $15+9=24$ numbers that are divisible by $3$ or $5$ , except we counted the multiples of $\\text{LCM}(3,5)=15$ twice, once for $3$ and o... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_3 | D | 5 | How many whole numbers lie in the interval between $\frac{5}{3}$ and $2\pi$
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}$ | [
"The smallest whole number in the interval is $2$ because $5/3$ is more than $1$ but less than $2$ . The largest whole number in the interval is $6$ because $2\\pi$ is more than $6$ but less than $7$ . There are five whole numbers in the interval. They are $2$ $3$ $4$ $5$ , and $6$ , so the answer is $\\boxed{5}$"
... |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_13 | C | 9 | How many zeros are at the end of the product \[25\times 25\times 25\times 25\times 25\times 25\times 25\times 8\times 8\times 8?\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$ | [
"\\begin{align*} (5^2)^7\\times (2^3)^3 &= 5^{14}\\times 2^{9} \\\\ &= 5^5\\times 10^9 \\end{align*} Since $5^5$ doesn't end in a $0$ , we can conclude that the product ends in $9$ zeroes $\\rightarrow \\boxed{9}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_21 | C | 240 | Hui is an avid reader. She bought a copy of the best seller Math is Beautiful . On the first day, Hui read $1/5$ of the pages plus $12$ more, and on the second day she read $1/4$ of the remaining pages plus $15$ pages. On the third day she read $1/3$ of the remaining pages plus $18$ pages. She then realized that there ... | [
"Let $x$ be the number of pages in the book. After the first day, Hui had $\\frac{4x}{5}-12$ pages left to read. After the second, she had $\\left(\\frac{3}{4}\\right)\\left(\\frac{4x}{5}-12\\right)-15 = \\frac{3x}{5}-24$ left. After the third, she had $\\left(\\frac{2}{3}\\right)\\left(\\frac{3x}{5}-24\\right)-18=... |
https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_2 | D | 8 | I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$ | [
"Let the numbers be $x$ and $y$ . Then we have $x+y=11$ and $xy=24$ . Solving for $x$ in the first equation yields $x=11-y$ , and substituting this into the second equation gives $(11-y)(y)=24$ . Simplifying this gives $-y^2+11y=24$ , or $y^2-11y+24=0$ . This factors as $(y-3)(y-8)=0$ , so $y=3$ or $y=8$ , and the ... |
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_10 | D | 128 | If $(3x-1)^7 = a_7x^7 + a_6x^6 + \cdots + a_0$ , then $a_7 + a_6 + \cdots + a_0$ equals
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 64 \qquad \text{(D)}\ -64 \qquad \text{(E)}\ 128$ | [
"Solution by e_power_pi_times_i\nNotice that if $x=1$ , then $a_7x^7 + a_6x^6 + \\cdots + a_0 = a_7 + a_6 + \\cdots + a_0$ . Therefore the answer is $(3(1)-1)^7) = \\boxed{128}$"
] |
https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_15 | null | 69 | If $(x, y)$ is a solution to the system $xy=6$ and $x^2y+xy^2+x+y=63$ ,
find $x^2+y^2$
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ \frac{1173}{32} \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 69 \qquad \textbf{(E)}\ 81$ | [
"First note that $x^2y+xy^2+x+y= (xy+1)(x+y)$ . Substituting $6$ for $xy$ gives $7(x+y)= 63$ , giving a result of $x+y=9$ . Squaring this equation and subtracting by $12$ , gives us $x^2+y^2= \\boxed{69}$"
] |
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_1 | B | 1 | If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad \textbf{(E) }-1\text{ or }-2$ | [
"By guessing and checking, 2 works. $\\frac{2}{x} = \\boxed{1}$ ~awin",
"Multiplying each side by $x^2$ , we get $x^2-4x+4 = 0$ . Factoring, we get $(x-2)(x-2) = 0$ . Therefore, $x = 2$ $\\frac{2}{x} = \\boxed{1}$ ~awin",
"Directly factoring, we get $(1-\\frac{2}{x})^2 = 0$ . Thus $\\frac{2}{x}$ must equal $\\b... |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_6 | C | 17 | If $1998$ is written as a product of two positive integers whose difference is as small as possible, then the difference is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }17 \qquad \mathrm{(D) \ }47 \qquad \mathrm{(E) \ } 93$ | [
"If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to $\\sqrt{1998}$ as possible.\nSince $45^2 = 2025$ , the factors should be as close to $44$ or $45$ as possible.\nBreaking down $1998$ into its prime factors gives $1998 = 2\\cdot 3^3 \\cdot 37$\n$37$ is... |
https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_1 | E | 9 | If $2$ is a solution (root) of $x^3+hx+10=0$ , then $h$ equals:
$\textbf{(A) }10\qquad \textbf{(B) }9 \qquad \textbf{(C) }2\qquad \textbf{(D) }-2\qquad \textbf{(E) }-9$ | [
"Substitute $2$ for $x$ . We are given that this equation is true. Solving for $h$ gives $h=-9$ . The answer is $\\boxed{9}$"
] |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_5 | B | 18 | If $20\%$ of a number is $12$ , what is $30\%$ of the same number?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 30$ | [
"$20\\%$ of a number is equal to $\\frac{1}{5}$ of that number. Let $n$ =the number\n$\\frac{1}{5}n = 12$ Multiply both sides by 5\n$n = 60$\n$30\\%$ of $n$ is equal to $\\frac{3}{10}n = \\frac{3}{10}\\cdot60 = 3\\cdot6 = \\boxed{18}$"
] |
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_28 | D | 7 | If $2137^{753}$ is multiplied out, the units' digit in the final product is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$ | [
"$7^1$ has a unit digit of $7$ $7^2$ has a unit digit of $9$ $7^3$ has a unit digit of $3$ $7^4$ has a unit digit of $1$ $7^5$ has a unit digit of $7$\nNotice that the unit digit eventually cycles to itself when the exponent is increased by $4$ . It also does not matter what the other digits are in the base becaus... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_13 | E | 0 | If $2^a+2^b=3^c+3^d$ , the number of integers $a,b,c,d$ which can possibly be negative, is, at most:
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 0$ | [
"Assume $c,d \\ge 0$ , and WLOG, assume $a<0$ and $a \\le b$ . This also takes into account when $b$ is negative. That means \\[\\frac{1}{2^{-a}} + 2^b = 3^c + 3^d\\] Multiply both sides by $2^{-a}$ to get \\[1 + 2^{-a+b} = 2^{-a} (3^c + 3^d)\\] Note that both sides are integers. If $a \\ne b$ , then the right s... |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_5 | C | 3 | If $2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},$ what is the value of $k$
$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ } 5$ | [
"Divide both sides of the original equation by $2^{1995}$ , giving:\n$2^3 - 2^2 - 2^1 + 1 = k$\n$k = 3$ , and the answer is $\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_1 | A | 15 | If $2x+1=8$ , then $4x+1=$
$\mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ }19$ | [
"We have \\begin{align*}2x+1 = 8 &\\iff 2x = 7 \\\\ &\\iff x = \\frac{7}{2},\\end{align*} so \\begin{align*}4x+1 &= 4\\left(\\frac{7}{2}\\right)+1 \\\\ &= 2(7)+1 \\\\ &= \\boxed{15}",
"From $2x = 7$ (as above), we can directly compute \\begin{align*}4x &= 2(2x) \\\\ &= 2(7) \\\\ &= 14,\\end{align*} so $4x+1 = 14+... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_25 | B | 2 | If $60^a=3$ and $60^b=5$ , then $12^{(1-a-b)/\left(2\left(1-b\right)\right)}$ is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad$ | [
"We have that $12=\\frac{60}{5}$ . We can substitute our value for 5, to get \\[12=\\frac{60}{60^b}=60\\cdot 60^{-b}=60^{1-b}.\\] Hence \\[12^{(1-a-b)/\\left(2\\left(1-b\\right)\\right)}=60^{(1-b)(1-a-b)/\\left(2\\left(1-b\\right)\\right)}=60^{(1-a-b)/2}.\\] Since $4=\\frac{60}{3\\cdot 5}$ , we have \\[4=\\frac{60}... |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_4 | E | 25 | If $991+993+995+997+999=5000-N$ , then $N=$
$\text{(A)}\ 5 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 25$ | [
"\\begin{align*} 991+993+995+997+999=5000-N \\\\ &\\Rightarrow (1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1) = 5000-N \\\\ &\\Rightarrow 5\\times 1000-(1+3+5+7+9) = 5000 -N \\\\ &\\Rightarrow 5000-25=5000-N \\\\ &\\Rightarrow N=25\\rightarrow \\boxed{25}"
] |
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_33 | E | 0.5 | If $9^{x + 2} = 240 + 9^x$ , then the value of $x$ is:
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2\qquad \textbf{(C)}\ 0.3\qquad \textbf{(D)}\ 0.4\qquad \textbf{(E)}\ 0.5$ | [
"$9^{x+2}$ can be rewritten as $9^x*9^2=9^x*81$ , which means the equation can be rewritten as $81(9^x)=240+9^x$ , or $80(9^x)=240$ , or $9^x=3$ . Therefore, $x=\\boxed{0.5}$"
] |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_25 | null | 20 | If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$ . For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$ , but if $\overline{AC}$ and $\overline{CB}$ are edges and $\... | [
"Since the icosahedron is symmetric polyhedron, we can rotate it so that R is on the topmost vertex. Since Q and \nS basically the same, we can first count the probability that $d(Q,R) = d(R,S)$\n$\\mathfrak{Case} \\ \\mathfrak{1}: d(Q,R) = d(R,S) = 1$\nThere are 5 points $P$ such that $d(Q,P) = 1$ . There is $5 \\... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_21 | null | 20 | If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$ . For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$ , but if $\overline{AC}$ and $\overline{CB}$ are edges and $\... | [
"Since the icosahedron is symmetric polyhedron, we can rotate it so that R is on the topmost vertex. Since Q and \nS basically the same, we can first count the probability that $d(Q,R) = d(R,S)$\n$\\mathfrak{Case} \\ \\mathfrak{1}: d(Q,R) = d(R,S) = 1$\nThere are 5 points $P$ such that $d(Q,P) = 1$ . There is $5 \\... |
https://artofproblemsolving.com/wiki/index.php/1966_AHSME_Problems/Problem_25 | D | 52 | If $F(n+1)=\frac{2F(n)+1}{2}$ for $n=1,2,\cdots$ and $F(1)=2$ , then $F(101)$ equals:
$\text{(A) } 49 \quad \text{(B) } 50 \quad \text{(C) } 51 \quad \text{(D) } 52 \quad \text{(E) } 53$ | [
"Notice that $\\frac{2F(n)+1}{2}=F(n)+\\frac{1}{2}.$\nThis means that for every single increment $n$ goes up from $1$ $F(n)$ will increase by $\\frac{1}{2}.$ Since $101$ is $100$ increments from $1$ $F(n)$ will increase $\\frac{1}{2}\\times100=50.$\nSince $F(1)=2,$ $F(101)$ will equal $2+50=\\boxed{52}.$"
] |
https://artofproblemsolving.com/wiki/index.php/1958_AHSME_Problems/Problem_12 | A | 1 | If $P = \frac{s}{(1 + k)^n}$ then $n$ equals:
$\textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 + k)}}\qquad \textbf{(B)}\ \log{\left(\frac{s}{P(1 + k)}\right)}\qquad \textbf{(C)}\ \log{\left(\frac{s - P}{1 + k}\right)}\qquad \\ \textbf{(D)}\ \log{\left(\frac{s}{P}\right)} + \log{(1 + k)}\qquad \textbf{(E... | [
"\\[P=\\frac{s}{(1+k)^n}\\]\n\\[(1+k)^n=\\frac{s}{P}\\]\nTake the $\\log$ of each side.\n\\[n \\log(1+k) = \\log\\left(\\frac{s}{P}\\right)\\]\n\\[n = \\frac{\\log\\left(\\frac{s}{P}\\right)}{\\log(1+k)} \\to \\boxed{1}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_42 | C | 3 | If $S = i^n + i^{-n}$ , where $i = \sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4}$ | [
"We first use the fact that $i^{-n}=\\frac1{i^n}=\\left(\\frac1i\\right)^n=(-i)^n$ . Note that $i^4=1$ and $(-i)^4=1$ , so $i^n$ and $(-i)^n$ have are periodic with periods at most 4. Therefore, it suffices to check for $n=0,1,2,3$\nFor $n=0$ , we have $i^0+(-i)^0=1+1=2$\nFor $n=1$ , we have $i^1+(-i)^1=i-i=0$\nFor... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_24 | D | 80 | If $\angle A = 20^\circ$ and $\angle AFG =\angle AGF$ , then $\angle B+\angle D =$
[asy] pair A,B,C,D,EE,F,G; A = (0,0); B = (9,4); C = (21,0); D = (13,-12); EE = (4,-16); F = (13/2,-6); G = (8,0); draw(A--C--EE--B--D--cycle); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,SE); label("$E$",EE,SW); ... | [
"As a strategy, think of how $\\angle B + \\angle D$ would be determined, particularly without determining either of the angles individually, since it may not be possible to determine $\\angle B$ or $\\angle D$ alone. If you see $\\triangle BFD$ , then you can see that the problem is solved quickly after determini... |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_7 | B | 50 | If $\angle A = 60^\circ$ $\angle E = 40^\circ$ and $\angle C = 30^\circ$ , then $\angle BDC =$
[asy] pair A,B,C,D,EE; A = origin; B = (2,0); C = (5,0); EE = (1.5,3); D = (1.75,1.5); draw(A--C--D); draw(B--EE--A); dot(A); dot(B); dot(C); dot(D); dot(EE); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,SE); label("$D$... | [
"The sum of the angles in a triangle is $180^\\circ$ . We can find $\\angle ABE = 80^\\circ$ , so $\\angle CBD = 180-80=100^\\circ$\n\\[\\angle BDC = 180-100-30=\\boxed{50}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_9 | D | 18 | If $\angle A$ is four times $\angle B$ , and the complement of $\angle B$ is four times the complement of $\angle A$ , then $\angle B=$
$\textbf{(A)}\ 10^{\circ} \qquad\textbf{(B)}\ 12^{\circ} \qquad\textbf{(C)}\ 15^{\circ} \qquad\textbf{(D)}\ 18^{\circ} \qquad\textbf{(E)}\ 22.5^{\circ}$ | [
"Let $\\angle A=x$ and $\\angle B=y$ . From the first condition, we have $x=4y$ . From the second condition, we have \\[90-y=4(90-x).\\] Substituting $x=4y$ into the previous equation and solving yields \\begin{align*}90-y=4(90-4y)&\\implies 90-y=360-16y\\\\&\\implies 15y=270\\\\&\\implies y=\\boxed{18}"
] |
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_5 | C | 50 | If $\angle \text{CBD}$ is a right angle , then this protractor indicates that the measure of $\angle \text{ABC}$ is approximately
[asy] unitsize(36); pair A,B,C,D; A=3*dir(160); B=origin; C=3*dir(110); D=3*dir(20); draw((1.5,0)..(0,1.5)..(-1.5,0)); draw((2.5,0)..(0,2.5)..(-2.5,0)--cycle); draw(A--B); draw(C--B); draw(D... | [
"We have that $20^{\\circ}+\\angle ABD +\\angle CBD=160^{\\circ}$ , or $\\angle ABD +\\angle CBD=140^{\\circ}$ . Since $\\angle CBD$ is a right angle, we have $\\angle ABD=140^{\\circ}-90^{\\circ}=50^{\\circ}\\Rightarrow \\boxed{50}$"
] |
https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_2 | E | 2 | If $\begin{tabular}{r|l}a&b \\ \hline c&d\end{tabular} = \text{a}\cdot \text{d} - \text{b}\cdot \text{c}$ , what is the value of $\begin{tabular}{r|l}3&4 \\ \hline 1&2\end{tabular}$
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | [
"Plugging in values for $a$ $b$ $c$ , and $d$ , we get\n$a=3$ $b=4$ $c=1$ $d=2$\n$a\\times d=3\\times2=6$\n$b\\times c=4\\times1=4$\n$6-4=2$\n$\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/1990_AHSME_Problems/Problem_1 | E | 8 | If $\dfrac{\frac{x}{4}}{2}=\dfrac{4}{\frac{x}{2}}$ , then $x=$
$\text{(A)}\ \pm\frac{1}{2}\qquad\text{(B)}\ \pm 1\qquad\text{(C)}\ \pm 2\qquad\text{(D)}\ \pm 4\qquad\text{(E)}\ \pm 8$ | [
"Cross-multiplying leaves\n\\begin{align*}\\dfrac{x^2}{8} &= 8\\\\ x^2 &= 64\\\\ \\sqrt{x^2} &= \\sqrt{64}\\\\ x &= \\pm 8\\end{align*}\nSo the answer is $\\boxed{8}$"
] |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_12 | D | 1,991 | If $\frac{2+3+4}{3}=\frac{1990+1991+1992}{N}$ , then $N=$
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 1990 \qquad \text{(D)}\ 1991 \qquad \text{(E)}\ 1992$ | [
"Note that for all integers $n\\neq 0$ \\[\\frac{(n-1)+n+(n+1)}{n}=3.\\] Thus, we must have $N=1991\\rightarrow \\boxed{1991}$",
"As we know that $\\frac{1990+1991+1992}{n}$ has to be some multiple of $\\frac{2+3+4}{3}$ , then we know that the first equation is $995$ (1990/2) times bigger than the second one(in m... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_7 | E | 127 | If $\frac{3}{5}=\frac{M}{45}=\frac{60}{N}$ , what is $M+N$
$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 105\qquad \textbf{(E)}\ 127$ | [
"Separate into two equations $\\frac35 = \\frac{M}{45}$ and $\\frac35 = \\frac{60}{N}$ and solve for the unknowns. $M=27$ and $N=100$ , therefore $M+N=\\boxed{127}$"
] |
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_44 | E | 2 | If $\frac{xy}{x+y}= a,\frac{xz}{x+z}= b,\frac{yz}{y+z}= c$ , where $a, b, c$ are other than zero, then $x$ equals:
$\textbf{(A)}\ \frac{abc}{ab+ac+bc}\qquad\textbf{(B)}\ \frac{2abc}{ab+bc+ac}\qquad\textbf{(C)}\ \frac{2abc}{ab+ac-bc}$ $\textbf{(D)}\ \frac{2abc}{ab+bc-ac}\qquad\textbf{(E)}\ \frac{2abc}{ac+bc-ab}$ | [
"Note that $\\frac{1}{a}=\\frac{1}{x}+\\frac{1}{y}$ $\\frac{1}{b}=\\frac{1}{x}+\\frac{1}{z}$ , and $\\frac{1}{c}=\\frac{1}{y}+\\frac{1}{z}$ . Therefore\n\\[\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=\\frac{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}}{2}\\]\nTherefore\n\\[\\frac{1}{x}=\\frac{1}{2a}+\\frac{1}{2b}+\\frac{1}{2c... |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_26 | C | 0 | If $\left(r+\frac1r\right)^2=3$ , then $r^3+\frac1{r^3}$ equals
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 6$ | [
"We know $r+\\frac1r=\\sqrt3$ . Cubing this gives $r^3+3r+\\frac3r+\\frac1{r^3}=3\\sqrt3$ . But $3r+\\frac3r=3\\left(r+\\frac1r\\right)=3\\sqrt3$ , so subtracting this from the first equation gives $r^3+\\frac1{r^3}=\\boxed{0}$ . \n(Actually, $r+\\frac1r$ could have been equal to $-\\sqrt3$ instead of $\\sqrt3$ , b... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_17 | D | 35 | If $\log (xy^3) = 1$ and $\log (x^2y) = 1$ , what is $\log (xy)$
$\mathrm{(A)}\ -\frac 12 \qquad\mathrm{(B)}\ 0 \qquad\mathrm{(C)}\ \frac 12 \qquad\mathrm{(D)}\ \frac 35 \qquad\mathrm{(E)}\ 1$ | [
"We rewrite the logarithms in the problem. \\[\\log(x) + 3\\log(y) = 1\\] \\[2\\log(x) + \\log(y) = 1\\] \\[\\log(x) + \\log(y) = c\\] where $c$ is the desired quantity. Set $u = \\log(x)$ and $v = \\log(y)$ . Then we have that \\[u + 3y = 1 \\textbf{(1)}\\] \\[2u + y = 1 \\textbf{(2)}\\] \\[u + v = c\\] . Notice t... |
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_38 | B | 1.47 | If $\log 2 = .3010$ and $\log 3 = .4771$ , the value of $x$ when $3^{x+3} = 135$ is approximately
$\textbf{(A) \ } 5 \qquad \textbf{(B) \ } 1.47 \qquad \textbf{(C) \ } 1.67 \qquad \textbf{(D) \ } 1.78 \qquad \textbf{(E) \ } 1.63$ | [
"Taking the logarithm in base $3$ of both sides, we get $x+3 = \\log_3 135$ . Using the property $\\log ab = \\log a + \\log b$ , we get $x+3 = \\log_3 5 + \\log_3 3^3$ , or $x = \\log_3 5$ . Converting into base $10$ gives $x = \\frac{\\log 5}{\\log 3} = \\frac{1 - \\log 2}{\\log 3}$ . Now, plugging in the values ... |
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_17 | C | 2.43 | If $\log x-5 \log 3=-2$ , then $x$ equals:
$\textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25}$ | [
"$\\log x-5 \\log 3=-2$\n$\\log x- \\log 3^5=-2$\n$\\log x- \\log 243 =-2$\n$\\log x / 243 = -2$\n$x/243 = 10^{-2}$\n$x=\\frac{243}{100}$ $x=\\boxed{2.43}$"
] |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_26 | E | 10 | If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$ | [
"We have $b=\\log_{10}{10^b}$ . Substituting, we find $\\log_{10}{m}= \\log_{10}{10^b}-\\log_{10}{n}$ . Using $\\log{a}-\\log{b}=\\log{\\dfrac{a}{b}}$ , the left side becomes $\\log_{10}{\\dfrac{10^b}{n}}$ . Because $\\log_{10}{m}=\\log_{10}{\\dfrac{10^b}{n}}$ $m=\\boxed{10}$",
"adding $\\log_{10} n$ to both side... |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_15 | A | 18 | If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$ , then one value for $x$ is
$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$ | [
"We divide both sides of the equation by $\\cos{2x}\\times\\cos{3x}$ to get $\\frac{\\sin{2x}\\times\\sin{3x}}{\\cos{2x}\\times\\cos{3x}}=1$ , or $\\tan{2x}\\times\\tan{3x}=1$\nThis looks a lot like the formula relating the slopes of two perpendicular lines , which is $m_1\\times m_2=-1$ , where $m_1$ and $m_2$ are... |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_15 | null | 18 | If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$ , then one value for $x$ is
$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$ | [
"Start by subtracting $\\sin{2x}\\sin{3x}$ from both sides to get $\\cos{2x}\\cos{3x}-\\sin{2x}\\sin{3x}=0$ . We recognize that this is of the form $\\cos{a}\\cos{b}-\\sin{a}\\sin{b}=\\cos{(a+b)}$ , so $\\cos{2x}\\cos{3x}-\\sin{2x}\\sin{3x}=\\cos{(2x+3x)}=0$ $\\cos{90^\\circ}=0$ , so $x=\\boxed{18}$"
] |
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_1 | E | 16 | If $\sqrt{x+2}=2$ , then $(x+2)^2$ equals:
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16$ | [
"If we square both sides of the $\\sqrt{x+2} = 2$ , we will get $x+2 = 4$ , if we square that again, we get $(x+2)^2 = \\boxed{16}$",
"We can immediately get that $x = 2$ , after we square $(2+2)$ , we get $\\boxed{16}$"
] |
https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_3 | null | 150 | If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$ | [
"Since $\\cot$ is the reciprocal function of $\\tan$\n$\\cot x + \\cot y = \\frac{1}{\\tan x} + \\frac{1}{\\tan y} = \\frac{\\tan x + \\tan y}{\\tan x \\cdot \\tan y} = 30$\nThus, $\\tan x \\cdot \\tan y = \\frac{\\tan x + \\tan y}{30} = \\frac{25}{30} = \\frac{5}{6}$\nUsing the tangent addition formula:\n$\\tan(x+... |
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_20 | C | 2 | If $\tan{\alpha}$ and $\tan{\beta}$ are the roots of $x^2 - px + q = 0$ , and $\cot{\alpha}$ and $\cot{\beta}$ are the roots of $x^2 - rx + s = 0$ , then $rs$ is necessarily
$\textbf{(A)} \ pq \qquad \textbf{(B)} \ \frac{1}{pq} \qquad \textbf{(C)} \ \frac{p}{q^2} \qquad \textbf{(D)}\ \frac{q}{p^2}\qquad \textbf{(E)}... | [
"By Vieta's Formulae, we have $\\tan(\\alpha)\\tan(\\beta)=q$ and $\\cot(\\alpha)\\cot(\\beta)=s$ . Recalling that $\\cot\\theta=\\frac{1}{\\tan\\theta}$ , we have $\\frac{1}{\\tan(\\alpha)\\tan(\\beta)}=\\frac{1}{q}=s$\nAlso by Vieta's Formulae, we have $\\tan(\\alpha)+\\tan(\\beta)=p$ and $\\cot(\\alpha)+\\cot(\\... |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_3 | D | 17 | If $\texttt{a,b,}$ and $\texttt{c}$ are digits for which
then $\texttt{a+b+c =}$
$\mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18$ | [
"Working from right to left, we see that $2 - b = 3$ . Clearly if $b$ is a single digit integer, this cannot be possible. Therefore, there must be some borrowing from $a$ . Borrow $1$ from the digit $a$ , and you get $12 - b = 3$ , giving $b = 9$\nSince $1$ was borrowed from $a$ , we have from the tens column $(... |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_1 | C | 7 | If $\texttt{a}$ and $\texttt{b}$ are digits for which
$\begin{array}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{array}$
then $\texttt{a+b =}$
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }12$ | [
"From the units digit calculation, we see that the units digit of $a\\times 3$ is $9$ . Since $0 \\le a \\le 9$ and $a$ is an integer, the only value of $a$ that works is is $a=3$ . As a double-check, that does work, since $23 \\times 3 = 69$ , which is the first line of the multiplication.\nThe second line of th... |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_11 | A | 6 | If $\text{A}*\text{B}$ means $\frac{\text{A}+\text{B}}{2}$ , then $(3*5)*8$ is
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16\qquad \text{(E)}\ 30$ | [
"We just plug in and evaluate: \\begin{align*} (3*5)*8 &= \\left( \\frac{3+5}{2}\\right) *8 \\\\ &= 4*8 \\\\ &= \\frac{4+8}{2} \\\\ &= 6 \\\\ \\end{align*}\n$\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_24 | null | 2 | If $\theta$ is a constant such that $0 < \theta < \pi$ and $x + \dfrac{1}{x} = 2\cos{\theta}$ , then for each positive integer $n$ $x^n + \dfrac{1}{x^n}$ equals
$\textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\... | [
"Multiply both sides by $x$ and rearrange to $x^2-2x\\cos(\\theta)+1=0$ . Using the quadratic equation, we can solve for $x$ . After some simplifying:\n\\[x=\\cos(\\theta) + \\sqrt{\\cos^2(\\theta)-1}\\] \\[x=\\cos(\\theta) + \\sqrt{(-1)(\\sin^2(\\theta))}\\] \\[x=\\cos(\\theta) + i\\sin(\\theta)\\]\nSubstituting t... |
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_17 | A | 1 | If $\theta$ is an acute angle, and $\sin 2\theta=a$ , then $\sin\theta+\cos\theta$ equals
$\textbf{(A) }\sqrt{a+1}\qquad \textbf{(B) }(\sqrt{2}-1)a+1\qquad \textbf{(C) }\sqrt{a+1}-\sqrt{a^2-a}\qquad\\ \textbf{(D) }\sqrt{a+1}+\sqrt{a^2-a}\qquad \textbf{(E) }\sqrt{a+1}+a^2-a$ | [
"Let $x = \\sin\\theta+\\cos\\theta$ , so we want to find $x$ . First, square the expression to get $x^2 = \\sin^2 \\theta + 2\\sin\\theta\\cos\\theta + \\cos^2 \\theta$ . Recall that $\\sin^2 \\theta + \\cos^2 \\theta = 1$ and $\\sin 2\\theta = 2\\sin\\theta\\cos\\theta$ . Plugging these in, the equation simplifie... |
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_32 | D | 2 | If $\triangle ABC$ is inscribed in a semicircle whose diameter is $AB$ , then $AC+BC$ must be
$\textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2}$ $\textbf{(E)}\ AB^{2}$ | [
"Because $AB$ is the diameter of the semi-circle, it follows that $\\angle C = 90$ . Now we can try to eliminate all the solutions except for one by giving counterexamples.\n$\\textbf{(A):}$ Set point $C$ anywhere on the perimeter of the semicircle except on $AB$ . By triangle inequality, $AC+BC>AB$ , so $\\textbf{... |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_16 | B | 2 | If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$ , then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none... | [
"Noting that $\\usepackage{gensymb} 25 \\degree + 20 \\degree = 45 \\degree$ , we apply the angle sum formula \\[\\tan(A+B) = \\frac{\\tan A+\\tan B}{1-\\tan A\\tan B},\\] giving \\begin{align*}1 &= \\tan 45^{\\circ} \\\\ &= \\tan(A+B) \\\\ &= \\frac{\\tan A+\\tan B}{1-\\tan A\\tan B},\\end{align*} so \\[\\tan A + ... |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_8 | A | 3 | If $a = - 2$ , the largest number in the set $\{ - 3a, 4a, \frac {24}{a}, a^2, 1\}$ is
$\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1$ | [
"Since all the numbers are small, we can just evaluate the set to be \\[\\{ (-3)(-2), 4(-2), \\frac{24}{-2}, (-2)^2, 1 \\}= \\{ 6, -8, -12, 4, 1 \\}\\]\nThe largest number is $6$ , which corresponds to $-3a$\n$\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_4 | C | 3,333 | If $a = 1,~ b = 10, ~c = 100$ , and $d = 1000$ , then $(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)$ is equal to
$\textbf{(A) }1111\qquad \textbf{(B) }2222\qquad \textbf{(C) }3333\qquad \textbf{(D) }1212\qquad \textbf{(E) }4242$ | [
"Adding all four of the equations up, we can see that it equals \\[3(a+b+c+d)\\] This is equal to $3(1111) = \\boxed{3333}$ ~awin"
] |
https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_15 | A | 2 | If $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1$ , then $b$ is
$\textbf{(A)}\ -2\qquad \textbf{(B)}\ -1\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2$ | [
"Using polynomial division, we find that the remainder is $(2a+b)x+(a+b+1)$ , so for the condition to hold, we need this remainder to be $0$ . This gives $2a+b=0$ and $a+b+1=0$ , so $b=-2a$ and $a-2a+1=0 \\implies a=1 \\implies b=-2$ , which is $\\boxed{2}.$"
] |
https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_24 | E | 6 | If $a$ and $b$ are positive real numbers and each of the equations $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real roots , then the smallest possible value of $a+b$ is
$\mathrm{(A) \ }2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ } 6$ | [
"Since both of the equations have real roots, both of their discriminants are nonnegative. Therefore, we have\n$a^2-4(1)(2b)=a^2-8b\\geq0\\implies a^2\\geq8b$ from the first equation, and\n$(2b)^2-4(1)(a)=4b^2-4a\\geq0\\implies 4b^2\\geq4a\\implies b^2\\geq a$ from the second.\nWe can square the second equation to ... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.