id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
2,801 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_9 | 1 | In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\te... | Let the number of attempted three-point shots be $x$ and the number of attempted two-point shots be $y$ . We know that $x+y=30$ , and we need to evaluate $3(0.2x) + 2(0.3y)$ , as we know that the three-point shots are worth $3$ points and that she made $20$ % of them and that the two-point shots are worth $2$ and that... | 18 |
2,802 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_9 | 2 | In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\te... | The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes $0.2 \cdot 30 = 6$ shots, which are worth $6 \cdot 3 = \boxed{18}$ points. If we assume Shenille only attempts two-pointers, then she ... | 18 |
2,803 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_10 | 1 | A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C... | Let the total amount of flowers be $x$ . Thus, the number of pink flowers is $0.6x$ , and the number of red flowers is $0.4x$ . The number of pink carnations is $\frac{2}{3}(0.6x) = 0.4x$ and the number of red carnations is $\frac{3}{4}(0.4x) = 0.3x$ . Summing these, the total number of carnations is $0.4x+0.3x=0.7x... | 70 |
2,804 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_10 | 2 | A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C... | We have $\dfrac15$ for pink roses, $1-\dfrac6{10}=\dfrac4{10}=\dfrac25$ red flowers, $\dfrac6{10}-\dfrac15=\dfrac35-\dfrac15=\dfrac25$ pink carnations, $\dfrac25\cdot \dfrac34=\dfrac6{20}=\dfrac3{10}$ red carnations we add them up to get $\dfrac25+\dfrac3{10}=\dfrac7{10}=70\%$ so our final answer is 70% or $\boxed{70}$ | 70 |
2,805 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_11 | 1 | A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person plannin... | Let the number of students on the council be $x$ . To select a two-person committee, we can select a "first person" and a "second person." There are $x$ choices to select a first person; subsequently, there are $x-1$ choices for the second person. This gives a preliminary count of $x(x-1)$ ways to choose a two-person ... | 10 |
2,806 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_11 | 2 | A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person plannin... | To choose $2$ people from $n$ total people and get $10$ as a result, we can establish the equation $\binom{n}{2}=10$ which we can easily see $n=5$ , so there are $5$ people. The question asks how many ways to choose $3$ people from the $5$ , so there are $\binom{5}{3}=\boxed{10}$ ways. | 10 |
2,807 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_13 | 1 | How many three-digit numbers are not divisible by $5$ , have digits that sum to less than $20$ , and have the first digit equal to the third digit?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$ | We use a casework approach to solve the problem. These three digit numbers are of the form $\overline{xyx}$ .( $\overline{abc}$ denotes the number $100a+10b+c$ ). We see that $x\neq 0$ and $x\neq 5$ , as $x=0$ does not yield a three-digit integer and $x=5$ yields a number divisible by 5.
The second condition is that t... | 60 |
2,808 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_14 | 1 | A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$ . How many edges does the remaining solid have?
$\textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad$ | We can use Euler's polyhedron formula that says that $F+V=E+2$ . We know that there are originally $6$ faces on the cube, and each corner cube creates $3$ more. $6+8(3) = 30$ . In addition, each cube creates $7$ new vertices while taking away the original $8$ , yielding $8(7) = 56$ vertices. Thus $E+2=56+30$ , so $E... | 84 |
2,809 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_14 | 2 | A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$ . How many edges does the remaining solid have?
$\textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad$ | The removal of each cube adds nine additional edges to the solid. Since a cube initially has $12$ edges and there are eight vertices, the number of edges will be $12 + 9 \times 8 = \boxed{84}$ | 84 |
2,810 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_15 | 1 | Two sides of a triangle have lengths $10$ and $15$ . The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$ | The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,
therefore the length of the side perpendicular to that altitude will be between $10$ and $15$ . The only answer choice that meets this requirement is $\boxed{12... | 12 |
2,811 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_15 | 2 | Two sides of a triangle have lengths $10$ and $15$ . The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$ | Let the height to the side of length $15$ be $h_{1}$ , the height to the side of length 10 be $h_{2}$ , the area be $A$ , and the height to the unknown side be $h_{3}$
Because the area of a triangle is $\frac{bh}{2}$ , we get that $15(h_{1}) = 2A$ and $10(h_{2}) = 2A$ , so, setting them equal, $h_{2} = \frac{3h_{1}}{2}... | 12 |
2,812 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_17 | 1 | Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next $365$ -day period will exactly two friends visit her?
$\textbf{... | The $365$ -day time period can be split up into $6$ $60$ -day time periods, because after $60$ days, all three of them visit again (Least common multiple of $3$ $4$ , and $5$ ).
You can find how many times each pair of visitors can meet by finding the LCM of their visiting days and dividing that number by 60.
Remember ... | 54 |
2,813 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_17 | 2 | Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next $365$ -day period will exactly two friends visit her?
$\textbf{... | From the information above, we get that $A=3x$ $B=4x$ $C=5x$
Now, we want the days in which exactly two of these people meet up
The three pairs are $(A,B)$ $(B,C)$ $(A,C)$
Notice that we are trying to find the LCM of each pair.
Hence, $LCM(A,B)=12x$ $LCM(B,C)=20x$ $LCM(A,C)=15x$
Notice that we want to eliminate when al... | 54 |
2,814 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_18 | 1 | Let points $A = (0, 0)$ $B = (1, 2)$ $C=(3, 3)$ , and $D = (4, 0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$ , where these fractions are in lowest terms. What is $p+q+r+s$
$\textbf{(A)}\ 5... | First, we shall find the area of quadrilateral $ABCD$ . This can be done in any of three ways:
Pick's Theorem $[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.$
Splitting: Drop perpendiculars from $B$ and $C$ to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area i... | 58 |
2,815 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_18 | 2 | Let points $A = (0, 0)$ $B = (1, 2)$ $C=(3, 3)$ , and $D = (4, 0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$ , where these fractions are in lowest terms. What is $p+q+r+s$
$\textbf{(A)}\ 5... | Let the point where the altitude from $E$ to $\overline{AD}$ be labeled $F$ .
Following the steps above, you can find that the height of $\triangle ADE$ is $\frac{15}{8}$ , and from there split the base into two parts, $x$ , and $4-x$ , such that $x$ is the segment from the origin to the point $F$ , and $4-x$ is the se... | 58 |
2,816 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_19 | 1 | In base $10$ , the number $2013$ ends in the digit $3$ . In base $9$ , on the other hand, the same number is written as $(2676)_{9}$ and ends in the digit $6$ . For how many positive integers $b$ does the base- $b$ -representation of $2013$ end in the digit $3$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\... | We want the integers $b$ such that $2013\equiv 3\pmod{b} \Rightarrow b$ is a factor of $2010$ . Since $2010=2 \cdot 3 \cdot 5 \cdot 67$ , it has $(1+1)(1+1)(1+1)(1+1)=16$ factors. Since $b$ cannot equal $1, 2,$ or $3$ , as these cannot have the digit $3$ in their base representations, our answer is $16-3=\boxed{13}$ | 13 |
2,817 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_21 | 1 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate... | Let $x$ be the number of coins. After the $k^{\text{th}}$ pirate takes his share, $\frac{12-k}{12}$ of the original amount is left. Thus, we know that
$x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot \frac{9}{12} \cdot \frac{8}{12} \cdot \frac{7}{12} \cdot \frac{6}{12} \cdot \frac{5}{12} \cdot \frac{4}{12} \cdot \fra... | 925 |
2,818 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_21 | 2 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate... | Solution $1$ mentioned the expression $x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot ... \cdot \frac{1}{12}$ . Note that this is equivalent to $\frac{x \cdot 11!}{12^{11}}$
We can compute the amount of factors of $2$ $3$ $5$ , etc. but this is not necessary. To minimize this expression, we must take out factors of $2... | 925 |
2,819 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_21 | 3 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate... | We know that the 11th pirate takes $\frac{11}{12}$ of what is left from the 10th pirate, so we have the 12th pirate taking \[\frac{12}{12}\cdot(1-\frac{11}{12})=1\cdot\frac{1}{12}\] of what is left from the 10th pirate. Similarly, since the 10th pirate takes $\frac{10}{12}$ of what is left from the 9th pirate, we have ... | 925 |
2,820 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | 1 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)... | Let $BX = q$ $CX = p$ , and $AC$ meets the circle at $Y$ and $Z$ , with $Y$ on $AC$ . Then $AZ = AY = 86$ . Using the Power of a Point (Secant-Secant Power Theorem), we get that $p(p+q) = 11(183) = 11 * 3 * 61$ . We know that $p+q>p$ , so $p$ is either $3$ $11$ , or $33$ . We also know that $p>11$ by the triangle i... | 61 |
2,821 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | 2 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)... | Let $x$ represent $CX$ , and let $y$ represent $BX$ . Since the circle goes through $B$ and $X$ $AB = AX = 86$ .
Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 y + xy^2 + 86^2 y = 97^2 y$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to... | 61 |
2,822 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | 3 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)... | Let $CX=x, BX=y$ . Let the circle intersect $AC$ at $D$ and the diameter including $AD$ intersect the circle again at $E$ .
Use power of a point on point C to the circle centered at A.
So $CX \cdot CB=CD \cdot CE \Rightarrow x(x+y)=(97-86)(97+86) \Rightarrow x(x+y)=3*11*61$
Obviously $x+y>x$ so we have three solution p... | 61 |
2,823 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | 4 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)... | [asy] unitsize(2); import olympiad; import graph; pair A,B,C,D,E; A = (0,0); B = (70,51); C = (97,0); D = (82,29); E = (76,40); draw(Circle((0,0),86.609)); draw(A--B--C--A); draw(A--B--E--A); draw(A--D); dot(A); dot(B,blue); dot(C); dot(D,blue); dot(E); label("A",A,S); label("B",B,NE); label("C",C,S); label("D",D,NE)... | 61 |
2,824 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_23 | 5 | In $\triangle ABC$ $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)... | Let $E$ be the foot of the altitude from $A$ to $BX.$ Since $\triangle ABX$ is isosceles $AX=AB=86,EB=EX,$ and the answer is $EC+EB=EC+EX.$ $(EC+EX)(EC-EX)=EC^2-EX^2=(97^2-AE^2)-(86^2-AE^2)=97^2-86^2=2013$ by the Pythagorean Theorem. Only $EC+EX=\boxed{61}.$ | 61 |
2,825 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_24 | 1 | Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how ... | Let us label the players of the first team $A$ $B$ , and $C$ , and those of the second team, $X$ $Y$ , and $Z$
$\textbf{1}$ . One way of scheduling all six distinct rounds could be:
Round 1: $AX$ $BY$ $CZ$
Round 2: $AX$ $BZ$ $CY$
Round 3: $AY$ $BX$ $CZ$
Round 4: $AY$ $BZ$ $CX$
Round 5: $AZ$ $BX$ $CY$
Round 6: $AZ$ $BY$... | 900 |
2,826 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_24 | 2 | Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how ... | Label the players of the first team $A$ $B$ , and $C$ , and those of the second team, $X$ $Y$ , and $Z$ . We can start by assigning an opponent to person $A$ for all $6$ games. Since $A$ has to play each of $X$ $Y$ , and $Z$ twice, there are $\frac{6!}{2!2!2!} = 90$ ways to do this. We can assume that the opponents for... | 900 |
2,827 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25 | 1 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | If you draw a clear diagram like the one below, it is easy to see that there are $\boxed{49}$ points. | 49 |
2,828 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25 | 2 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | Let the number of intersections be $x$ . We know that $x\le \dbinom{8}{4} = 70$ , as every $4$ vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. However, four diagonals intersect in the center, so we need to subtract $\dbinom{4}{2} -1 = 5$ from this count, $70-5... | 49 |
2,829 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25 | 4 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | Like solution one, we may draw. Except note that the octagon has eight regions, and each region has an equal number of points, so drawing only one of the eight regions and the intersection points suffices. One of the eight regions contains $8$ points (not including the octagon center). However each adjacent region shar... | 49 |
2,830 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_25 | 5 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | [asy] size(8cm); pathpen = black; // draw the circle pair[] A; for (int i=0; i<8; ++i) { A[i] = dir(45*i); } D(CR((0,0), 1)); // draw the octagon and diagonals // choose pen colors pen[] colors; colors[1] = yellow; colors[2] = purple; colors[3] = green; colors[4] = orange; for (int d=1; d<=4; ++d) { pathpen = colors[d]... | 49 |
2,831 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_2 | 1 | Mr. Green measures his rectangular garden by walking two of the sides and finds that it is $15$ steps by $20$ steps. Each of Mr. Green's steps is $2$ feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?
$\textbf{(A)}\... | Since each step is $2$ feet, his garden is $30$ by $40$ feet. Thus, the area of $30(40) = 1200$ square feet. Since he is expecting $\frac{1}{2}$ of a pound per square foot, the total amount of potatoes expected is $1200 \times \frac{1}{2} = \boxed{600}$ | 600 |
2,832 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_3 | 1 | On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3$ . In degrees, what was the low temperature in Lincoln that day?
$\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \te... | Let $L$ be the low temperature. The high temperature is $L+16$ . The average is $\frac{L+(L+16)}{2}=3$ . Solving for $L$ , we get $L=\boxed{5}$ | 5 |
2,833 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_4 | 1 | When counting from $3$ to $201$ $53$ is the $51^{st}$ number counted. When counting backwards from $201$ to $3$ $53$ is the $n^{th}$ number counted. What is $n$
$\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150$ | Note that $n$ is equal to the number of integers between $53$ and $201$ , inclusive. Thus, $n=201-53+1=\boxed{149}$ | 149 |
2,834 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_5 | 1 | Positive integers $a$ and $b$ are each less than $6$ . What is the smallest possible value for $2 \cdot a - a \cdot b$
$\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2$ | Factoring the equation gives $a(2 - b)$ . From this we can see that to obtain the least possible value, $2 - b$ should be negative, and should be as small as possible. To do so, $b$ should be maximized. Because $2 - b$ is negative, we should maximize the positive value of $a$ as well. The maximum values of both $a$ and... | 15 |
2,835 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_8 | 1 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qqu... | Let Ray and Tom drive 40 miles. Ray's car would require $\frac{40}{40}=1$ gallon of gas and Tom's car would require $\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles, on $1+4=5$ gallons of gas, for a combined rate of $\frac{80}{5}=$ $\boxed{16}$ | 16 |
2,836 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_8 | 2 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qqu... | Taking the harmonic mean of the two rates, we get \[\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{16}.\] | 16 |
2,837 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_8 | 3 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qqu... | Let the number of miles that Ray and Tom each drive be denoted by $m$ . Thus the number of gallons Ray's car uses can be represented by $\frac{m}{40}$ and the number of gallons that Tom's car uses can likewise be expressed as $\frac{m}{10}$ . Thus the total amount of gallons used by both cars can be expressed as $\frac... | 16 |
2,838 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_9 | 1 | Three positive integers are each greater than $1$ , have a product of $27000$ , and are pairwise relatively prime. What is their sum?
$\textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165$ | The prime factorization of $27000$ is $2^3*3^3*5^3$ . These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $\boxed{160}$ | 160 |
2,839 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_10 | 1 | 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\textb... | 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 two poi... | 20 |
2,840 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_11 | 2 | Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$ . What is $x+y$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | If we move every term including $x$ or $y$ to the LHS, we get \[x^2 - 10x + y^2 + 6y = -34.\] We can complete the square to find that this equation becomes \[(x - 5)^2 + (y + 3)^2 = 0.\] Since the square of any real number is nonnegative, we know that the sum is greater than or equal to $0$ . Equality holds when the v... | 2 |
2,841 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_12 | 1 | Let $S$ be the set of sides and diagonals of a regular pentagon. A pair of elements of $S$ are selected at random without replacement. What is the probability that the two chosen segments have the same length?
$\textbf{(A) }\frac{2}5\qquad\textbf{(B) }\frac{4}9\qquad\textbf{(C) }\frac{1}2\qquad\textbf{(D) }\frac{5}9\qq... | The problem is simply asking how many ways are there to choose two sides or two diagonals. Hence, the probability is \[\dfrac{\binom{5}{2} + \binom{5}{2}}{\binom{10}{2}} = \dfrac{10+10}{45} = \dfrac{20}{45}=\boxed{49}\] | 49 |
2,842 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_13 | 1 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad ... | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", $1+2+3+4+5+6+7+8+9=45$ numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because $53-... | 8 |
2,843 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_13 | 2 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad ... | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. We notice that the number of numbers is $1 + 2 + 3 + 4 ...$ every time we finish a "turn" we notice the sum of these would be the largest number $\frac{n... | 8 |
2,844 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_13 | 3 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad ... | Let $T(n)$ denote the $n$ th triangle number. Then, observe that the $T(n)$ th number said is $n$ . It follows that the $55$ th number is $10$ (as $55 = T(10)$ ). Thus, the $53$ rd number is $10 - 2 = 8$ , which is answer choice $\boxed{8}$ | 8 |
2,845 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_17 | 1 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges... | If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first ... | 103 |
2,846 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_17 | 2 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges... | We can approach this problem by assuming he goes to the red booth first. You start with $75 \text{R}$ and $75 \text{B}$ and at the end of the first booth, you will have $1 \text{R}$ and $112 \text{B}$ and $37 \text{S}$ . We now move to the blue booth, and working through each booth until we have none left, we will end ... | 103 |
2,847 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_17 | 3 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges... | Let $x$ denote the number of visits to the first booth and $y$ denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows: \[R(x,y)=-2x+y+75\] \[B(x,y)=x-3y+75\] There are no legal exchanges when he has fewer than $2$ red coins and fewer than $3$ blue coins... | 103 |
2,848 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_18 | 1 | The number $2013$ has the property that its units digit is the sum of its other digits, that is $2+0+1=3$ . How many integers less than $2013$ but greater than $1000$ have this property?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58$ | We take cases on the thousands digit, which must be either $1$ or $2$ :
If the number is of the form $\overline{1bcd},$ where $b, c, d$ are digits, then we must have $d = 1 + b + c.$ Since $d \le 9,$ we must have $b + c \le 9 - 1 = 8.$ By casework on the value of $b$ , we find that there are $1 + 2 + \dots + 9 = 45$ po... | 46 |
2,849 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_18 | 2 | The number $2013$ has the property that its units digit is the sum of its other digits, that is $2+0+1=3$ . How many integers less than $2013$ but greater than $1000$ have this property?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58$ | Let's start with the case that starts with $200$ . We have only one number, which is $2002$ . If we look at the $1900s$ , we have no solutions because $1+9 = 10$ , and because we can only use digits from $1$ through $9$ , it is impossible. If we looks at the $1800s$ , we do have one solution, which is $1809$ . If we lo... | 46 |
2,850 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_20 | 1 | The number $2013$ is expressed in the form
where $a_1 \ge a_2 \ge \cdots \ge a_m$ and $b_1 \ge b_2 \ge \cdots \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | The prime factorization of $2013$ is $61\cdot11\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and will le... | 2 |
2,851 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_21 | 1 | Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$ . What is the smallest possible value of $N$
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 8... | Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$ . Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$ . Note that this means that $x_{1}$ and $y_{1}$ must have the same value modulo $8$ . To minimize, let one... | 104 |
2,852 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_21 | 2 | Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$ . What is the smallest possible value of $N$
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 8... | WLOG, let $a_i$ $b_i$ be the sequences with $a_1<b_1$ . Then \[N=5a_1+8a_2=5b_1+8b_2\] or \[5a_1+8a_2=5(a_1+c)+8(a_2-d)\] for some natural numbers $c$ $d$ . Thus $5c=8d$ . To minimize $c$ and $d$ , we have $(c,d)=(8,5)$ , or \[5a_1+8a_2=5(a_1+8)+8(a_2-5).\] To minimize $a_1$ and $b_1$ , we have $(a_1,b_1)=(0,0+c)=(0,8)... | 104 |
2,853 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_22 | 1 | The regular octagon $ABCDEFGH$ has its center at $J$ . Each of the vertices and the center are to be associated with one of the digits $1$ through $9$ , with each digit used once, in such a way that the sums of the numbers on the lines $AJE$ $BJF$ $CJG$ , and $DJH$ are all equal. In how many ways can this be done?
$\... | First of all, note that $J$ must be $1$ $5$ , or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:
[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2)... | 152 |
2,854 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_22 | 2 | The regular octagon $ABCDEFGH$ has its center at $J$ . Each of the vertices and the center are to be associated with one of the digits $1$ through $9$ , with each digit used once, in such a way that the sums of the numbers on the lines $AJE$ $BJF$ $CJG$ , and $DJH$ are all equal. In how many ways can this be done?
$\... | As in solution 1, $J$ must be $1$ $5$ , or $9$ giving us 3 choices. Additionally $A+E = B+F = C+G = D+H$ . This means once we choose $J$ there are $8$ remaining choices. Going clockwise from $A$ we count, $8$ possibilities for $A$ . Choosing $A$ also determines $E$ which leaves $6$ choices for $B$ , once $B$ is chosen... | 152 |
2,855 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_23 | 1 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overli... | Since $\angle{AFB}=\angle{ADB}=90^{\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ , so $\triangle ABF \sim \triangle ADE$ are similar. In addition, $\triangle ADE \sim \triangle ACD$ . We can easily find $AD=12$ $BD = 5$ , and $DC=9$ using Pythagorean triples.
So, the ratio of the l... | 21 |
2,856 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_23 | 2 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overli... | From solution 1, we know that $AD = 12$ and $DC = 9$ . Since $\triangle ADC \sim \triangle DEC$ , we can figure out that $DE = \frac{36}{5}$ . We also know what $AC$ is so we can figure what $AE$ is: $AE = 15 - \frac{27}{5} = \frac{48}{5}$ . Quadrilateral $ABDF$ is cyclic, implying that $\angle{B} + \angle{DFA}$ = 180°... | 21 |
2,857 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_23 | 3 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overli... | If we draw a diagram as given, but then add point $G$ on $\overline{BC}$ such that $\overline{FG}\perp\overline{BC}$ in order to use the Pythagorean theorem, we end up with similar triangles $\triangle{DFG}$ and $\triangle{DCE}$ . Thus, $FG=\tfrac35x$ and $DG=\tfrac45x$ , where $x$ is the length of $\overline{DF}$ . Us... | 21 |
2,858 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_23 | 4 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overli... | First, we find $BD = 5$ $DC = 9$ , and $AD = 12$ via the Pythagorean Theorem or by using similar triangles. Next, because $DE$ is an altitude of triangle $ADC$ $DE = \frac{AD\cdot DC}{AC} = \frac{36}{5}$ . Using that, we can use the Pythagorean Theorem and similar triangles to find $EC = \frac{27}{5}$ and $AE = \frac{4... | 21 |
2,859 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_24 | 1 | A positive integer $n$ is nice if there is a positive integer $m$ with exactly four positive divisors (including $1$ and $m$ ) such that the sum of the four divisors is equal to $n$ . How many numbers in the set $\{ 2010,2011,2012,\dotsc,2019 \}$ are nice?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \q... | A positive integer with only four positive divisors has its prime factorization in the form of $a \cdot b$ , where $a$ and $b$ are both prime positive integers or $c^3$ where $c$ is a prime. One can easily deduce that none of the numbers are even near a cube so the second case is not possible. We now look at the case o... | 1 |
2,860 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_24 | 2 | A positive integer $n$ is nice if there is a positive integer $m$ with exactly four positive divisors (including $1$ and $m$ ) such that the sum of the four divisors is equal to $n$ . How many numbers in the set $\{ 2010,2011,2012,\dotsc,2019 \}$ are nice?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \q... | After deducing that $2012$ and case $1$ is impossible, and since there is no option for $0$ $2016$ is obviously a solution and the answer is $\boxed{1}$ | 1 |
2,861 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_25 | 1 | 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.
Say that $N \equiv a \pmod{6}$
also that $N \equiv b \pmod{5}$
Substituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b < 5$ (because ot... | 25 |
2,862 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_25 | 2 | 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 ... | Notice that there are exactly $1000-100=900=5^2\cdot 6^2$ possible values of $N$ . This means, in $100\le N\le 999$ , every possible combination of $2$ digits will happen exactly once. We know that $N=900,901,902,903,904$ works because $900\equiv\dots00_5\equiv\dots00_6$
We know for sure that the units digit will add p... | 25 |
2,863 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_25 | 3 | 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 ... | Notice that $N_5$ ranges from $3$ to $5$ digits and $N_6$ ranges from $3$ to $4$ digits.
Then let $a_i$ $b_i$ denotes the digits of $N_5$ $N_6$ , respectively such that \[0\le a_i<5,0\le b_i<6\] Thus we have \[N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4\] \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\... | 25 |
2,864 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_25 | 4 | 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 ... | Observe that the maximum possible value of the sum of the last two digits of the base $5$ number and the base $6$ number is $44+55=99$ .
Let $N \equiv a \pmod {25}$ and $N \equiv b \pmod {36}$
If $a < \frac{25}{2}$ $2N \equiv 2a \pmod {25}$ and if $a > \frac{25}{2}$ $2N \equiv 2a - 25 \pmod {25}$
Using the same logic ... | 25 |
2,865 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_1 | 1 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | Cagney can frost one in $20$ seconds, and Lacey can frost one in $30$ seconds. Working together, they can frost one in $\frac{20\cdot30}{20+30} = \frac{600}{50} = 12$ seconds. In $300$ seconds ( $5$ minutes), they can frost $\boxed{25}$ cupcakes. | 25 |
2,866 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_1 | 2 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | In $300$ seconds ( $5$ minutes), Cagney will frost $\dfrac{300}{20} = 15$ cupcakes, and Lacey will frost $\dfrac{300}{30} = 10$ cupcakes. Therefore, working together they will frost $15 + 10 = \boxed{25}$ cupcakes. | 25 |
2,867 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_1 | 3 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | Since Cagney frosts $3$ cupcakes a minute, and Lacey frosts $2$ cupcakes a minute, they together frost $3+2=5$ cupcakes a minute. Therefore, in $5$ minutes, they frost $5\times5 = 25 \Rightarrow \boxed{25}$ | 25 |
2,868 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_3 | 1 | 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?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$ | [asy] draw((-2,1)--(-6,1),red+dashed,EndArrow); draw((-6,2)--(5,2),blue+dashed,EndArrow); dot((-2,0)); dot((-6,0)); dot((5,0)); label("$-2$",(-2,0),dir(270)); label("$-6$",(-6,0),dir(270)); label("$5$",(5,0),dir(270)); label("$4$",(-4,0.9),dir(270)); label("$11$",(-1.5,2.5),dir(90)); [/asy]
Crawling from $-2$ to $-6$ t... | 15 |
2,869 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_4 | 1 | Let $\angle ABC = 24^\circ$ and $\angle ABD = 20^\circ$ . What is the smallest possible degree measure for $\angle CBD$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 12$ | $\angle ABD$ and $\angle ABC$ share ray $AB$ . In order to minimize the value of $\angle CBD$ $D$ should be located between $A$ and $C$
$\angle ABC = \angle ABD + \angle CBD$ , so $\angle CBD = 4$ . The answer is $\boxed{4}$ | 4 |
2,870 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_5 | 1 | Last year 100 adult cats, half of whom were female, were brought into the Smallville Animal Shelter. Half of the adult female cats were accompanied by a litter of kittens. The average number of kittens per litter was 4. What was the total number of cats and kittens received by the shelter last year?
$\textbf{(A)}\ 150\... | Half of the 100 adult cats are female, so there are $\frac{100}{2}$ $50$ female cats. Half of those female adult cats have a litter of kittens, so there would be $\frac{50}{2}$ $25$ litters. Since the average number of kittens per litter is 4, this implies that there are $25 \times 4$ $100$ kittens. So the total num... | 200 |
2,871 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_8 | 1 | The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Let the three numbers be equal to $a$ $b$ , and $c$ . We can now write three equations:
$a+b=12$
$b+c=17$
$a+c=19$
Adding these equations together, we get that
$2(a+b+c)=48$ and
$a+b+c=24$
Substituting the original equations into this one, we find
$c+12=24$
$a+17=24$
$b+19=24$
Therefore, our numbers are 12, 7, and 5. T... | 7 |
2,872 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_8 | 2 | The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Let the three numbers be $a$ $b$ and $c$ and $a<b<c$ . We get the three equations:
$a+b=12$
$a+c=17$
$b+c=19$
To isolate $b$ , We add the first and last equations and then subtract the second one.
$(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7$
Because $b$ is the middle number, the middle number is $... | 7 |
2,873 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_9 | 1 | A pair of six-sided dice are labeled so that one die has only even numbers (two each of 2, 4, and 6), and the other die has only odd numbers (two of each 1, 3, and 5). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 7?
$\textbf{(A)}\ \frac{1}{6}\qquad\textb... | The total number of combinations when rolling two dice is $6*6 = 36$
There are three ways that a sum of 7 can be rolled. $2+5$ $4+3$ , and $6+1$ . There are two 2's on one die and two 5's on the other, so there are a total of 4 ways to roll the combination of 2 and 5. There are two 4's on one die and two 3's on the oth... | 13 |
2,874 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_9 | 2 | A pair of six-sided dice are labeled so that one die has only even numbers (two each of 2, 4, and 6), and the other die has only odd numbers (two of each 1, 3, and 5). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 7?
$\textbf{(A)}\ \frac{1}{6}\qquad\textb... | Assume we roll the die with only evens first. For whatever value rolled, there are exactly 2 faces on the odd die that makes the sum 7. The odd die has 6 faces, so our probability is $\boxed{13}$ | 13 |
2,875 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_10 | 1 | Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12... | Let $a_1$ be the first term of the arithmetic progression and $a_{12}$ be the last term of the arithmetic progression. From the formula of the sum of an arithmetic progression (or arithmetic series), we have $12*\frac{a_1+a_{12}}{2}=360$ , which leads us to $a_1 + a_{12} = 60$ $a_{12}$ , the largest term of the progres... | 8 |
2,876 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_10 | 2 | Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12... | If we let $a$ be the smallest sector angle and $r$ be the difference between consecutive sector angles, then we have the angles $a, a+r, a+2r, \cdots. a+11r$ . Use the formula for the sum of an arithmetic sequence and set it equal to 360, the number of degrees in a circle.
\begin{align*} \frac{a+a+11r}{2}\cdot 12 &= 36... | 8 |
2,877 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_10 | 3 | Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12... | Starting with the smallest term, $a - 5x \cdots a, a + x \cdots a + 6x$ where $a$ is the sixth term and $x$ is the difference. The sum becomes $12a + 6x = 360$ since there are $360$ degrees in the central angle of the circle. The only condition left is that the smallest term in greater than zero. Therefore, $a - 5x > 0... | 8 |
2,878 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_11 | 1 | Externally tangent circles with centers at points $A$ and $B$ have radii of lengths $5$ and $3$ , respectively. A line externally tangent to both circles intersects ray $AB$ at point $C$ . What is $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4$ | Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$ $AB=8$ . Also, let $BC=x$ . As $\angle ADC$ and $\angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $\angle ACD$ $\triangle ADC \sim \triangle BEC$ . From this we can... | 12 |
2,879 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_14 | 1 | Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?
$\textbf{(A)}\ 480 \qquad\textbf{(B)}\ 481 \qquad\textbf{(C)}\ 482 ... | There are 15 rows with 15 black tiles, and 16 rows with 16 black tiles, so the answer is $15^2+16^2 =225+256= \boxed{481}$ | 481 |
2,880 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_14 | 2 | Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?
$\textbf{(A)}\ 480 \qquad\textbf{(B)}\ 481 \qquad\textbf{(C)}\ 482 ... | We build the $31 \times 31$ checkerboard starting with a board of $30 \times 30$ that is exactly half black. There are $15 \cdot 30$ black tiles in this region.
Add to this $30 \times 30$ checkerboard a $1 \times 30$ strip on the bottom that has $15$ black tiles.
Add to this $31 \times 30$ checkerboard a $31 \times 1$... | 481 |
2,881 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_14 | 3 | Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?
$\textbf{(A)}\ 480 \qquad\textbf{(B)}\ 481 \qquad\textbf{(C)}\ 482 ... | Drawing smaller scale sketches, we notice that the odd columns of an $n \times n$ (where $n$ is odd) board have $\left \lceil \frac{n}{2} \right \rceil$ black tiles, while the even columns have $\left \lfloor \frac{n}{2} \right \rfloor$ black tiles. In our case, we have a $31 \times 31$ board. We have $16$ odd columns,... | 481 |
2,882 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_15 | 1 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | $AC$ intersects $BC$ at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so $\triangle ABC \sim \triangle BED$ . The hypotenuse of right triangle BED is $\sqrt{1^2+2^2}=\sqrt{5}$
\[\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rig... | 15 |
2,883 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_15 | 2 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | Let $\text{E}$ be the origin. Then, $\text{D}=(1, 0)$ $\text{A}=(0, 2)$ $\text{B}=(1, 2)$ $\text{F}=(2, 1)$
${EB}$ can be represented by the line $y=2x$ Also, ${AF}$ can be represented by the line $y=-\frac{1}{2}x+2$
Subtracting the second equation from the first gives us $\frac{5}{2}x-2=0$ .
Thus, $x=\frac{4}{5}$ .
P... | 15 |
2,884 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_15 | 3 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | Triangle $EAB$ is similar to triangle $EHI$ ; line $HI = 1/2$
Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \frac{3}{2} = 2: 3$
Based on similarity the length of the height of $GC$ is thus $\frac{2}{5}\cdot1 = \frac{2}{5}$
Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1... | 15 |
2,885 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_15 | 4 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | Let $L$ be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let $K$ be the top right corner of the top right unit square, where segment $ABL$ is 2 units in length.
Because of the Pythagorean Theorem, since $AC = 2$ and $LK$ = 1, the diagonal of triangle $ALK$ is $\sqrt... | 15 |
2,886 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | 1 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Since $a$ and $b$ are relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $\frac{73}{3}$ , the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$ . Looking at the answer choices, the only multiple of $3$ is $\boxed{3}$ | 3 |
2,887 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | 2 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Using difference of cubes in the numerator and cancelling out one $(a-b)$ in the numerator and denominator gives $\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}$
Set $x = a^2 + b^2$ , and $y = ab$ . Then $\frac{x + y}{x - 2y} = \frac{73}{3}$ . Cross multiplying gives $3x + 3y = 73x - 146y$ , and simplifying gi... | 3 |
2,888 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | 3 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | The first step is the same as above which gives $\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}$
Then we can subtract $3ab$ and then add $3ab$ to get $\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}$ , which gives $1+\frac{3ab}{(a-b)^2}=\frac{73}{3}$ $\frac{3ab}{(a-b)^2}=\frac{70}{3}$ .
Cross multiply $9ab=70(a-b)^2$ . S... | 3 |
2,889 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | 4 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Slightly expanding, we have that $\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$
Canceling the $(a-b)$ , cross multiplying, and simplifying, we obtain that
$0=70a^2-149ab+70b^2$ .
Dividing everything by $b^2$ , we get that
$0=70(\frac{a}{b})^2-149(\frac{a}{b})+70$
Applying the quadratic formula....and followin... | 3 |
2,890 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | 6 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Let us rewrite the expression as $\frac{(a-b)^2 + 3ab}{(a-b)^2}$ . Now letting $x = a - b$ , we simplify the expression to $\frac{70x^2 + 3ab}{x^2} = \frac{73}{3}$ . Cross multiplying and doing a bit of simplification, we obtain that $ab = \frac{70x^2}{9}$ . Since $a$ and $b$ are both integers, we know that $\frac{70x^... | 3 |
2,891 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_19 | 1 | Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24%... | Let Paula work at a rate of $p$ , the two helpers work at a combined rate of $h$ , and the time it takes to eat lunch be $L$ , where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:
\[(8-L)(p+h)=50\]
\[(6.2-L)h=24\]
\... | 48 |
2,892 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_19 | 2 | Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24%... | Because Paula worked from \[8:00 \text{A.M.}\] to \[7:12 \text{P.M.}\] , she worked for 11 hours and 12 minutes = 672 minutes. Since there is $100-50-24=26$ % of the house left, we get the equation $26a=672$ . Because $672$ is $22$ mod $26$ , looking at our answer choices, the only answer that is $22$ $\text{mod}$ $26$... | 48 |
2,893 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_22 | 1 | The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$
$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$ | The sum of the first $m$ odd integers is given by $m^2$ . The sum of the first $n$ even integers is given by $n(n+1)$
Thus, $m^2 = n^2 + n + 212$ . Since we want to solve for n, rearrange as a quadratic equation: $n^2 + n + (212 - m^2) = 0$
Use the quadratic formula: $n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}$ . Since... | 255 |
2,894 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_22 | 2 | The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$
$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$ | As above, start off by noting that the sum of the first $m$ odd integers $= m^2$ and the sum of the first $n$ even integers $= n(n+1)$ . Clearly $m > n$ , so let $m = n + a$ , where $a$ is some positive integer. We have:
$(n+a)^2 = n(n+1) + 212$ .
Expanding, grouping like terms and factoring, we get: $n = \frac{(212 - ... | 255 |
2,895 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_22 | 3 | The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$
$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$ | Using the closed forms for the sums, we get $m^2=n(n+1)+212$ , or $m^2=n^2+n+212$ . We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to let us complete the square. Our equation is now $4m^2=4n^2+4n+848$ . Complete the square on the right hand side: ... | 255 |
2,896 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_23 | 1 | Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?
$\text{(A)}\ 60\qquad\... | Note that if $n$ is the number of friends each person has, then $n$ can be any integer from $1$ to $4$ , inclusive.
One person can have at most 4 friends since they cannot be all friends (stated in the problem).
Also note that the cases of $n=1$ and $n=4$ are the same, since a map showing a solution for $n=1$ can corre... | 170 |
2,897 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_24 | 1 | Let $a$ $b$ , and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$
What is $a$
$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$ | Add the two equations.
$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$
Now, this can be rearranged and factored.
$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$
$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$
$a$ $b$ , and $c$ are all integers, so the three terms on the left side of the equation must all be perfe... | 253 |
2,898 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_25 | 1 | Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\... | Since $x,y,z$ are all reals located in $[0, n]$ , the number of choices for each one is continuous so we use geometric probability.
WLOG( Without loss of generality ), assume that $n\geq x \geq y \geq z \geq 0$ . Then the set of points $(x,y,z)$ is a tetrahedron, or a triangular pyramid. The point $(x,y,z)$ distributes... | 10 |
2,899 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_25 | 2 | Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\... | Because $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ , which means that $x$ $y$ , and $z$ distributes uniformly and independently in the interval $[0,n]$ . So the point $(x, y, z)$ distributes uniformly in the cubic $0\leqslant x, y, z \leqslant n$ , as shown in the figure below. T... | 10 |
2,900 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_25 | 3 | Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\... | If $x$ $y$ , and $z$ are separated by at least one, then by subtracting the minimum space between the three variables (which is $2$ ), $x$ $y$ , and $z$ can be chosen randomly in the interval $[0,n-2]$ . The probability is hence $\dfrac{(n-2)^3}{n^3} > \dfrac{1}{2}.$ $\boxed{10}$ is the minimum value in the answer choi... | 10 |
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