link stringlengths 75 84 | letter stringclasses 5
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https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_10 | A | 1.5 | An inverted cone with base radius $12 \mathrm{cm}$ and height $18 \mathrm{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has radius of $24 \mathrm{cm}$ . What is the height in centimeters of the water in the cylinder?
$\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ... | [
"The volume of a cone is $\\frac{1}{3} \\cdot\\pi \\cdot r^2 \\cdot h$ where $r$ is the base radius and $h$ is the height. The water completely fills up the cone so the volume of the water is $\\frac{1}{3}\\cdot18\\cdot144\\pi = 6\\cdot144\\pi$\nThe volume of a cylinder is $\\pi \\cdot r^2 \\cdot h$ so the volume o... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_6 | A | 1.5 | An inverted cone with base radius $12 \mathrm{cm}$ and height $18 \mathrm{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has radius of $24 \mathrm{cm}$ . What is the height in centimeters of the water in the cylinder?
$\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ... | [
"The volume of a cone is $\\frac{1}{3} \\cdot\\pi \\cdot r^2 \\cdot h$ where $r$ is the base radius and $h$ is the height. The water completely fills up the cone so the volume of the water is $\\frac{1}{3}\\cdot18\\cdot144\\pi = 6\\cdot144\\pi$\nThe volume of a cylinder is $\\pi \\cdot r^2 \\cdot h$ so the volume o... |
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_10 | B | 20 | An isosceles right triangle with legs of length $8$ is partitioned into $16$ congruent triangles as shown. The shaded area is
[asy] for (int a=0; a <= 3; ++a) { for (int b=0; b <= 3-a; ++b) { fill((a,b)--(a,b+1)--(a+1,b)--cycle,grey); } } for (int c=0; c <= 3; ++c) { draw((c,0)--(c,4-c),linewid... | [
"Because the smaller triangles are congruent, the shaded area take $\\frac{10}{16}$ of the largest triangles area, which is $\\frac{8 \\times 8}{2}=32$ , so the shaded area is $\\frac{10}{16} \\times 32= \\boxed{20}$",
"Each of the triangle has side length of $\\frac{1}{4} \\times 8=2$ , so the area is $\\frac{1}... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_18 | B | 121 | An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?
$\mathrm{(A)}\ 120 \qquad \mathr... | [
"$10$ moves results in a lot of possible endpoints, so we try small cases first.\nIf the object only makes $1$ move, it is obvious that there are only 4 possible points that the object can move to.\nIf the object makes $2$ moves, it can move to $(0, 2)$ $(1, 1)$ $(2, 0)$ $(1, -1)$ $(0, -2)$ $(-1, -1)$ $(-2, 0)$ as ... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_18 | null | 121 | An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?
$\mathrm{(A)}\ 120 \qquad \mathr... | [
"Let the starting point be $(0,0)$ . After $10$ steps we can only be in locations $(x,y)$ where $|x|+|y|\\leq 10$ . Additionally, each step changes the parity of exactly one coordinate. Hence after $10$ steps we can only be in locations $(x,y)$ where $x+y$ is even. It can easily be shown that each location that sat... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_12B_Problems/Problem_21 | D | 13 | An object moves $8$ cm in a straight line from $A$ to $B$ , turns at an angle $\alpha$ , measured in radians and chosen at random from the interval $(0,\pi)$ , and moves $5$ cm in a straight line to $C$ . What is the probability that $AC < 7$
$\mathrm{(A)}\ \frac{1}{6} \qquad\mathrm{(B)}\ \frac{1}{5} \qquad\mathrm{(C)}... | [
"$WLOG$ , let the object turn clockwise.\nLet $B = (0, 0)$ $A = (0, -8)$\nNote that the possible points of $C$ create a semi-circle of radius $5$ and center $B$ . The area where $AC < 7$ is enclosed by a circle of radius $7$ and center $A$ . The probability that $AC < 7$ is $\\frac{\\angle ABO}{180 ^\\circ}$\nThe f... |
https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_1 | null | 300 | An ordered pair $(m,n)$ of non-negative integers is called "simple" if the addition $m+n$ in base $10$ requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to $1492$ | [
"Since no carrying over is allowed, the range of possible values of any digit of $m$ is from $0$ to the respective digit in $1492$ (the values of $n$ are then fixed). Thus, the number of ordered pairs will be $(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\\cdot 5\\cdot 10\\cdot 3 = \\boxed{300}$"
] |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_16 | B | 191 | An organization has $30$ employees, $20$ of whom have a brand A computer while the other $10$ have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if th... | [
"We claim that to maximize the number of cables used, we isolate one computer and connect all cables for the remaining $29$ computers, then connect one more cable for the isolated computer.\nIf a brand A computer is isolated, then the technician can use at most $19\\cdot10+1=191$ cables. If a brand B computer is is... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_2 | null | 144 | An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$ . Find $N$ | [
"First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to $0.58$\nThe probability both are green is $\\frac{4}{10}\\cdot\\frac{16}{16+N}$ , and the probability both are blue is $\\frac{6}{10}\\cdot\\frac{N}{16+N}$ , so \\[\\frac{4}{... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_26 | B | 21 | An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:
(a) the selection of four red marbles;
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbles; and
(... | [
"Let the bag contain $n$ marbles total, with $r, w, b, g$ representing the number of red, white, blue, and green marbles, respectively. Note that $r + w + b + g = n$\nThe number of ways to select four red marbles out of the set of marbles without replacement is:\n\\[\\binom{r}{4} = \\frac{r!}{24\\cdot (r -4)!}\\]\... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | B | 15 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | [
"Suppose that we have a deck, currently containing just one black card. We then insert $n$ red cards one-by-one into the deck at random positions. It is easy to see using induction, that the black card is randomly situated in the deck.\nNow, suppose that we have this deck again, with only one black card. Each time ... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | B | 15 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | [
"Suppose that we have a deck, currently containing just one black card. We then insert $n$ red cards one-by-one into the deck at random positions. It is easy to see using induction, that the black card is randomly situated in the deck.\nNow, suppose that we have this deck again, with only one black card. Each time ... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_3 | D | 12 | Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$ | [
"Let $A$ be the age of Ana and $B$ be the age of Bonita. Then,\n\\[A-1 = 5(B-1)\\] and \\[A = B^2.\\]\nSubstituting the second equation into the first gives us\n\\[B^2-1 = 5(B-1).\\]\nBy using difference of squares and dividing, $B=4.$ Moreover, $A=B^2=16.$\nThe answer is $16-4 = 12 \\implies \\boxed{12}$",
"Simp... |
https://artofproblemsolving.com/wiki/index.php/1996_AJHSME_Problems/Problem_18 | A | 1,920 | Ana's monthly salary was $$2000$ in May. In June she received a 20% raise. In July she received a 20% pay cut. After the two changes in June and July, Ana's monthly salary was
$\text{(A)}\ 1920\text{ dollars} \qquad \text{(B)}\ 1980\text{ dollars} \qquad \text{(C)}\ 2000\text{ dollars} \qquad \text{(D)}\ 2020\text{ ... | [
"Notice that a 20% raise translates to Ana's monthly salary multiplied by $(100 + 10)\\% = 120\\%$ , and a 20% pay cut translates to her salary multiplied by $100\\% - 20\\% = 80\\%$ , so\n\\[2000 \\cdot 120\\% = 2400\\] \\[2400 \\cdot 80\\% = 1920 \\rightarrow \\boxed{1920}\\]"
] |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_4 | null | 61 | Ana, Bob, and CAO bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading... | [
"\nLet $a,b,c$ be the labeled lengths as shown in the diagram. Also, assume WLOG the time taken is $1$ second.\nObserve that $\\dfrac{2a+b+c}{8.6}=1$ or $2a+b+c=8.6$ , and $\\dfrac{b+c}{6.2}=1$ or $b+c=6.2$ . Subtracting the second equation from the first gives $2a=2.4$ , or $a=1.2$\nNow, let us solve $b$ and $c$ .... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_20 | B | 65 | Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from t... | [
"Since the distance between them decreases at a rate of $1$ kilometer per minute when they are both biking, their combined speed is $1$ kilometer per minute. Andrea travels three times as fast as Lauren, so they travel at speeds of $\\frac{3}{4}$ kilometers per minute and $\\frac{1}{4}$ kilometers per minute, respe... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_20 | D | 65 | Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from t... | [
"Because the speed of Andrea is 3 times as fast as Lauren and the distance between them is decreasing at a rate of 1 kilometer per minute, Andrea's speed is $\\frac{3}{4} \\textbf{km/min}$ , and Lauren's $\\frac{1}{4} \\textbf{km/min}$ . Therefore, after 5 minutes, Andrea will have biked $\\frac{3}{4} \\cdot 5 = \\... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_20 | null | 65 | Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from t... | [
"Let their speeds in kilometers per hour be $v_A$ and $v_L$ . We know that $v_A=3v_L$ and that $v_A+v_L=60$ . (The second equation follows from the fact that $1\\mathrm km/min = 60\\mathrm km/h$ .) This solves to $v_A=45$ and $v_L=15$\nAs the distance decreases at a rate of $1$ kilometer per minute, after $5$ minut... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | D | 471 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac... | [
"Let our denominator be $(5!)^3$ , so we consider all possible distributions.\nWe use PIE (Principle of Inclusion and Exclusion) to count the successful ones.\nWhen we have at $1$ box with all $3$ balls the same color in that box, there are $_{5} C _{1} \\cdot _{5} P _{1} \\cdot (4!)^3$ ways for the distributions t... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_12 | B | 13 | Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?
$\textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34$ | [
"If we designate a person to be on a certain side, then all placements of the other people can be considered unique. WLOG, assign Angie to be on the side. There are then $3!=6$ total seating arrangements. If Carlos is across from Angie, there are only $2!=2$ ways to fill the remaining two seats. Then the probabilit... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_5 | null | 53 | Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she comp... | [
"We could see that both $374$ and $319$ are divisible by $11$ in the outset, and that $34$ and $29$ , the quotients, are relatively prime. Both are the $average$ number of minutes across the $11$ days, so we need to subtract $\\left \\lfloor{\\frac{11}{2}}\\right \\rfloor=5$ from each to get $(n,t)=(29,24)$ and $29... |
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_29 | B | 24 | Ann and Barbara were comparing their ages and found that Barbara is as old as Ann was when Barbara was as old as
Ann had been when Barbara was half as old as Ann is. If the sum of their present ages is $44$ years, then Ann's age is
$\textbf{(A) }22\qquad \textbf{(B) }24\qquad \textbf{(C) }25\qquad \textbf{(D) }26\qqua... | [
"This problem is very wordy. Nonetheless, let $a$ and $b$ be Ann and Barbara's current ages, respectively. We are given that $a+b=44$ . Let $y$ equal the difference between their ages, so $y=a-b$ . Know that $y$ is constant because the difference between their ages will always be the same.\nNow, let's tackle the eq... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_3 | D | 22 | Ann made a $3$ -step staircase using $18$ toothpicks as shown in the figure. How many toothpicks does she need to add to complete a $5$ -step staircase?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$
[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse... | [
"We can see that a $1$ -step staircase requires $4$ toothpicks and a $2$ -step staircase requires $10$ toothpicks. Thus, to go from a $1$ -step to $2$ -step staircase, $6$ additional toothpicks are needed and to go from a $2$ -step to $3$ -step staircase, $8$ additional toothpicks are needed. Applying this pattern,... |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_5 | C | 3 | Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned $6$ years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is $30$ years. How many years older than Bella is Anna?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qqu... | [
"Five years ago, Bella was $6$ years old, and the kitten was $0$ years old.\nToday, Bella is $11$ years old, and the kitten is $5$ years old. It follows that Anna is $30-11-5=14$ years old.\nTherefore, Anna is $14-11=\\boxed{3}$ years older than Bella."
] |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_5 | D | 22 | Anna enjoys dinner at a restaurant in Washington, D.C., where the sales tax on meals is 10%. She leaves a 15% tip on the price of her meal before the sales tax is added, and the tax is calculated on the pre-tip amount. She spends a total of 27.50 dollars for dinner. What is the cost of her dinner without tax or tip in ... | [
"Let $x$ be the cost of her dinner.\n$27.50=x+\\frac{1}{10}*x+\\frac{3}{20}*x$\n$27+\\frac{1}{2}=\\frac{5}{4}*x$\n$\\frac{55}{2}=\\frac{5}{4}x$\n$\\frac{55}{2}*\\frac{4}{5}=x$\n$x=22$\n$\\boxed{22}$"
] |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_16 | D | 5 | Annie and Bonnie are running laps around a $400$ -meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{... | [
"Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run $\\boxed{5}$ laps.",
"Call $x$ the distance Annie runs. If Annie is $25... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_12 | D | 83 | Antonette gets $70 \%$ on a 10-problem test, $80 \%$ on a 20-problem test and $90 \%$ on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 77\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 83\qquad\textbf{(E)}\ 87$ | [
"$70 \\% \\cdot 10=7$\n$80 \\% \\cdot 20=16$\n$90 \\% \\cdot 30=27$\nAdding them up gets $7+16+27=50$ . The overall percentage correct would be $\\frac{50}{60}=\\frac{5}{6}=5 \\cdot 16.\\overline{6}=83.\\overline{3} \\approx \\boxed{83}$"
] |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_20 | D | 3 | Any three vertices of the cube $PQRSTUVW$ , shown in the figure below, can be connected to form a triangle. (For example, vertices $P$ $Q$ , and $R$ can be connected to form isosceles $\triangle PQR$ .) How many of these triangles are equilateral and contain $P$ as a vertex?
[asy] unitsize(4); pair P,Q,R,S,T,U,V,W; P=(... | [
"The only equilateral triangles that can be formed are through the diagonals of the faces of the square. From P you have $3$ possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are $3$ possible triangles. So the answer is $\\boxed{3}$ ~Math645\n~andliu766\n~e___",
"Ea... |
https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_3 | B | 144 | Applied to a bill for $\textdollar{10,000}$ the difference between a discount of $40$ % and two successive discounts of $36$ % and $4$ %,
expressed in dollars, is:
$\textbf{(A)}0\qquad \textbf{(B)}144\qquad \textbf{(C)}256\qquad \textbf{(D)}400\qquad \textbf{(E)}416$ | [
"Taking the discount of $40$ % means you're only paying $60$ % of the bill. That results in $10,000\\cdot0.6=\\textdollar{6,000}$\nLikewise, taking two discounts of $36$ % and $4$ % means taking $64$ % of the original amount and then $96$ % of the result. That results in $10,000\\cdot0.64\\cdot0.96=\\textdollar{6... |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_10 | E | 55 | Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?
$\text{(A)}\ 48 \qquad \text{(B)}\ 51 \qquad \text{(C)}\ 52 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$ | [
"Shea has grown $20\\%$ , if x was her original height, then $1.2x = 60$ , so she was originally $\\frac{60}{1.2}=50$ inches tall which is a $60 - 50 = 10$ inch increase. Ara also started off at $50$ inches. Since Ara grew half as much as Shea, Ara grew $\\frac{10}{2} = 5$ inches. Therefore, Ara is now $50+5=55$... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_20 | C | 8 | Arithmetic sequences $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1<a_2 \le b_2$ and $a_n b_n = 2010$ for some $n$ . What is the largest possible value of $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 288 \qquad \textbf{(E)}\ 2009$ | [
"Since $\\left(a_n\\right)$ and $\\left(b_n\\right)$ have integer terms with $a_1=b_1=1$ , we can write the terms of each sequence as\n\\begin{align*}&\\left(a_n\\right) \\Rightarrow \\{1, x+1, 2x+1, 3x+1, ...\\}\\\\ &\\left(b_n\\right) \\Rightarrow \\{1, y+1, 2y+1, 3y+1, ...\\}\\end{align*}\nwhere $x$ and $y$ $x\\... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_24 | B | 621 | Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move... | [
"We say that a game state is an N-position if it is winning for the next player (the player to move), and a P-position if it is winning for the other player. We are trying to find which of the given states is a P-position.\nFirst we note that symmetrical positions are P-positions, as the second player can win by mi... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_22 | B | 621 | Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move... | [
"We say that a game state is an N-position if it is winning for the next player (the player to move), and a P-position if it is winning for the other player. We are trying to find which of the given states is a P-position.\nFirst we note that symmetrical positions are P-positions, as the second player can win by mi... |
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_2 | null | 76 | Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability t... | [
"We first assume a population of $100$ to facilitate solving. Then we simply organize the statistics given into a Venn diagram.\n\nLet $x$ be the number of men with all three risk factors. Since \"the probability that a randomly selected man has all three risk factors, given that he has A and B is $\\frac{1}{3}$ ,... |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_19 | E | 64 | Around the outside of a $4$ by $4$ square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters. Another square, $ABCD$ , has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles. The area... | [
"The radius of each semicircle is $2$ , half the sidelength of the square. The line straight down the middle of square $ABCD$ is the sum of two radii and the length of the smaller square, which is equivalent to its side length. The area of $ABCD$ is $(4+2+2)^2 = \\boxed{64}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_8 | D | 15 | As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $1/2$ mile in front of her. After she passes him, she can see him in her rear mirror until he is $1/2$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of ... | [
"Because they are both moving in the same direction, Emily is riding relative to Emerson $12-8=4$ mph. Now we can look at it as if Emerson is not moving at all [on his skateboard] and Emily is riding at $4$ mph. It takes her\n\\[\\frac12 \\ \\text{mile} \\cdot \\frac{1\\ \\text{hour}}{4\\ \\text{miles}} = \\frac18\... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_13 | C | 20,090 | As shown below, convex pentagon $ABCDE$ has sides $AB=3$ $BC=4$ $CD=6$ $DE=3$ , and $EA=7$ . The pentagon is originally positioned in the plane with vertex $A$ at the origin and vertex $B$ on the positive $x$ -axis. The pentagon is then rolled clockwise to the right along the $x$ -axis. Which side will touch the point ... | [
"The perimeter of the polygon is $3+4+6+3+7 = 23$ . Hence as we roll the polygon to the right, every $23$ units the side $\\overline{AB}$ will be the bottom side.\nWe have $2009 = 23 \\times 87 + 8$ . Thus at some point in time we will get the situation when $A=(2001,0)$ and $\\overline{AB}$ is the bottom side. Obv... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_19 | E | 90 | As shown in the figure below, a regular dodecahedron (the polyhedron consisting of $12$ congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are... | [
"From the top, we can go down in five different ways to the five faces underneath the first face. From here we can go down or go to the adjacent faces. From the face you went down from the top face, you can either go clockwise or counterclockwise $1$ $2$ $3$ ,or $4$ times, or you can go straight down. Then from the... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_19 | E | 810 | As shown in the figure below, a regular dodecahedron (the polyhedron consisting of $12$ congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are... | [
"Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.\nWe have $5$ choices for which... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_7 | D | 170 | As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$ . Point $F$ lies on $\overline{AD}$ so that $DE=DF$ , and $ABCD$ is a square. What is the degree measure of $\angle AFE$
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); p... | [
"By angle subtraction, we have $\\angle ADE = 360^\\circ - \\angle ADC - \\angle CDE = 160^\\circ.$ Note that $\\triangle DEF$ is isosceles, so $\\angle DFE = \\frac{180^\\circ - \\angle ADE}{2}=10^\\circ.$ Finally, we get $\\angle AFE = 180^\\circ - \\angle DFE = \\boxed{170}$ degrees.",
"We can extend $\\overli... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_6 | D | 170 | As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$ . Point $F$ lies on $\overline{AD}$ so that $DE=DF$ , and $ABCD$ is a square. What is the degree measure of $\angle AFE$
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); p... | [
"By angle subtraction, we have $\\angle ADE = 360^\\circ - \\angle ADC - \\angle CDE = 160^\\circ.$ Note that $\\triangle DEF$ is isosceles, so $\\angle DFE = \\frac{180^\\circ - \\angle ADE}{2}=10^\\circ.$ Finally, we get $\\angle AFE = 180^\\circ - \\angle DFE = \\boxed{170}$ degrees.",
"We can extend $\\overli... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_20 | E | 17 | As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\overarc{AEB},\overarc{BFC},$ and $\overarc{CGD},$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$... | [
"Divide the circle into four parts: the top semicircle by connecting E, F, and G( $A$ ); the bottom sector ( $B$ ), whose arc angle is $120^{\\circ}$ because the large circle's radius is $2$ and the short length (the radius of the smaller semicircles) is $1$ , giving a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle... |
https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_15 | E | 17 | As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\overarc{AEB},\overarc{BFC},$ and $\overarc{CGD},$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$... | [
"Divide the circle into four parts: the top semicircle by connecting E, F, and G( $A$ ); the bottom sector ( $B$ ), whose arc angle is $120^{\\circ}$ because the large circle's radius is $2$ and the short length (the radius of the smaller semicircles) is $1$ , giving a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle... |
https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_6 | null | 315 | As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png | [
"Let the interior point be $P$ , let the points on $\\overline{BC}$ $\\overline{CA}$ and $\\overline{AB}$ be $D$ $E$ and $F$ , respectively. Let $x$ be the area of $\\triangle APE$ and $y$ be the area of $\\triangle CPD$ . Note that $\\triangle APF$ and $\\triangle BPF$ share the same altitude from $P$ , so the r... |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_22 | E | 10 | Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had $56$ dollars. The absolute difference between the amounts Ashley and Betty had to spend was $19$ dollars. The absolute difference between the amounts Betty and Carlos had was $7$ dollars, between Car... | [
"Working backwards, if $6 \\le E \\le 10$ , then $6 \\pm 11 \\le A \\le 10 \\pm 11$ . Since $A$ is a positive integer, $17 \\le A \\le 21$\nSince $17 \\le A \\le 21$ , we know that $17 \\pm 19 \\le B \\le 21 \\pm 19$ . But if $B=36$ , which is the smallest possible \"plus\" value, then $E + A + B = 6 + 17 + 36 = ... |
https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_7 | null | 757 | Assume that $a$ $b$ $c$ , and $d$ are positive integers such that $a^5 = b^4$ $c^3 = d^2$ , and $c - a = 19$ . Determine $d - b$ | [
"It follows from the givens that $a$ is a perfect fourth power $b$ is a perfect fifth power, $c$ is a perfect square and $d$ is a perfect cube . Thus, there exist integers $s$ and $t$ such that $a = t^4$ $b = t^5$ $c = s^2$ and $d = s^3$ . So $s^2 - t^4 = 19$ . We can factor the left-hand side of this equation a... |
https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8 | null | 334 | Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+8... | [
"Note that each given equation is of the form \\[f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7\\] for some $k\\in\\{1,2,3\\}.$\nWhen we expand $f(k)$ and combine like terms, we obtain a quadratic function of $k:$ \\[f(k)=ak^2+bk+c,\\] where $a,b,$ and $c$ are linear combinations of $... |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_23 | D | 56 | Assume the adjoining chart shows the $1980$ U.S. population, in millions, for each region by ethnic group. To the nearest percent , what percent of the U.S. Black population lived in the South?
\[\begin{tabular}[t]{c|cccc} & NE & MW & South & West \\ \hline White & 42 & 52 & 57 & 35 \\ Black & 5 & 5 & 15 & 2 \\ Asian... | [
"There are $5+5+15+2=27$ million Blacks living in the U.S. Out of these, $15$ of them live in the South, so the percentage is $\\frac{15}{27}\\approx \\frac{60}{108}\\approx 56\\%$\n$\\boxed{56}$"
] |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_3 | A | 1 | Assuming $a\neq3$ $b\neq4$ , and $c\neq5$ , what is the value in simplest form of the following expression? \[\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}\]
$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) ... | [
"If $x\\neq y,$ then $\\frac{x-y}{y-x}=-1.$ We use this fact to simplify the original expression: \\[\\frac{\\color{red}\\overset{-1}{\\cancel{a-3}}}{\\color{blue}\\underset{1}{\\cancel{5-c}}} \\cdot \\frac{\\color{green}\\overset{-1}{\\cancel{b-4}}}{\\color{red}\\underset{1}{\\cancel{3-a}}} \\cdot \\frac{\\color{b... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_18 | E | 11 | At Central Middle School the $108$ students who take the AMC8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of $15$ cookies, lists this items: $1\frac{1}{2}$ cups flour, $2$ eggs, $3$... | [
"If $108$ students eat $2$ cookies on average, there will need to be $108\\cdot 2 = 216$ cookies. But with the smaller attendance, you will only need $100\\% - 25\\% = 75\\%$ of these cookies, or $75\\% \\cdot 216 = 0.75\\cdot 216 = 162$ cookies.\n$162$ cookies requires $\\frac{162}{15} = 10.8$ batches. However, ... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_17 | C | 5 | At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: $1\frac{1}{2}$ cups of flour, $2$ eggs, $3$... | [
"If $108$ students eat $2$ cookies on average, there will need to be $108\\cdot 2 = 216$ cookies. There are $15$ cookies per pan, meaning there needs to be $\\frac{216}{15} = 14.4$ pans. However, since half-recipes are forbidden, we need to round up and make $\\lceil \\frac{216}{15}\\rceil = 15$ pans.\n$1$ pan re... |
https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_19 | B | 6 | At Central Middle School, the 108 students who take the AMC 8 meet in the evening to talk about food and eat an average of two cookies apiece. Hansel and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists these items: $1\frac{1}{2}$ cups flour, $2$ eggs, $3$ tabl... | [
"For $216$ cookies, you need to make $\\frac{216}{15} = 14.4$ pans. Since fractional pans are forbidden, round up to make $\\lceil \\frac{216}{15} \\rceil = 15$ pans.\nThere are $3$ tablespoons of butter per pan, meaning $3 \\cdot 15 = 45$ tablespoons of butter are required for $15$ pans.\nEach stick of butter has... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_1 | C | 28 | At Euclid Middle School the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Mrs. Germain's class, $8$ students in Mr. Newton's class, and $9$ students in Mrs. Young's class taking the AMC $8$ this year. How many mathematics students at Euclid Middle School are taking the co... | [
"Given that these are the only math teachers at Euclid Middle School and we are told how many from each class are taking the AMC 8, we simply add the three numbers to find the total. $11+8+9=\\boxed{28}$"
] |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_15 | D | 99 | At Euler Middle School, $198$ students voted on two issues in a school referendum with the following results: $149$ voted in favor of the first issue and $119$ voted in favor of the second issue. If there were exactly $29$ students who voted against both issues, how many students voted in favor of both issues?
$\textbf... | [
"We can see that this is a Venn Diagram Problem.\nFirst, we analyze the information given. There are $198$ students. Let's use A as the first issue and B as the second issue.\n$149$ students were for A, and $119$ students were for B. There were also $29$ students against both A and B.\nSolving this without a Venn D... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10A_Problems/Problem_18 | D | 51 | At Jefferson Summer Camp, $60\%$ of the children play soccer, $30\%$ of the children swim, and $40\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?
$\mathrm{(A)}\ 30\% \qquad \mathrm{(B)}\ 40\% \qquad \mathrm{(C)}\ 49\% \qquad \mathrm{(D)}\ 51\% \qquad \mathrm{(... | [
"Out of the soccer players, $40\\%$ swim. As the soccer players are $60\\%$ of the whole, the swimming soccer players are $0.4 \\cdot 0.6 = 0.24 = 24\\%$ of all children.\nThe non-swimming soccer players then form $60\\% - 24\\% = 36\\%$ of all the children.\nOut of all the children, $30\\%$ swim. We know that $24\... |
https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_13 | D | 100 | At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000... | [
"We can set up a system of equations where $a$ is the sets of twins, $b$ is the sets of triplets, and $c$ is the sets of quadruplets. \\[\\begin{split} 2a + 3b + 4c & = 1000 \\\\ b & = 4c \\\\ a & = 3b \\end{split}\\]\nSolving for $c$ and $a$ in the second and third equations and substituting into the first equatio... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_11 | D | 25 | At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they d... | [
"$60\\% \\cdot 20\\% = 12\\%$ of the people that claim that they like dancing actually dislike it, and $40\\% \\cdot 90\\% = 36\\%$ of the people that claim that they dislike dancing actually dislike it. Therefore, the answer is $\\frac{12\\%}{12\\%+36\\%} = \\boxed{25}$",
"Assume WLOG that there are 100 people. ... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_3 | null | 88 | At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, and computer science. There are two male and two female professors in each department. A committee of six professors is to contain three men and three women and must also contain two professors from ea... | [
"There are two cases:\nCase 1: One man and one woman is chosen from each department.\nCase 2: Two men are chosen from one department, two women are chosen from another department, and one man and one woman are chosen from the third department.\nFor the first case, in each department there are ${{2}\\choose{1}} \\ti... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_19 | C | 154 | At a competition with $N$ players, the number of players given elite status is equal to $2^{1+\lfloor \log_{2} (N-1) \rfloor}-N$ . Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N$
$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 90 \qquad \textbf{(C)}\ 154 \qquad \textb... | [
"We start with $2^{1+\\lfloor\\log_{2}(N-1)\\rfloor}-N = 19$ . After rearranging, we get $\\lfloor\\log_{2}(N-1)\\rfloor = \\log_{2} \\left(\\frac{N+19}{2}\\right)$\nSince $\\lfloor\\log_{2}(N-1)\\rfloor$ is a positive integer, $\\frac{N+19}{2}$ must be in the form of $2^{m}$ for some positive integer $m$ . From th... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_8 | B | 245 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\... | [
"Each one of the ten people has to shake hands with all the $20$ other people they don’t know. So $10\\cdot20 = 200$ . From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands from $10$ , or $\\binom... |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_5 | B | 245 | At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf... | [
"All of the handshakes will involve at least one person from the $10$ who knows no one. Label these ten people $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ $I$ $J$\nPerson $A$ from the group of $10$ will initiate a handshake with everyone else ( $29$ people). Person $B$ initiates $28$ handshakes plus the one already counted f... |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_4 | B | 32 | At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The $132$ students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?
$\textbf{(A)} ~23 \qquad\textb... | [
"There are $46$ students paired with a blue partner. The other $11$ students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are $64$ students remaining. Therefore the requested number of pairs is $\\tfrac{64}{2}=\\boxed{32}$ ~Punxsutawney Phil"
] |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_2 | B | 32 | At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The $132$ students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?
$\textbf{(A)} ~23 \qquad\textb... | [
"There are $46$ students paired with a blue partner. The other $11$ students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are $64$ students remaining. Therefore the requested number of pairs is $\\tfrac{64}{2}=\\boxed{32}$ ~Punxsutawney Phil"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_22 | B | 38 | At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is $\frac25$ . What fraction of the people in the room are married men?
$\textbf{(A)}\ \frac13\qquad \textbf{(B)}\ \frac38\qquad \textbf{(C)}\ \frac25\qquad \textbf{(D)}\ \frac{5}{12}\qquad ... | [
"Assume arbitrarily (and WLOG) there are $5$ women in the room, of which $5 \\cdot \\frac25 = 2$ are single and $5-2=3$ are married. Each married woman came with her husband, so there are $3$ married men in the room as well for a total of $5+3=8$ people. The fraction of the people that are married men is $\\boxed{... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_13 | D | 18 | At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?
$\mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24$ | [
"If each man danced with $3$ women, then there will be a total of $3\\times12=36$ pairs of men and women. However, each woman only danced with $2$ men, so there must have been $\\frac{36}2 \\Longrightarrow \\boxed{18}$ women.",
"Consider drawing out a diagram. Let a circle represent a man, and let a shaded circl... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_3 | A | 3.75 | At a store, when a length or a width is reported as $x$ inches that means it is at least $x - 0.5$ inches and at most $x + 0.5$ inches. Suppose the dimensions of a rectangular tile are reported as $2$ inches by $3$ inches. In square inches, what is the minimum area for the rectangle?
$\textbf{(A)}\ 3.75 \qquad\textbf{(... | [
"The minimum dimensions of the rectangle are $1.5$ inches by $2.5$ inches. The minimum area is $1.5\\times2.5=\\boxed{3.75}$ square inches."
] |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_9 | B | 441 | At a twins and triplets convention, there were $9$ sets of twins and $6$ sets of triplets, all from different families. Each twin shook hands with all the twins except his/her siblings and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many ha... | [
"There are $18$ total twins and $18$ total triplets. Each of the twins shakes hands with the $16$ twins not in their family and $9$ of the triplets, a total of $25$ people. Each of the triplets shakes hands with the $15$ triplets not in their family and $9$ of the twins, for a total of $24$ people. Dividing by two ... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_3 | A | 3 | At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made $48$ free throws. How many free throws did she make at the first practice?
$\mathrm{(A) \ } 3 \qquad \mathrm{(B) \ } 6 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 12\qquad... | [
"At the fourth practice she made $48/2=24$ throws, at the third one it was $24/2=12$ , then we get $12/2=6$ throws for the second practice, and finally $6/2=3\\Rightarrow\\boxed{3}$ throws at the first one."
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_1 | A | 3 | At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 48 free throws. How many free throws did she make at the first practice?
$(\mathrm {A}) 3\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm ... | [
"Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made $\\frac{1}{2} \\cdot 48 = 24$ free throws, on the third $12$ , on the second $6$ , and on the first $3 \\Rightarrow \\mathrm{(A)}$\nBecause there are five days, or four transformations between days (day... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_7 | null | 280 | At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number ... | [
"Say the student in the center starts out with $a$ coins, the students neighboring the center student each start with $b$ coins, and all other students start out with $c$ coins. Then the $a$ -coin student has five neighbors, all the $b$ -coin students have three neighbors, and all the $c$ -coin students have four n... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_4 | C | 60 | At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values... | [
"At noon on a certain day, let $M$ and $L$ be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that $M=L+N.$\nAt $4{:}00,$ we get \\begin{align*} |(M-5)-(L+3)| &= 2 \\\\ |M-L-8| &= 2 \\\\ |N-8| &= 2. \\end{align*} We have two cases:\nTogether, the product of all possible values o... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_3 | C | 60 | At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values... | [
"At noon on a certain day, let $M$ and $L$ be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that $M=L+N.$\nAt $4{:}00,$ we get \\begin{align*} |(M-5)-(L+3)| &= 2 \\\\ |M-L-8| &= 2 \\\\ |N-8| &= 2. \\end{align*} We have two cases:\nTogether, the product of all possible values o... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_12 | B | 30 | At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage ... | [
"First, find the amount of money one will pay for three sandals without the discount. We have $\\textdollar 50\\times 3 \\text{ sandals} = \\textdollar 150$\nThen, find the amount of money using the discount: $50 + 0.6 \\times 50 + \\frac{1}{2} \\times 50 = \\textdollar 105$\nFinding the percentage yields $\\frac{1... |
https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_19 | D | 26.9 | At the beginning of a trip, the mileage odometer read $56,200$ miles. The driver filled the gas tank with $6$ gallons of gasoline. During the trip, the driver filled his tank again with $12$ gallons of gasoline when the odometer read $56,560$ . At the end of the trip, the driver filled his tank again with $20$ gallo... | [
"The first six gallons are irrelevant. We start with the odometer at $56,200$ miles, and a full gas tank. The total gas consumed by the car during the trip is equal to the total gas the driver had to buy to make the tank full again, i.e., $12+20=32$ gallons. The distance covered is $57,060 - 56,200 = 860$ miles. He... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_12 | D | 60 | At the beginning of the school year, $50\%$ of all students in Mr. Well's class answered "Yes" to the question "Do you love math", and $50\%$ answered "No." At the end of the school year, $70\%$ answered "Yes" and $30\%$ answered "No." Altogether, $x\%$ of the students gave a different answer at the beginning and end o... | [
"Clearly, the minimum possible value would be $70 - 50 = 20\\%$ . The maximum possible value would be $30 + 50 = 80\\%$ . The difference is $80 - 20 = \\boxed{60}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_6 | D | 60 | At the beginning of the school year, $50\%$ of all students in Mr. Well's class answered "Yes" to the question "Do you love math", and $50\%$ answered "No." At the end of the school year, $70\%$ answered "Yes" and $30\%$ answered "No." Altogether, $x\%$ of the students gave a different answer at the beginning and end o... | [
"Clearly, the minimum possible value would be $70 - 50 = 20\\%$ . The maximum possible value would be $30 + 50 = 80\\%$ . The difference is $80 - 20 = \\boxed{60}$"
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_6 | B | 2 | At the beginning of the school year, Lisa's goal was to earn an $A$ on at least $80\%$ of her $50$ quizzes for the year. She earned an $A$ on $22$ of the first $30$ quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an $A$
$\textbf{(A) }\ 1 \qquad... | [
"Lisa's goal was to get an $A$ on $80\\% \\cdot 50 = 40$ quizzes. She already has $A$ 's on $22$ quizzes, so she needs to get $A$ 's on $40-22=18$ more. There are $50-30=20$ quizzes left, so she can afford to get less than an $A$ on $20-18=\\boxed{2}$ of them."
] |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_4 | B | 2 | At the beginning of the school year, Lisa's goal was to earn an $A$ on at least $80\%$ of her $50$ quizzes for the year. She earned an $A$ on $22$ of the first $30$ quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an $A$
$\textbf{(A) }\ 1 \qquad... | [
"Lisa's goal was to get an $A$ on $80\\% \\cdot 50 = 40$ quizzes. She already has $A$ 's on $22$ quizzes, so she needs to get $A$ 's on $40-22=18$ more. There are $50-30=20$ quizzes left, so she can afford to get less than an $A$ on $20-18=\\boxed{2}$ of them."
] |
https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_8 | D | 55 | At the end of $1994$ , Walter was half as old as his grandmother. The sum of the years in which they were born was $3838$ . How old will Walter be at the end of $1999$
$\textbf{(A)}\ 48 \qquad \textbf{(B)}\ 49\qquad \textbf{(C)}\ 53\qquad \textbf{(D)}\ 55\qquad \textbf{(E)}\ 101$ | [
"In $1994$ , if Water is $x$ years old, then Walter's grandmother is $2x$ years old.\nThis means that Walter was born in $1994 - x$ , and Walter's grandmother was born in $1994 - 2x$\nThe sum of those years is $3838$ , so we have:\n$1994 - x + 1994 - 2x = 3838$\n$3988 - 3x = 3838$\n$x = 50$\nIf Walter is $50$ years... |
https://artofproblemsolving.com/wiki/index.php/1997_AJHSME_Problems/Problem_18 | B | 35 | At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for $$5$ . This week they are on sale at 5 boxes for $$4$ . The percent decrease in the price per box during the sale was closest to
$\text{(A)}\ 30\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 45\% \qquad \tex... | [
"Last week, each box was $\\frac{5}{4} = 1.25$\nThis week, each box is $\\frac{4}{5} = 0.80$\nPercent decrease is given by $\\frac{X_{old} - X_{new}}{X_{old}} \\cdot 100\\%$\nThis, the percent decrease is $\\frac{1.25 - 0.8}{1.25}\\cdot 100\\% = \\frac{45}{125} \\cdot 100\\% = 36\\%$ , which is closest to $\\boxed{... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_2 | B | 38.5 | At the theater children get in for half price. The price for $5$ adult tickets and $4$ child tickets is $$24.50$ . How much would $8$ adult tickets and $6$ child tickets cost?
$\textbf{(A) }$35\qquad \textbf{(B) }$38.50\qquad \textbf{(C) }$40\qquad \textbf{(D) }$42\qquad \textbf{(E) }$42.50$ | [
"Suppose $x$ is the price of an adult ticket. The price of a child ticket would be $\\frac{x}{2}$\n\\begin{eqnarray*} 5x + 4(x/2) = 7x &=& 24.50\\\\ x &=& 3.50\\\\ \\end{eqnarray*}\nPlug in for 8 adult tickets and 6 child tickets.\n\\begin{eqnarray*} 8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\\\ &=&\\boxed{38.50}"
] |
https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_1 | B | 16 | Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?
$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37$ | [
"If Brianna is half as old as Aunt Anna, then Brianna is $\\frac{42}{2}$ years old, or $21$ years old.\nIf Caitlin is $5$ years younger than Brianna, she is $21-5$ years old, or $16$\nSo, the answer is $\\boxed{16}$",
"Since Brianna is half of Aunt Anna's age this means that Brianna is $21$ years old.\nNow we jus... |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_14 | B | 48 | Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed ... | [
"The way to Temple took $\\frac{50}{60}=\\frac56$ hours, and the way back took $\\frac{50}{40}=\\frac54$ for a total of $\\frac56 + \\frac54 = \\frac{25}{12}$ hours. The trip is $50\\cdot2=100$ miles. The average speed is $\\frac{100}{25/12} = \\boxed{48}$ miles per hour.",
"We calculate the harmonic mean of Aust... |
https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems/Problem_22 | D | 148 | Azar and Carl play a game of tic-tac-toe. Azar places an in $X$ one of the boxes in a $3$ -by- $3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all boxes are filled or one of the players has of their symbols in a row... | [
"We need to find out the number of configurations with 3 $O$ and 3 $X$ with 3 $O$ in a row, and 3 $X$ not in a row.\n$\\textbf{Case 1}$ : 3 $O$ are in a horizontal row or a vertical row.\nStep 1: We determine the row that 3 $O$ occupy.\nThe number of ways is 6.\nStep 2: We determine the configuration of 3 $X$\nThe ... |
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_2 | null | 125 | Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probabili... | [
"Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label.\n2022AIMEIIP2.png\nThis problem can be solved by using $2$ cases.\n$\\textbf{Case 1:}$ $C$ 's opponent for the semifinal i... |
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_6 | A | 0.21 | Back in 1930, Tillie had to memorize her multiplication facts from $0 \times 0$ to $12 \times 12$ . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd... | [
"There are a total of $(12+1) \\times (12+1) = 169$ products, and a product is odd if and only if both its factors are odd. There are $6$ odd numbers between $0$ and $12$ , namely $1, 3, 5, 7, 9, 11,$ hence the number of odd products is $6 \\times 6 = 36$ . Therefore the answer is $36/169 \\doteq \\boxed{0.21}$",
... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_8 | B | 5 | Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?
$\textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9$ | [
"By adding a number from Bag A and a number from Bag B together, the values we can get are $3, 5, 7, 5, 7, 9, 7, 9, 11.$ Therefore the number of different values is $\\boxed{5}$",
"The sum of an even number added to an odd number is always odd. The smallest possible sum is $3$ , and the largest possible sum is $1... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | D | 37.5 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. B... | [
"To solve this question, you can use $y = mx + b$ where the $x$ is Fahrenheit and the $y$ is Breadus. We have $(110,0)$ and $(350,100)$ . We want to find the value of $y$ in $(200,y)$ that falls on this line. The slope for these two points is $\\frac{5}{12}$ $y = \\frac{5}{12}x + b$ . Solving for $b$ using $(110, 0... |
https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_5 | E | 22 | Barney Schwinn notices that the odometer on his bicycle reads $1441$ , a palindrome, because it reads the same forward and backward. After riding $4$ more hours that day and $6$ the next, he notices that the odometer shows another palindrome, $1661$ . What was his average speed in miles per hour?
$\textbf{(A)}\ 15\qqua... | [
"Barney travels $1661-1441=220$ miles in $4+6=10$ hours for an average of $220/10=\\boxed{22}$ miles per hour."
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_25 | B | 14 | Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?
[asy] path card=((0,0)--(0,3)--(2,3)--(2,0)--... | [
"Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number(common sum) would be to add another even number to 44, and a different one to 38. Since there is only one even prime (2), the middle card's hidden... |
https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13 | null | 371 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is th... | [
"We casework to find the number of ways to get each possible score. Note that the lowest possible score is $2$ and the highest possible score is $7$ . Let the bijective function $f(x)=\\{1,2,3,4,5,6\\} \\to \\{1,2,3,4,5,6\\}$ denote the row number of the rook for the corresponding column number.\nThus, the expected... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_1 | null | 114 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When ... | [
"Let $x$ be the amount of paint in a strip. The total amount of paint is $130+164+188=482$ . After painting the wall, the total amount of paint is $482-4x$ . Because the total amount in each color is the same after finishing, we have\n\\[\\frac{482-4x}{3} = 130-x\\] \\[482-4x=390-3x\\] \\[x=92\\]\nOur answer is $48... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20 | A | 48 | Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\... | [
"At the beginning of the problem, the Unicorns had played $y$ games and they had won $x$ of these games. From the information given in the problem, we can say that $\\frac{x}{y}=0.45.$ Next, the Unicorns win 6 more games and lose 2 more, for a total of $6+2=8$ games played during district play. We are told that the... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20 | null | 48 | Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\... | [
"Simplifying 45% to $\\frac{9}{20}$ , we see that the numbers of games before district play are a multiple of 20. After that the Aces played 8 more games to the total number of games is in the form of 20x+8 where x is any positive integer. The only answer choice is $\\boxed{48}$ , which is 20(2)+8.",
"First we si... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_17 | A | 704 | Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \t... | [
"Every 10 feet Bella goes, Ella goes 50 feet, which means a total of 60 feet. They need to travel that 60 feet $10560\\div60=176$ times to travel the entire 2 miles. Since Bella goes 10 feet 176 times, this means that she travels a total of 1760 feet. And since she walks 2.5 feet each step, $1760\\div2.5=\\boxed{70... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_20 | A | 7 | Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last ... | [
"Work backwards. The last number Bernardo produces must be in the range $[950,999]$ . That means that before this, Silvia must produce a number in the range $[475,499]$ . Before this, Bernardo must produce a number in the range $[425,449]$ . Before this, Silvia must produce a number in the range $[213,224]$ . Befor... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_20 | null | 7 | Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last ... | [
"The last number that Bernardo says has to be between 950 and 999. Note that $1\\rightarrow 2\\rightarrow 52\\rightarrow 104\\rightarrow 154\\rightarrow 308\\rightarrow 358\\rightarrow 716\\rightarrow 766$ contains 4 doubling actions. Thus, we have $x \\rightarrow 2x \\rightarrow 2x+50 \\rightarrow 4x+100 \\rightar... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_14 | A | 7 | Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last ... | [
"Work backwards. The last number Bernardo produces must be in the range $[950,999]$ . That means that before this, Silvia must produce a number in the range $[475,499]$ . Before this, Bernardo must produce a number in the range $[425,449]$ . Before this, Silvia must produce a number in the range $[213,224]$ . Befor... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12B_Problems/Problem_14 | null | 7 | Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last ... | [
"The last number that Bernardo says has to be between 950 and 999. Note that $1\\rightarrow 2\\rightarrow 52\\rightarrow 104\\rightarrow 154\\rightarrow 308\\rightarrow 358\\rightarrow 716\\rightarrow 766$ contains 4 doubling actions. Thus, we have $x \\rightarrow 2x \\rightarrow 2x+50 \\rightarrow 4x+100 \\rightar... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_25 | E | 25 | Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers ... | [
"First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.\nSay that $N \\equiv a \\pmod{6}$\nalso that $N \\equiv b \\pmod{5}$\nSubstituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b < 5$... |
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