link stringlengths 75 84 | letter stringclasses 5
values | answer float64 0 2,935,363,332B | problem stringlengths 14 5.33k | solution sequencelengths 1 13 |
|---|---|---|---|---|
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_20 | A | 9 | "If a whole number $n$ is not prime, then the whole number $n-2$ is not prime." A value of $n$ which shows this statement to be false is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 23$ | [
"To show this statement to be false, we need a non-prime value of $n$ such that $n-2$ is prime. Since $13$ and $23$ are prime, they won't prove anything relating to the truth of the statement.\nNow we just check the statement for $n=9,12,16$ . If $n=12$ or $n=16$ , then $n-2$ is $10$ or $14$ , which aren't prime.... |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_1 | E | 250 | $(1+11+21+31+41)+(9+19+29+39+49)=$
$\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250$ | [
"We make use of the associative and commutative properties of addition to rearrange the sum as \\begin{align*} (1+49)+(11+39)+(21+29)+(31+19)+(41+9) &= 50+50+50+50+50 \\\\ &= 250 \\Longrightarrow \\boxed{250}",
"Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This giv... |
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_19 | A | 167,400 | $(1901+1902+1903+\cdots + 1993) - (101+102+103+\cdots + 193) =$
$\text{(A)}\ 167,400 \qquad \text{(B)}\ 172,050 \qquad \text{(C)}\ 181,071 \qquad \text{(D)}\ 199,300 \qquad \text{(E)}\ 362,142$ | [
"We see that $1901=1800+101$ $1902=1800+102$ , etc. Each term in the first set of numbers is $1800$ more than the corresponding term in the second set; Because there are $93$ terms in the first set, the expression can be paired up as follows and simplified:\n\\[(1901-101) + (1902-102) + (1903-103) + \\cdots + (1993... |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_8 | E | 26 | $(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$ | [
"We use the distributive property to get \\[3\\times 4+2\\times 4+2\\times 3 = 26 \\rightarrow \\boxed{26}\\]",
"Since $\\frac12+\\frac13+\\frac14 > \\frac12+\\frac14+\\frac14 = 1$ , we have \\[(2\\times 3\\times 4)\\left(\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}\\right) > 2\\times 3\\times 4 \\times 1 = 24\\] The o... |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_5 | D | 3 | $-15+9\times (6\div 3) =$
$\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12$ | [
"We use the order of operations here to get\n\\begin{align*} -15+9\\times (6\\div 3) &= -15+9\\times 2 \\\\ &= -15+18 \\\\ &= 3 \\rightarrow \\boxed{3}"
] |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_1 | E | 0.426 | $.4+.02+.006=$
$\text{(A)}\ .012 \qquad \text{(B)}\ .066 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .24 \qquad \text{(E)} .426$ | [
"$.4+.02+.006 = .400 + .020 + .006 = .426\\rightarrow \\boxed{.426}$"
] |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_1 | B | 222,222,222,223 | $1,000,000,000,000-777,777,777,777=$
$\text{(A)}\ 222,222,222,222 \qquad \text{(B)}\ 222,222,222,223 \qquad \text{(C)}\ 233,333,333,333 \qquad \\ \text{(D)}\ 322,222,222,223 \qquad \text{(E)}\ 333,333,333,333$ | [
"\\begin{align*} 1,000,000,000,000-777,777,777,777 &= 999,999,999,999-777,777,777,777+1 \\\\ &= 222,222,222,222+1 \\\\ &= 222,222,222,223 \\rightarrow \\boxed{222,222,222,223}"
] |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_16 | C | 0 | $1-2-3+4+5-6-7+8+9-10-11+\cdots + 1992+1993-1994-1995+1996=$
$\text{(A)}\ -998 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 998$ | [
"Put the numbers in groups of $4$\n$(1-2-3+4)+(5-6-7+8)+(9-10-11+ 12) + \\cdots + (1993-1994-1995+1996)$\nThe first group has a sum of $0$\nThe second group increases the two positive numbers on the end by $1$ , and decreases the two negative numbers in the middle by $1$ . Thus, the second group also has a sum of ... |
https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_14 | null | 168 | $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$ , where $i$ $j$ , and $k$ are integers between $1$ and $10$ , inclusive. Find the number of different lines that contain exactly $8$ of these points. | [
"$Case \\textrm{ } 1:$ The lines are not parallel to the faces\nA line through the point $(a,b,c)$ must contain $(a \\pm 1, b \\pm 1, c \\pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.\nWe look at the one from $(1,1,1)$ to $(1... |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_14 | null | 768 | $150\times 324\times 375$ rectangular solid is made by gluing together $1\times 1\times 1$ cubes. An internal diagonal of this solid passes through the interiors of how many of the $1\times 1\times 1$ cubes | [
"Place one corner of the solid at $(0,0,0)$ and let $(a,b,c)$ be the general coordinates of the diagonally opposite corner of the rectangle, where $a, b, c \\in \\mathbb{Z_{+}}$\nWe consider the vector equation of the diagonal segment represented by: $(x, y, z) =m (a, b, c)$ , where $m \\in \\mathbb{R}, 0 < m \\le ... |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_16 | D | 1,000 | $1990-1980+1970-1960+\cdots -20+10 =$
$\text{(A)}\ -990 \qquad \text{(B)}\ -10 \qquad \text{(C)}\ 990 \qquad \text{(D)}\ 1000 \qquad \text{(E)}\ 1990$ | [
"In the middle, we have $\\cdots + 1010-1000+990 -\\cdots$\nIf we match up the back with the front, and then do the same for the rest, we get pairs with $2000$ and $-2000$ , so these will cancel out. In the middle, we have $2000-1000$ which doesn't cancel, but gives us $1000 \\rightarrow \\boxed{1000}$",
"We can... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_5 | B | 25 | $2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?
$\textbf{(A) } 16\qquad \textbf{(B) } 25\qquad \textbf{(C) } 36\qquad \textbf{(D) } 49\qquad \textbf{(E) } 64$ | [
"By placing the $2 \\times 3$ rectangle adjacent to the $3 \\times 4$ rectangle with the 3 side of the $2 \\times 3$ rectangle next to the 4 side of the $3 \\times 4$ rectangle, we get a figure that can be completely enclosed in a square with a side length of 5. The area of this square is $5^2 = 25$\nSince placing ... |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_24 | B | 6 | $2$ by $2$ square is divided into four $1$ by $1$ squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares.
$\text{(A)}\ 4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 16$ | [
"If a green square cannot share its top or right side with a red square, then a red square can not share its bottom or left side with a green square. Let us split this up into several cases.\nCase 1: There are no green squares. This can be done in $1$ way.\nCase 2: There is one green square and three red squares. T... |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_3 | E | 1,800 | $2(81+83+85+87+89+91+93+95+97+99)=$
$\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$ | [
"Find that \\[(81+83+85+87+89+91+93+95+97+99) = 5 \\cdot 180\\] Which gives us \\begin{align*} 2(5 \\cdot 180) &= 10 \\cdot 180\\\\ &= 1800 & \\text{ Thus \\boxed{1800}",
"$2(81+83+85+87+89+91+93+95+97+99)$ Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, ea... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_32 | D | 8 | $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textbf{(D)}\ 8\text{ ft} \qquad \textbf{(E)}\ 4\text{ ft}$ | [
"By the Pythagorean triple $(7,24,25)$ , the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \\boxed{8}$",
... |
https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_12 | A | 45 | $2\left(1-\dfrac{1}{2}\right) + 3\left(1-\dfrac{1}{3}\right) + 4\left(1-\dfrac{1}{4}\right) + \cdots + 10\left(1-\dfrac{1}{10}\right)=$
$\text{(A)}\ 45 \qquad \text{(B)}\ 49 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$ | [
"Taking the first product, we have\n$\\left(1-\\frac{1}{2}\\right)=\\frac{1}{2}$\n$\\frac{1}{2}\\times2=1$\nLooking at the second, we get\n$\\left(1-\\frac{1}{3}\\right)=\\frac{2}{3}$\n$\\frac{2}{3}\\times3=2$\nWe seem to be going up by $1$\nJust to check,\n$1-\\frac{1}{n}=\\frac{n-1}{n}$\n$\\frac{n-1}{n}\\times n=... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_7 | E | 5 | $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?
$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$ | [
"We can eliminate B, C, and D, because they are not $21$ subtracted by any multiple of $4$ . Finally, we see that there is no way to have A, so the solution is $\\boxed{5}$",
"Let $x$ be the number of $1x1$ tiles. There are $21$ squares and each $2x2$ or $1x4$ tile takes up 4 squares, so $x \\equiv 1 \\pmod{4}$ ,... |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_10 | B | 2,989 | $4(299)+3(299)+2(299)+298=$
$\text{(A)}\ 2889 \qquad \text{(B)}\ 2989 \qquad \text{(C)}\ 2991 \qquad \text{(D)}\ 2999 \qquad \text{(E)}\ 3009$ | [
"We can make use of the distributive property as follows: \\begin{align*} 4(299)+3(299)+2(299)+298 &= 4(299)+3(299)+2(299)+1(299)-1 \\\\ &= (4+3+2+1)(299)-1 \\\\ &= 10(299)-1 \\\\ &= 2989 \\\\ \\end{align*}\n$\\boxed{2989}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_2 | B | 4 | $4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$ | [
"After reversing the numbers on the second and fourth rows, the block will look like this:\n$\\begin{tabular}[t]{|c|c|c|c|} \\multicolumn{4}{c}{}\\\\\\hline 1&2&3&4\\\\\\hline 11&10&9&8\\\\\\hline 15&16&17&18\\\\\\hline 25&24&23&22\\\\\\hline \\end{tabular}$\nThe positive difference between the two diagonal sums is... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_2 | B | 4 | $4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$ | [
"After reversing the numbers on the second and fourth rows, the block will look like this:\n$\\begin{tabular}[t]{|c|c|c|c|} \\multicolumn{4}{c}{}\\\\\\hline 1&2&3&4\\\\\\hline 11&10&9&8\\\\\\hline 15&16&17&18\\\\\\hline 25&24&23&22\\\\\\hline \\end{tabular}$\nThe positive difference between the two diagonal sums is... |
https://artofproblemsolving.com/wiki/index.php/1998_AJHSME_Problems/Problem_21 | B | 52 | $4\times 4\times 4$ cubical box contains 64 identical small cubes that exactly fill the box. How many of these small cubes touch a side or the bottom of the box?
$\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$ | [
"Each small cube would have dimensions $1\\times 1\\times 1$ making each cube a unit cube.\nIf there are $16$ cubes per face and there are $5$ faces we are counting, we have $16\\times 5= 80$ cubes.\nSome cubes are on account of overlap between different faces.\nWe could reduce this number by subtracting the overla... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_9 | null | 106 | $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$ | [
"Firstly, we consider how many different ways possible to divide the $7\\times 1$ board.\nWe ignore the cases of 1 or 2 pieces since we need at least one tile of each color.\nSecondly, we use Principle of Inclusion-Exclusion to consider how many ways to color them:\nFinally, we combine them together: $15\\times 6+2... |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2 | B | 945 | $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$ | [
"To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \\cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].\n[mathjax]90\\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.\nRearranging the numbers so each pair sums up to 10, we have:\n[mathjax... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_16 | E | 59 | $A$ $B$ $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles $B$ and $C$
$\textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59$ | [
"Let the number of rocks in $A$ be $a$ $B$ be $b$ $C$ be $c$ . The total weight of $A$ be $40a$ $B$ be $50b$ $C$ be $kc$\nWe can write the information given as, $\\frac{40a + 50b}{a+b} = 43$ $\\frac{40a + kc}{a+c} = 44$ $\\frac{50b + kc}{b+c} = ?$\n$40a + 50b = 43 a + 43 b$ $3a = 7b$\n$40a + kc = 44a + 44 c$ $kc = ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_16 | null | 59 | $A$ $B$ $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles $B$ and $C$
$\textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59$ | [
"Let the total number of rocks in pile $A$ be $A_n$ , and the total number of rocks in pile $B$ be $B_n$ . Then, by restriction 3 (the average of $A$ and $B$ ), we can establish the equation: \\[\\frac{40A_n+50B_n}{A_n+B_n}=43\\] .\nCross-multiplying, we get: \\[40A_n+50B_n=43A_n+43B_n \\implies 3A_n=7B_n\\] .\nLet... |
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_30 | B | 3 | $A$ and $B$ together can do a job in $2$ days; $B$ and $C$ can do it in four days; and $A$ and $C$ in $2\frac{2}{5}$ days.
The number of days required for A to do the job alone is:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8$ | [
"Let $A$ do $r_A$ of the job per day, $B$ do $r_B$ of the job per day, and $C$ do $r_C$ of the job per day. These three quantities have unit $\\frac{\\text{job}}{\\text{day}}$ . Therefore our three conditions give us the three equations: \\begin{align*} (2\\text{ days})(r_A+r_B)&=1\\text{ job},\\nonumber\\\\ (4\\te... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_7 | null | 293 | $ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy] | [
"Since $EF$ is the perpendicular bisector of $\\overline{BB'}$ , it follows that $BE = B'E$ (by SAS). By the Pythagorean Theorem , we have $AB' = 15$ . Similarly, from $BF = B'F$ , we have \\begin{align*} BC^2 + CF^2 = B'D^2 + DF^2 &\\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\\\ BC &= \\frac{70}{3} \\end{align... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_23 | C | 19 | $ABCD$ is a square of side length $\sqrt{3} + 1$ . Point $P$ is on $\overline{AC}$ such that $AP = \sqrt{2}$ . The square region bounded by $ABCD$ is rotated $90^{\circ}$ counterclockwise with center $P$ , sweeping out a region whose area is $\frac{1}{c} (a \pi + b)$ , where $a$ $b$ , and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a + b + c$
$\textbf{(A)} \ 15 \qquad \textbf{(B)} \ 17 \qquad \textbf{(C)} \ 19 \qquad \textbf{(D)} \ 21 \qquad \textbf{(E)} \ 23$ | [
"We first note that diagonal $\\overline{AC}$ is of length $\\sqrt{6} + \\sqrt{2}$ . It must be that $\\overline{AP}$ divides the diagonal into two segments in the ratio $\\sqrt{3}$ to $1$ . It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular regi... |
https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_28 | C | 4 | $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended,
respectively. Let $O$ be the center of the pentagon. If $OP = 1$ , then $AO + AQ + AR$ equals
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label("$O$",O,dir(B)); label("$1$",(O+P)/2,W); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$E$",E,dir(E)); label("$P$",P,dir(P)); label("$Q$",Q,dir(Q)); label("$R$",R,dir(R)); [/asy]
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$ | [
"To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles. \nIf $s$ is the side length of the regular pentagon, then each of the triangles $AOB$ $BOC$ $COD$ $DOE$ , and $EOA$ has base $s$ and height 1, so... |
https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_28 | null | 4 | $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended,
respectively. Let $O$ be the center of the pentagon. If $OP = 1$ , then $AO + AQ + AR$ equals
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label("$O$",O,dir(B)); label("$1$",(O+P)/2,W); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$E$",E,dir(E)); label("$P$",P,dir(P)); label("$Q$",Q,dir(Q)); label("$R$",R,dir(R)); [/asy]
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$ | [
"Now, we know that angle $D$ has measure $\\frac{180 \\cdot 3}{5} = 108$ . Since \\[\\sin 54 = \\frac{OP}{DO} = \\frac{1}{DO}, DO = \\frac{1}{\\sin 54}\\] \\[\\tan 54 = \\frac{OP}{DP} = \\frac{1}{DP}, DP = \\frac{1}{\\tan 54}\\] Therefore, $AB = 2DP = \\frac{2}{\\tan 54}$ \\[\\sin 72 = \\frac{AQ}{AB} = AQ \\tan 54 ... |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_13 | B | 30 | $R$ varies directly as $S$ and inversely as $T$ . When $R = \frac{4}{3}$ and $T = \frac {9}{14}$ $S = \frac37$ . Find $S$ when $R = \sqrt {48}$ and $T = \sqrt {75}$
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60$ | [
"\\[R=c\\cdot\\frac{S}T\\]\nfor some constant $c$\nYou know that\n\\[\\frac43=c\\cdot\\frac{3/7}{9/14}=c\\cdot\\frac37\\cdot\\frac{14}9=c\\cdot\\frac23\\,,\\]\nso\n\\[c=\\frac{4/3}{2/3}=2\\,.\\]\nWhen $R=\\sqrt{48}$ and $T=\\sqrt{75}$ we have\n\\[\\sqrt{48}=\\frac{2S}{\\sqrt{75}}\\,,\\]\nso\n\\[S=\\frac12\\sqrt{48\... |
https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_1 | B | 2 | $[x-(y-z)] - [(x-y) - z] =$
$\textbf{(A)}\ 2y \qquad \textbf{(B)}\ 2z \qquad \textbf{(C)}\ -2y \qquad \textbf{(D)}\ -2z \qquad \textbf{(E)}\ 0$ | [
"The expression becomes $(x-y+z)-(x-y-z) = x-y+z-x+y+z = 2z$ , which is $\\boxed{2}$"
] |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_12 | D | 35 | $\angle 1 + \angle 2 = 180^\circ$
$\angle 3 = \angle 4$
Find $\angle 4.$
[asy] pair H,I,J,K,L; H = (0,0); I = 10*dir(70); J = I + 10*dir(290); K = J + 5*dir(110); L = J + 5*dir(0); draw(H--I--J--cycle); draw(K--L--J); draw(arc((0,0),dir(70),(1,0),CW)); label("$70^\circ$",dir(35),NE); draw(arc(I,I+dir(250),I+dir(290),CCW)); label("$40^\circ$",I+1.25*dir(270),S); label("$1$",J+0.25*dir(162.5),NW); label("$2$",J+0.25*dir(17.5),NE); label("$3$",L+dir(162.5),WNW); label("$4$",K+dir(-52.5),SE); [/asy] $\text{(A)}\ 20^\circ \qquad \text{(B)}\ 25^\circ \qquad \text{(C)}\ 30^\circ \qquad \text{(D)}\ 35^\circ \qquad \text{(E)}\ 40^\circ$ | [
"Using the left triangle, we have:\n$\\angle 1 + 70 + 40 = 180$\n$\\angle 1 = 180 - 110$\n$\\angle 1 = 70$\nUsing the given fact that $\\angle 1 + \\angle 2 = 180$ , we have $\\angle 2 = 180 - 70 = 110$\nFinally, using the right triangle, and the fact that $\\angle 3 = \\angle 4$ , we have:\n$\\angle 2 + \\angle 3 ... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_1 | B | 1 | $\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}=$
$\text{(A)}\ -1 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$ | [
"\\begin{align*} \\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \\dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\\\ &= \\dfrac{1+1+1+1+1}{1+1+1+1+1} \\\\ &= 1 \\rightarrow \\boxed{1}"
] |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_1 | C | 0.1997 | $\dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000} =$
$\text{(A)}\ 0.0026 \qquad \text{(B)}\ 0.0197 \qquad \text{(C)}\ 0.1997 \qquad \text{(D)}\ 0.26 \qquad \text{(E)}\ 1.997$ | [
"Convert each fraction to a decimal.\n$\\dfrac{1}{10} + \\dfrac{9}{100} + \\dfrac{9}{1000} + \\dfrac{7}{10000}$\n$0.1+0.09+0.009+0.0007=0.1997$\n$\\boxed{0.1997}$"
] |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_2 | D | 10 | $\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}+\dfrac{55}{10}=$
$\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$ | [
"$1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \\dfrac{(9)(10)}{2} = 45$\n$\\frac{45+55}{10} = \\dfrac{100}{10} = \\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_14 | E | 37 | $\diamondsuit$ and $\Delta$ are whole numbers and $\diamondsuit \times \Delta =36$ . The largest possible value of $\diamondsuit + \Delta$ is
$\text{(A)}\ 12 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20\ \qquad \text{(E)}\ 37$ | [
"Since it doesn't take too long, we can just make a table with all the possible values of the sum \\[\\begin{array}{|c|c|c|} \\multicolumn{3}{}{} \\\\ \\hline \\diamondsuit & \\Delta & \\diamondsuit + \\Delta \\\\ \\hline 36 & 1 & 37 \\\\ \\hline 18 & 2 & 20 \\\\ \\hline 12 & 3 & 15 \\\\ \\hline 9 & 4 & 13 \\\\ \\h... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_16 | D | 18 | $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
$\textbf{(A) } 6 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 27$ | [
"We have $t_n = \\frac{n (n+1)}{2}$ .\nIf $t_n$ is a perfect square, then it can be written as $\\frac{n (n+1)}{2} = k^2$ ,\nwhere $k$ is a positive integer.\nThus, $n (n+1) = 2 k^2$ . Rearranging, we get $(2n+1)^2-2(2k)^2=1$ , a Pell equation. So $\\frac{2n+1}{2k}$ must be a truncation of the continued fraction fo... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_16 | null | 18 | $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
$\textbf{(A) } 6 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 27$ | [
"According to the problem, we want to find integer $p$ such $\\frac{n(n+1)}{2}=p^2$ , after expanding, we have $n^2+n=2p^2, 4n^2+4n=8p^2, (2n+1)^2-8p^2=1$ , we call $2n+1=q$ , the equation becomes $q^2-8p^2=1$ , obviously $(q,p)=(3,1)$ is the elementary solution for this pell equation, thus the forth smallest solut... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_3 | E | 120 | $\frac{(3!)!}{3!}=$
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120$ | [
"The numerator is $(3!)! = 6!$\nThe denominator is $3! = 6$\nUsing the property that $6! = 6 \\cdot 5!$ in the numerator, the sixes cancel, leaving $5! = 120$ , which is answer $\\boxed{120}$"
] |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_3 | D | 200 | $\frac{10^7}{5\times 10^4}=$
$\text{(A)}\ .002 \qquad \text{(B)}\ .2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$ | [
"We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: \\[\\frac{10^7}{5 \\times 10^4} = \\frac{10^3}{5}\\]\nWe know that $10^3 = 10 \\times 10 \\times 10$ , so\n\\begin{align*} \\frac{10^3}{5} &= \\frac{10\\times 10\\times 10}{5} \\\\ ... |
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_2 | C | 12 | $\frac{16+8}{4-2}=$
$\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$ | [
"\\begin{align*} \\frac{16+8}{4-2} &= \\frac{24}{2} \\\\ &= 12\\rightarrow \\boxed{12}"
] |
https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_3 | D | 0.3 | $\frac{1}{10}+\frac{2}{20}+\frac{3}{30} =$
$\text{(A)}\ .1 \qquad \text{(B)}\ .123 \qquad \text{(C)}\ .2 \qquad \text{(D)}\ .3 \qquad \text{(E)}\ .6$ | [
"Each of the fractions simplify to $\\frac{1}{10}$ , so this sum is \\begin{align*} \\frac{1}{10}+\\frac{1}{10}+\\frac{1}{10} &= \\frac{3}{10} \\\\ &= .3 \\rightarrow \\boxed{.3}"
] |
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_2 | D | 0.246 | $\frac{2}{10}+\frac{4}{100}+\frac{6}{1000}=$
$\text{(A)}\ .012 \qquad \text{(B)}\ .0246 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .246 \qquad \text{(E)}\ 246$ | [
"\\begin{align*} \\frac{2}{10}+\\frac{4}{100}+\\frac{6}{1000} &= \\frac{200}{1000}+\\frac{40}{1000}+\\frac{6}{1000} \\\\ &= \\frac{246}{1000} \\\\ &= .246 \\rightarrow \\boxed{.246}"
] |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_2 | B | 0.08 | $\frac{2}{25}=$
$\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 12.5$ | [
"$\\frac{2}{25}=\\frac{2\\cdot 4}{25\\cdot 4} = \\frac{8}{100} = 0.08\\rightarrow \\boxed{.08}$"
] |
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_40 | C | 1 | $\left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2}$ equals:
$\textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2}$ $\textbf{(E)}\ [(x^{3}-1)^{2}]^{2}$ | [
"First, note that we can pull the exponents out of every factor, since they are all squared. This results in $\\left(\\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\\right)^{4}\\cdot\\left(\\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\\right)^{4}$ Now, multiplying the numerators together gives $\\left(\\frac{x^3+1}{x^3+1}\\right)^{4}\\cdot\... |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_18 | D | 1 | $\log p+\log q=\log(p+q)$ only if:
$\textbf{(A) \ }p=q=\text{zero} \qquad \textbf{(B) \ }p=\frac{q^2}{1-q} \qquad \textbf{(C) \ }p=q=1 \qquad$
$\textbf{(D) \ }p=\frac{q}{q-1} \qquad \textbf{(E) \ }p=\frac{q}{q+1}$ | [
"$\\log p+\\log q=\\log(p+q)\\implies \\log pq=\\log(p+q)\\implies pq=p+q\\implies \\boxed{1}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_9 | E | 24 | $\text{palindrome}$ , such as $83438$ , is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24$ | [
"$x$ is at most $999$ , so $x+32$ is at most $1031$ . The minimum value of $x+32$ is $1000$ . However, the only palindrome between $1000$ and $1032$ is $1001$ , which means that $x+32$ must be $1001$\nIt follows that $x$ is $969$ , so the sum of the digits is $\\boxed{24}$",
"For $x+32$ to be a four-digit number,... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_6 | E | 24 | $\text{palindrome}$ , such as $83438$ , is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24$ | [
"$x$ is at most $999$ , so $x+32$ is at most $1031$ . The minimum value of $x+32$ is $1000$ . However, the only palindrome between $1000$ and $1032$ is $1001$ , which means that $x+32$ must be $1001$\nIt follows that $x$ is $969$ , so the sum of the digits is $\\boxed{24}$",
"For $x+32$ to be a four-digit number,... |
https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_3 | D | 55 | $\triangle ABC$ has a right angle at $C$ and $\angle A = 20^\circ$ . If $BD$ $D$ in $\overline{AC}$ ) is the bisector of $\angle ABC$ , then $\angle BDC =$
$\textbf{(A)}\ 40^\circ \qquad \textbf{(B)}\ 45^\circ \qquad \textbf{(C)}\ 50^\circ \qquad \textbf{(D)}\ 55^\circ\qquad \textbf{(E)}\ 60^\circ$ | [
"Since $\\angle C = 90^{\\circ}$ and $\\angle A = 20^{\\circ}$ , we have $\\angle ABC = 70^{\\circ}$ . Thus $\\angle DBC = 35^{\\circ}$ . It follows that $\\angle BDC = 90^{\\circ} - 35^{\\circ} = 55^{\\circ}$ , which is $\\boxed{55}$"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_23 | D | 84 | $\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$ -by- $3$ grid. Shown below is a sample configuration with three $\triangle$ s in a line. [asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy] How many configurations will have three $\triangle$ s in a line and three $\bigcirc$ s in a line?
$\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$ | [
"Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.\nWe take casework:\nCase... |
https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_7 | C | 36 | A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is
[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]
$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$ | [
"The best way to solve this problem is to find patterns and to utilize them to our advantage. For example, we can't really do anything without knowing how many squares there are in the 37th row. But who wants to continue the diagram for 37 rows? And what if the problem said 100,000th row? It'll still be possible - ... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_2 | null | 52 | A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person. | [
"Clearly we have people moving at speeds of $6,8$ and $10$ feet/second. Notice that out of the three people, Cy is at the largest disadvantage to begin with and since all speeds are close, it is hardest for him to catch up. Furthermore, Bob is clearly the farthest along. Thus it is reasonable to assume that there i... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_18 | D | 1,698 | A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$ -coordinate or the $y$ -coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$ $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$ | [
"As stated in the solution, there are $6$ points along the line $y=-x$ that constitute a sort of \"boundary\". Once the ant reaches one of these $6$ points, it is exactly halfway to $(4, 4)$ . Also notice that the ant will only cross one of the $6$ points during any one of its paths. Therefore we can divide the pro... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_18 | null | 1,698 | A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$ -coordinate or the $y$ -coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$ $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$ | [
"Each path must go through either the second or the fourth quadrant.\nEach path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$ $(-3,3)$ , and $(-2,2)$\nThere is $1$ path of the first kind, ${8\\choose 1}^2=64$ paths of the second kind, and ${8\\choose 2}^2=28^2=784$ paths... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_18 | D | 90 | A 3x3x3 cube is made of $27$ normal dice. Each die's opposite sides sum to $7$ . What is the smallest possible sum of all of the values visible on the $6$ faces of the large cube?
$\text{(A)}\ 60 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 84 \qquad \text{(D)}\ 90 \qquad \text{(E)} 96$ | [
"In a 3x3x3 cube, there are $8$ cubes with three faces showing, $12$ with two faces showing and $6$ with one face showing. The smallest sum with three faces showing is $1+2+3=6$ , with two faces showing is $1+2=3$ , and with one face showing is $1$ . Hence, the smallest possible sum is $8(6)+12(3)+6(1)=48+36+6=90$ ... |
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_14 | B | 18 | A bag contains only blue balls and green balls. There are $6$ blue balls. If the probability of drawing a blue ball at random from this bag is $\frac{1}{4}$ , then the number of green balls in the bag is
$\text{(A)}\ 12 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36$ | [
"The total number of balls in the bag must be $4\\times 6=24$ , so there are $24-6=18$ green balls $\\rightarrow \\boxed{18}$"
] |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_12 | C | 5 | A ball is dropped from a height of $3$ meters. On its first bounce it rises to a height of $2$ meters. It keeps falling and bouncing to $\frac{2}{3}$ of the height it reached in the previous bounce. On which bounce will it not rise to a height of $0.5$ meters?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$ | [
"Each bounce is $2/3$ times the height of the previous bounce. The first bounce reaches $2$ meters, the second $4/3$ , the third $8/9$ , the fourth $16/27$ , and the fifth $32/81$ . Half of $81$ is $40.5$ , so the ball does not reach the required height on bounce $\\boxed{5}$"
] |
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_33 | A | 5 | A bank charges $\textdollar{6}$ for a loan of $\textdollar{120}$ . The borrower receives $\textdollar{114}$ and repays the loan in $12$ easy installments of $\textdollar{10}$ a month. The interest rate is approximately:
$\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \%$ | [
"The borrower pays $\\textdollar{120}$ in a single year for a loan of $\\textdollar{114}$ . This means that the bank charges an interest of $\\textdollar{6}$ for a loan of $\\textdollar{114}$ over a single year, so that the annual interest rate is $100*\\frac{\\textdollar{6}}{\\textdollar{114}} = \\frac{100}{19} \\... |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_16 | A | 2.5 | A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar indicating the number sold during the winter is covered by a smudge. If exactly $25\%$ of the chain's hamburgers are sold in the fall, how many million hamburgers are sold in the winter?
[asy] size(250); void bargraph(real X, real Y, real ymin, real ymax, real ystep, real tickwidth, string yformat, Label LX, Label LY, Label[] LLX, real[] height,pen p=nullpen) { draw((0,0)--(0,Y),EndArrow); draw((0,0)--(X,0),EndArrow); label(LX,(X,0),plain.SE,fontsize(9)); label(LY,(0,Y),plain.NW,fontsize(9)); real yscale=Y/(ymax+ystep); for(real y=ymin; y<ymax; y+=ystep) { draw((-tickwidth,yscale*y)--(0,yscale*y)); label(format(yformat,y),(-tickwidth,yscale*y),plain.W,fontsize(9)); } int n=LLX.length; real xscale=X/(2*n+2); for(int i=0;i<n;++i) { real x=xscale*(2*i+1); path P=(x,0)--(x,height[i]*yscale)--(x+xscale,height[i]*yscale)--(x+xscale,0)--cycle; fill(P,p); draw(P); label(LLX[i],(x+xscale/2),plain.S,fontsize(10)); } for(int i=0;i<n;++i) draw((0,height[i]*yscale)--(X,height[i]*yscale),dashed); } string yf="%#.1f"; Label[] LX={"Spring","Summer","Fall","Winter"}; for(int i=0;i<LX.length;++i) LX[i]=rotate(90)*LX[i]; real[] H={4.5,5,4,4}; bargraph(60,50,1,5.1,0.5,2,yf,"season","hamburgers (millions)",LX,H,yellow); fill(ellipse((45,30),7,10),brown); [/asy]
$\text{(A)}\ 2.5 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 3.5 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 4.5$ | [
"What we want to find is the number of hamburgers sold in the winter. Since we don't know what it is, let's call it $x$ . From the graph, we know that in Spring, 4.5 million hamburgers were sold, in the Summer was 5 million and in the Fall was 4 million. We know that the number of hamburgers sold in Fall is exactly... |
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_4 | D | 590 | A barn with a roof is rectangular in shape, $10$ yd. wide, $13$ yd. long and $5$ yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:
$\mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720$ | [
"The walls are $13*5=65$ and $10*5=50$ in area, and the ceiling has an area of $10*13=130$\n$((65+50)2)2+130=590 \\Rightarrow \\boxed{590}$"
] |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_20 | E | 0.7 | A base-10 three digit number $n$ is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of $n$ are both three-digit numerals?
$\mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7$ | [
"To be a three digit number in base-10:\n$10^{2} \\leq n \\leq 10^{3}-1$\n$100 \\leq n \\leq 999$\nThus there are $900$ three-digit numbers in base-10\nTo be a three-digit number in base-9:\n$9^{2} \\leq n \\leq 9^{3}-1$\n$81 \\leq n \\leq 728$\nTo be a three-digit number in base-11:\n$11^{2} \\leq n \\leq 11^{3}-1... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_24 | B | 48 | A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division?
$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$ | [
"On one team they play $3N$ games in their division and $4M$ games in the other. This gives $3N+4M=76$\nSince $M>4$ we start by trying M=5. This doesn't work because $56$ is not divisible by $3$\nNext, $M=6$ does not work because $52$ is not divisible by $3$\nWe try $M=7$ does work by giving $N=16$ $~M=7$ and thus... |
https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_12 | null | 660 | A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$ | [
"We graph the $10$ shots on a grid. Suppose that a made shot is represented by a step of $(0,1)$ , and a missed shot is represented by $(1,0)$ . Then the basketball player's shots can be represented by the number of paths from $(0,0)$ to $(6,4)$ that always stay below the line $y=\\frac{2x}{3}$ . We can find the nu... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_1 | E | 6 | A basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player?
$\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$ | [
"The number of points could have been 10, 11, 12, 13, 14, or 15. This is because the minimum is 2*5=10 and the maximum is 3*5=15. The numbers between 10 and 15 are possible as well. Thus, the answer is $\\boxed{6}$"
] |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_10 | C | 20 | A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?
$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }30$ | [
"Let $x$ be the number of two point shots attempted and $y$ the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, $0.5(2x)+0.4(3y)=54$ or $x+1.2y=54$ . Because the team attempted 50% more tw... |
https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_14 | null | 71 | A beam of light strikes $\overline{BC}\,$ at point $C\,$ with angle of incidence $\alpha=19.94^\circ\,$ and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments $\overline{AB}\,$ and $\overline{BC}\,$ according to the rule: angle of incidence equals angle of reflection. Given that $\beta=\alpha/10=1.994^\circ\,$ and $AB=BC,\,$ determine the number of times the light beam will bounce off the two line segments. Include the first reflection at $C\,$ in your count.
AIME 1994 Problem 14.png | [
"At each point of reflection, we pretend instead that the light continues to travel straight. Note that after $k$ reflections (excluding the first one at $C$ ) the extended line will form an angle $k \\beta$ at point $B$ . For the $k$ th reflection to be just inside or at point $B$ , we must have $k\\beta \\le 180... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | B | 2,024 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$ | [
"Let $x=e^{i\\pi/6}$ , a $30^\\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is:\n\\[1+2x+3x^2+\\cdots+(k+1)x^k\\]\nWe need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetico-geometric series\n\\begin{align*} S &=1+2x+3x^2+\\cdots+2015x^... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | null | 2,024 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$ | [
"We first notice that if the bee is turning 30 degrees each turn, it will take 12 turns to be looking in the same direction when the bee initially left. This means we simply need to answer the question; how far will the bee be when the bee is facing in the same direction?\nFirst we use the fact that after 3 turns, ... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | A | 109 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | [
"We see that $a \\, \\diamondsuit \\, a = 1$ , and think of division. Testing, we see that the first condition $a \\, \\diamondsuit \\, (b \\, \\diamondsuit \\, c) = (a \\, \\diamondsuit \\, b) \\cdot c$ is satisfied, because $\\frac{a}{\\frac{b}{c}} = \\frac{a}{b} \\cdot c$ . Therefore, division can be the operati... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | A | 109 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$ | [
"We see that $a \\, \\diamondsuit \\, a = 1$ , and think of division. Testing, we see that the first condition $a \\, \\diamondsuit \\, (b \\, \\diamondsuit \\, c) = (a \\, \\diamondsuit \\, b) \\cdot c$ is satisfied, because $\\frac{a}{\\frac{b}{c}} = \\frac{a}{b} \\cdot c$ . Therefore, division can be the operati... |
https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_6 | null | 840 | A biologist wants to calculate the number of fish in a lake. On May 1 she catches a random sample of 60 fish, tags them, and releases them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate the number of fish in the lake on May 1, she assumes that 25% of these fish are no longer in the lake on September 1 (because of death and emigrations), that 40% of the fish were not in the lake May 1 (because of births and immigrations), and that the number of untagged fish and tagged fish in the September 1 sample are representative of the total population. What does the biologist calculate for the number of fish in the lake on May 1? | [
"Of the $70$ fish caught in September, $40\\%$ were not there in May, so $42$ fish were there in May. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, $\\frac{3}{42} = \\frac{60}{x} \\Longrightarrow \\boxed{840}$",
"First, we notice that there are 45 tags l... |
https://artofproblemsolving.com/wiki/index.php/2008_AIME_II_Problems/Problem_3 | null | 729 | A block of cheese in the shape of a rectangular solid measures $10$ cm by $13$ cm by $14$ cm. Ten slices are cut from the cheese. Each slice has a width of $1$ cm and is cut parallel to one face of the cheese. The individual slices are not necessarily parallel to each other. What is the maximum possible volume in cubic cm of the remaining block of cheese after ten slices have been cut off? | [
"Let the lengths of the three sides of the rectangular solid after the cutting be $a,b,c$ , so that the desired volume is $abc$ . Note that each cut reduces one of the dimensions by one, so that after ten cuts, $a+b+c = 10 + 13 + 14 - 10 = 27$ . By AM-GM $\\frac{a+b+c}{3} = 9 \\ge \\sqrt[3]{abc} \\Longrightarrow ab... |
https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_15 | null | 53 | A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$ , and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overarc{AB}$ on that face measures $120^\circ$ . The block is then sliced in half along the plane that passes through point $A$ , point $B$ , and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\cdot\pi + b\sqrt{c}$ , where $a$ $b$ , and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$
[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,"$A$",(-3,3*sqrt(3),0),W); label(rpic,"$B$",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy] | [
"Label the points where the plane intersects the top face of the cylinder as $C$ and $D$ , and the center of the cylinder as $O$ , such that $C,O,$ and $A$ are collinear. Let $N$ be the center of the bottom face, and $M$ the midpoint of $\\overline{AB}$ . Then $ON=4$ $MN=3$ (because of the 120 degree angle), and so... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_12 | D | 353 | A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wall?
[asy] draw((0,0)--(6,0)--(6,1)--(5,1)--(5,2)--(0,2)--cycle); draw((0,1)--(5,1)); draw((1,1)--(1,2)); draw((3,1)--(3,2)); draw((2,0)--(2,1)); draw((4,0)--(4,1)); [/asy]
$\text{(A)}\ 344\qquad\text{(B)}\ 347\qquad\text{(C)}\ 350\qquad\text{(D)}\ 353\qquad\text{(E)}\ 356$ | [
"Since the bricks are $1$ foot high, there will be $7$ rows. To minimize the number of blocks used, rows $1, 3, 5,$ and $7$ will look like the bottom row of the picture, which takes $\\frac{100}{2} = 50$ bricks to construct. Rows $2, 4,$ and $6$ will look like the upper row pictured, which has $49$ 2-foot bricks ... |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_12 | B | 16 | A board game spinner is divided into three regions labeled $A$ $B$ and $C$ . The probability of the arrow stopping on region $A$ is $\frac{1}{3}$ and on region $B$ is $\frac{1}{2}$ . The probability of the arrow stopping on region $C$ is:
$\text{(A)}\ \frac{1}{12}\qquad\text{(B)}\ \frac{1}{6}\qquad\text{(C)}\ \frac{1}{5}\qquad\text{(D)}\ \frac{1}{3}\qquad\text{(E)}\ \frac{2}{5}$ | [
"Since the arrow must land in one of the three regions, the sum of the probabilities must be 1. Thus the answer is $1-\\frac{1}{2}-\\frac{1}{3}=\\boxed{16}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_4 | B | 51.5 | A book that is to be recorded onto compact discs takes $412$ minutes to read aloud. Each disc can hold up to $56$ minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?
$\mathrm{(A)}\ 50.2 \qquad \mathrm{(B)}\ 51.5 \qquad \mathrm{(C)}\ 52.4 \qquad \mathrm{(D)}\ 53.8 \qquad \mathrm{(E)}\ 55.2$ | [
"Assuming that there are fractions of compact discs, it would take $412/56 ~= 7.357$ CDs to have equal reading time. However, since the number of discs must be a whole number, there are at least 8 CDs, in which case there would be $412/8 = 51.5$ minutes of reading on each of the 8 discs. The answer is $\\boxed{51.5... |
https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_9 | null | 342 | A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$ . He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens? | [
"On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$ , leaving only lockers $2 \\pmod{8}$ and $6 \\pmod{8}$ . Then he goes ahead and opens all lockers $2 \\pmod {8}$ , leaving lockers either $6 \\pmod {16}$ or $14 \\pmod {16}... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_21 | B | 7 | A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$ . The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl? [asy] import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); //Comment two lines below to remove red edges //draw(lVs[1]--lVs[3],red+2bp); //draw(uVsl[0]--uVsr[0],red+2bp); [/asy] $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | [
"We extend line segments $\\ell,m,$ and $n$ to their point of concurrency, as shown below: We claim that lines $\\ell,m,$ and $n$ are concurrent: In the lateral faces of the bowl, we know that lines $\\ell$ and $m$ must intersect, and lines $\\ell$ and $n$ must intersect. In addition, line $\\ell$ intersects the t... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_3 | D | 240 | A box $2$ centimeters high, $3$ centimeters wide, and $5$ centimeters long can hold $40$ grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold $n$ grams of clay. What is $n$
$\textbf{(A)}\ 120\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 200\qquad\textbf{(D)}\ 240\qquad\textbf{(E)}\ 280$ | [
"The first box has volume $2\\times3\\times5=30\\text{ cm}^3$ , and the second has volume $(2\\times2)\\times(3\\times3)\\times(5)=180\\text{ cm}^3$ . The second has a volume that is $6$ times greater, so it holds $6\\times40=\\boxed{240}$ grams."
] |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_4 | B | 76 | A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn?
$\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91$ | [
"We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee. \nNamely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_3 | B | 76 | A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn?
$\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91$ | [
"We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee. \nNamely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_2 | D | 9 | A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$ | [
"Let $a$ be the amount of triangular tiles and $b$ be the amount of square tiles.\nTriangles have $3$ edges and squares have $4$ edges, so we have a system of equations.\nWe have $a + b$ tiles total, so $a + b = 25$\nWe have $3a + 4b$ edges total, so $3a + 4b = 84$\nMultiplying the first equation by $3$ on both sid... |
https://artofproblemsolving.com/wiki/index.php/1971_AHSME_Problems/Problem_27 | null | 57 | A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips, and at most one third the number of red chips. The number which are white or blue is at least $55$ . The minimum number of red chips is
$\textbf{(A) }24\qquad \textbf{(B) }33\qquad \textbf{(C) }45\qquad \textbf{(D) }54\qquad \textbf{(E) }57$ | [
"Let the number of white be $2x$ . The number of blue is then $x-y$ for some constant $y$ . So we want $2x+x-y=55\\rightarrow 3x-y=55$ . We take mod 3 to find y. $55=1\\pmod{3}$ , so blue is 1 more than half white. The number of whites is then 36, and the number of blues is 19. So $19*3=\\boxed{57}$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_23 | A | 0 | A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5$ | [
"The counting numbers that leave a remainder of $4$ when divided by $6$ are $4, 10, 16, 22, 28, 34, \\cdots$ The counting numbers that leave a remainder of $3$ when\ndivided by $5$ are $3,8,13,18,23,28,33, \\cdots$ So $28$ is the smallest possible number\nof coins that meets both conditions. Because 28 is divisible... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_1 | B | 150 | A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $$1.00$ , he must sell:
$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\ \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$ | [
"The boy buys $3$ oranges for $10$ cents or $1$ orange for $\\frac{10}{3}$ cents. He sells them at $\\frac{20}{5}=4$ cents each. \nThat means for every orange he sells, he makes a profit of $4-\\frac{10}{3}=\\frac{2}{3}$ cents.\nTo make a profit of $100$ cents, he needs to sell $\\frac{100}{\\frac{2}{3}}=\\boxed{... |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_1 | null | 150 | A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $$1.00$ , he must sell:
$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\ \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$ | [
"The boy buys $3$ oranges for $10$ cents. He sells them at $5$ for $20$ cents. So, he buys $15$ for $50$ cents and sells them at $15$ for $60$ cents, so he makes $10$ cents of profit on every $15$ oranges. To make $100$ cents of profit, he needs to sell $15 \\cdot \\frac{100}{10} = \\boxed{150}$ oranges."
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_3 | E | 15 | A bug crawls along a number line, starting at $-2$ . It crawls to $-6$ , then turns around and crawls to $5$ . How many units does the bug crawl altogether?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$ | [
"\nCrawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\\boxed{15}$"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_1 | E | 15 | A bug crawls along a number line, starting at $-2$ . It crawls to $-6$ , then turns around and crawls to $5$ . How many units does the bug crawl altogether?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$ | [
"\nCrawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\\boxed{15}$"
] |
https://artofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_13 | null | 683 | A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$ | [
"Let $P_n$ represent the probability that the bug is at its starting vertex after $n$ moves. If the bug is on its starting vertex after $n$ moves, then it must be not on its starting vertex after $n-1$ moves. At this point it has $\\frac{1}{2}$ chance of reaching the starting vertex in the next move. Thus $P_n=\\fr... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_25 | E | 2,400 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
[asy] size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544)); draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767)); draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0)); draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)); draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080)); draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0)); dot((0,0)); dot((22,0)); label("$A$",(0,0),WNW); label("$B$",(22,0),E); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black); filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black); filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black); filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black); filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white); filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black); filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black); filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black); filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black); filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white); filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black); filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black); filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black); filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black); filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);[/asy]
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400$ | [
"\nThere is $1$ way to get to any of the red arrows. From the first (top) red arrow, there are $2$ ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_25 | null | 2,400 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
[asy] size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544)); draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767)); draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0)); draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)); draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080)); draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0)); dot((0,0)); dot((22,0)); label("$A$",(0,0),WNW); label("$B$",(22,0),E); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black); filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black); filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black); filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black); filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white); filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black); filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black); filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black); filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black); filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white); filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black); filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black); filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black); filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black); filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);[/asy]
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400$ | [
"We use casework. \nThe main thing to notice is that, if the bug does not go backwards, then every vertical set of arrows can be used, as the bug could travel straight downwards and then use any arrow to continue right.\nNote: The motivation is quite weird so follow my numbers as they start from the left (point A) ... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_22 | E | 2,400 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
[asy] size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544)); draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767)); draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0)); draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)); draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080)); draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0)); dot((0,0)); dot((22,0)); label("$A$",(0,0),WNW); label("$B$",(22,0),E); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black); filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black); filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black); filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black); filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white); filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black); filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black); filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black); filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black); filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white); filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black); filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black); filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black); filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black); filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);[/asy]
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400$ | [
"\nThere is $1$ way to get to any of the red arrows. From the first (top) red arrow, there are $2$ ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get... |
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