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 |
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2,701 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_18 | 1 | Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$ . Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$
$\textbf{(A) ... | Notice that $1000$ is $3E8$ when converted to hexadecimal ( $3 \cdot 16^2 + 14 \cdot 16^1 + 8 \cdot 16^0$ ). We will proceed by constructing numbers that consist of only numeric digits in hexadecimal.
The first digit could be $0,$ $1,$ $2,$ or $3,$ and the second two could be any digit $0 - 9$ , giving $4 \cdot 10 \cdo... | 21 |
2,702 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_18 | 2 | Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$ . Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$
$\textbf{(A) ... | First, we set a bound by writing $1000$ in base- $16$ $1000_{10}=3E8_{16}$ . Therefore, we are considering numbers with a maximum of $3$ digits, and a maximum of $3$ in the $256$ ths-place (the first place in a $3$ -digit number).
Case $1$ $1$ -digit numbers:
There are evidently $9$ numbers that fit this category.
Case... | 21 |
2,703 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_18 | 3 | Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$ . Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$
$\textbf{(A) ... | We can quickly see that $400$ in hexadecimal = $0+0+256*4$ = 1024. If we go down to 399 in hexadecimal, we have $9+9*16+3*256$ which is $921$ , which is obviously less than 1000. Therefore, the answer is $3+9+9$ $\boxed{21}$ | 21 |
2,704 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_20 | 1 | A rectangle with positive integer side lengths in $\text{cm}$ has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$ . Which of the following numbers cannot equal $A+P$
$\textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108$ | Let the rectangle's length be $a$ and its width be $b$ . Its area is $ab$ and the perimeter is $2a+2b$
Then $A + P = ab + 2a + 2b$ . Factoring, we have $(a + 2)(b + 2) - 4$
The only one of the answer choices that cannot be expressed in this form is $102$ , as $102 + 4$ is twice a prime. There would then be no way to ex... | 102 |
2,705 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_23 | 1 | The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$ | By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.
Because the zeros are integral, the discriminant of the function, $a^2 - 8a$ , is a perfect square, say $k^2$ . Then adding 16 to both sides and completing the square yields \[(a - 4)^2 = k^2 + 16.\] Therefore $(a-4)^2 - k... | 16 |
2,706 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_23 | 2 | The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$ | Let $r_1$ and $r_2$ be the integer zeroes of the quadratic. Since the coefficient of the $x^2$ term is $1$ , the quadratic can be written as \[(x - r_1)(x - r_2)=x^2 - (r_1 + r_2)x + r_1r_2\]
By comparing this with $x^2 - ax + 2a$ \[r_1 + r_2 = a\text{ and }r_1r_2 = 2a.\]
Plugging the first equation in the second, \[r_... | 16 |
2,707 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_24 | 1 | For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ $AB=2$ , and $CD=AD$ . How many different values of $p<2015$ are possible?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }6... | Let $BC = x$ and $CD = AD = y$ be positive integers. Drop a perpendicular from $A$ to $CD$ to show that, using the Pythagorean Theorem, that \[x^2 + (y - 2)^2 = y^2.\] Simplifying yields $x^2 - 4y + 4 = 0$ , so $x^2 = 4(y - 1)$ . Thus, $y$ is one more than a perfect square.
The perimeter $p = 2 + x + 2y = 2y + 2\sqrt{y... | 31 |
2,708 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_24 | 2 | For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ $AB=2$ , and $CD=AD$ . How many different values of $p<2015$ are possible?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }6... | Let $BC = x$ and $CD = AD = z$ be positive integers. Drop a perpendicular from $A$ to $CD$ . Denote the intersection point of the perpendicular and $CD$ as $E$
$AE$ 's length is $x$ , as well.
Call $ED$ $y$ .
By the Pythagorean Theorem, $x^2 + y^2 = (y + 2)^2$ .
And so: $x^2 = 4y + 4$ , or $y = (x^2-4)/4$
Writing this ... | 31 |
2,709 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_25 | 1 | Let $S$ be a square of side length $1$ . Two points are chosen independently at random on the sides of $S$ . The probability that the straight-line distance between the points is at least $\dfrac{1}{2}$ is $\dfrac{a-b\pi}{c}$ , where $a$ $b$ , and $c$ are positive integers with $\gcd(a,b,c)=1$ . What is $a+b+c$
$\te... | Divide the boundary of the square into halves, thereby forming $8$ segments. Without loss of generality, let the first point $A$ be in the bottom-left segment. Then, it is easy to see that any point in the $5$ segments not bordering the bottom-left segment will be distance at least $\dfrac{1}{2}$ apart from $A$ . Now, ... | 59 |
2,710 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_25 | 2 | Let $S$ be a square of side length $1$ . Two points are chosen independently at random on the sides of $S$ . The probability that the straight-line distance between the points is at least $\dfrac{1}{2}$ is $\dfrac{a-b\pi}{c}$ , where $a$ $b$ , and $c$ are positive integers with $\gcd(a,b,c)=1$ . What is $a+b+c$
$\te... | Let our points be called Point A and Point B. Let us first choose Point A to be on some side of the square. We have three cases:
Case 1: Point B is on the same side as Point A: This setup occurs with probability $\dfrac{1}{4}.$ This is the standard geometric probability problem. Since Point A and Point B can be anywher... | 59 |
2,711 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_3 | 1 | Isaac has written down one integer two times and another integer three times. The sum of the five numbers is $100$ , and one of the numbers is $28.$ What is the other number?
$\textbf{(A) }8\qquad\textbf{(B) }11\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }18$ | Let the first number be $x$ and the second be $y$ . We have $2x+3y=100$ . We are given one of the numbers is $28$ . If $x$ were to be $28$ $y$ would not be an integer, thus $y=28$ $2x+3(28)=100$ , which gives $x=\boxed{8}$ | 8 |
2,712 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_15 | 1 | The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
$\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66... | Let the amount of people be $p$ , horses be $h$ , sheep be $s$ , cows be $c$ , and ducks be $d$ .
We know \[3h=p\] \[4c=s\] \[3p=d\] Then the total amount of people, horses, sheep, cows, and ducks may be written as $p+h+s+c+d = 3h+h+4c+c+(3\times3h)$ . This is equivalent to $13h+5c$ . Looking through the options, we s... | 47 |
2,713 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_15 | 2 | The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
$\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66... | As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as $13d+5s$ . However, instead of going through each of the solutions and testing the options, you can use the Chicken McNugget Theorem to find the greatest number of people, horses, sheep, cows, and ducks that cannot ... | 47 |
2,714 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_17 | 1 | The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?
[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)... | The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals $4$ and $5$ , for an area $A=10$ . The height $h$ , of one pyramid, is $\frac{3}{2}$ , so the volume of one pyramid is $\frac{Ah}{3}=5$ . Thus, the octahedron has volume $2\cdot5=\boxed{10}$ | 10 |
2,715 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_17 | 2 | The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?
[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)... | The "base" of the octahedron is half the base of the rectangular prism because it is connected by the midpoints. Additionally, the volume of an octahedron is $\dfrac{1}{3}$ of its respective prism. Thus, the octahedron's volume is $\dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6}$ of the rectangular prism's volume, meani... | 10 |
2,716 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_18 | 1 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\te... | We can simplify the problem first, then apply reasoning to the original problem. Let's say that there are $8$ coins. Shaded coins flip heads, and blank coins flip tails. So, after the first flip;
[asy] filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),white); filldraw(circle((-3,0),0.35),white); filldra... | 56 |
2,717 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_18 | 2 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\te... | Every time the coins are flipped, half of them are expected to turn up heads. The expected number of heads on the first flip is $32$ , on the second flip is $16$ , and on the third flip, it is $8$ . Adding these gives $\boxed{56}$ | 56 |
2,718 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_18 | 3 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\te... | Every time the coins are flipped, each of them has a $1/2$ probability of being tails. Doing this $3$ times, $1/8$ of them will be tails. $64-64*1/8=$ $\boxed{56}$ | 56 |
2,719 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_18 | 4 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\te... | (Similar to solution 2)
Notice how:
$E(\text{Heads on 1st flip, 2nd flip, 3rd flip}) = E(\text{Heads on 1st flip}) + E(\text{Heads on 2nd flip}) + E(\text{Heads on 3rd flip})$
The expected number of heads for the first flip is simply $64 \cdot \frac{1}{2}$ , since each coin has a 1 in 2 chance of being heads. Then, we ... | 56 |
2,720 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_20 | 1 | problem_id
18bf638a0cb797d222d008b988c16b6d Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object | [asy]import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); label("2",(0,0,0),S); label("A",(1,0,0),W); label("B",(0,0,1),N); label("1",(1,0,1),NW); label("3",(1,1,0),S); label("C",(0,1,0),E);... | 6 |
2,721 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_20 | 2 | problem_id
18bf638a0cb797d222d008b988c16b6d Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object | Lets say that this cube is an unit cube and the given corner is $(0,0,0)$ . Because Erin cannot return back to her starting point, she cannot be on $(0,0,1)$ $(0,1,0)$ , or $(1,0,0)$ . She cannot be on $(1,1,0)$ $(1,0,1)$ , or $(0,1,1)$ , because after $7$ moves, the sum of all the coordinates has to be odd. Thus, Erin... | 6 |
2,722 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_20 | 3 | problem_id
18bf638a0cb797d222d008b988c16b6d Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object | Let's suppose the given corner on the cube is $(0,0,0)$ . Erin has 3 identical ways to proceed. Suppose she goes to $(0,1,0)$ . She now has two more identical ways to go. Let's say she goes to $(0,1,1)$ . She has to go to $(0,0,1)$ , otherwise, she will end up on that point after 7 moves. This is because once if she ch... | 6 |
2,723 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_21 | 1 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if neces... | Let $n$ be the number of steps. We have
\[\left\lceil \frac{n}{2} \right\rceil - 19 = \left\lceil \frac{n}{5} \right\rceil\]
We will proceed to solve this equation via casework.
Case $1$ $\left\lceil \frac{n}{2} \right\rceil = \frac{n}{2}$
Our equation becomes $\frac{n}{2} - 19 = \frac{n}{5} + \frac{j}{5}$ , where $j \... | 13 |
2,724 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_21 | 2 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if neces... | We know from the problem that Dash goes $3$ steps further than Cozy per jump (assuming they aren't within $4$ steps from the top). That means that if Dash takes $19$ fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least $3 \cdot 19=57$ steps high. We then start using guess-and-check:
... | 13 |
2,725 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_21 | 3 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if neces... | We're looking for natural numbers $x$ such that $\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil$
Let's call $x = 10a + b$ . We now have $2a + \left \lceil{\frac{b}{5}}\right \rceil + 19 = 5a + \left \lceil{\frac{b}{2}}\right \rceil$ , or
$19 - 3a = \left \lceil{\frac{b}{2}}\right... | 13 |
2,726 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_21 | 4 | Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if neces... | Translate the problem into following equation:
$n = 5D - \{0 \sim 4\} = 2C - \{0 \sim 1\}$
Since $C = D + 19$ , we have
$5D - \{0 \sim 4\} = 2D + 38 - \{0 \sim 1\}$
i.e.,
$3D = 38 + \{0 \sim 4\} - \{0 \sim 1\}$
We then have $D = 13$ when $\{1\} - \{0\}$ or $\{2\} - \{1\}$ (the dog's last jump has $2$ steps and the cat'... | 13 |
2,727 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | 1 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textb... | A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have:
\[\begin{array}{c|c|c|c|c|c|c} \mathrm{Factorial}&0!-4!&5!-9!&10!-14!&15!-19!&20!-24!&25!-29!\\\hlin... | 8 |
2,728 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | 2 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textb... | By Legendre's Formula and the information given, we have that $3\left(\left\lfloor{\frac{n}{5}}\right\rfloor+\left\lfloor{\frac{n}{25}}\right\rfloor\right)=\left\lfloor{\frac{2n}{5}}\right\rfloor+\left\lfloor{\frac{2n}{25}}\right\rfloor$
We have $n<100$ as there is no way that if $n>100$ $(2n)!$ would have $3$ times as... | 8 |
2,729 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | 3 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textb... | Let $n=5m+k$ for some natural numbers $m$ $k$ such that $k\in\{0,1,2,3,4\}$ . Notice that $n<5^3=125$ . Thus \[3(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor+\left\lfloor\frac{2n}{125}\right\rfloor\] For smaller ... | 8 |
2,730 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | 4 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textb... | We first note that the number of 0's in $n!$ is determined by how many 5's are in the prime factorization. We use Legendre's Formula and split up into two cases:
$\textbf{CASE ONE: } 5\leq 2n < 25.$
The only way we can fulfill the requirements is if $\lfloor{\dfrac{n}{5}}\rfloor = 1$ and $\lfloor{\dfrac{2n}{5}}\rfloor=... | 8 |
2,731 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_23 | 5 | Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textb... | We will use trial and error to determine the answer to this problem. If $n = 5$ , then $n!$ has $1$ zero, and $2n!$ will have $2$ zeros. But $3 \cdot 1 \neq 2$ so $n = 5$ does not work. Similarly $n = 6, 7$ do not work either. But $n = 8$ works because $8!$ has $1$ zero and $16!$ has $3$ zeros. Note that $n = 9$ also w... | 8 |
2,732 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_25 | 1 | A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; ... | We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\] Since $a\le b, ac \le bc$ , from the first equation we get $abc \le 6bc$ . Thus $a\le 6$ . From the second equation we see that $a > 2$ . Thus $a\in \{3, 4, 5, 6\}$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions. | 10 |
2,733 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10B_Problems/Problem_25 | 2 | A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; ... | The surface area is $2(ab+bc+ca)$ , and the volume is $abc$ , so equating the two yields
\[2(ab+bc+ca)=abc.\]
Divide both sides by $2abc$ to obtain \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.\]
First consider the bound of the variable $a$ . Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we ... | 10 |
2,734 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_3 | 1 | Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $\textdollar 2.50$ each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf co... | She first sells one-half of her $48$ loaves, or $\frac{48}{2}=24$ loaves. Each loaf sells for $\textdollar 2.50$ , so her total earnings in the morning is equal to \[24\cdot \textdollar 2.50 = \textdollar 60\]
This leaves 24 loaves left, and Bridget will sell $\dfrac{2}{3}\times 24=16$ of them for a price of $\textdoll... | 52 |
2,735 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | 1 | problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object | Let's use casework on the yellow house. The yellow house $(\text{Y})$ is either the $3^\text{rd}$ house or the last house.
Case 1: $\text{Y}$ is the $3^\text{rd}$ house.
The only possible arrangement is $\text{B}-\text{O}-\text{Y}-\text{R}$
Case 2: $\text{Y}$ is the last house.
There are two possible arrangements:
$\te... | 3 |
2,736 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | 2 | problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object | There are $4!=24$ arrangements without restrictions. There are $3!\cdot2!=12$ arrangements such that the blue house neighboring the yellow house (calculating the arrangments of [ $\text{BY}$ ], $\text{O}$ , and $\text{R}$ ). Hence, there are $24-12=12$ arrangements with the blue and yellow houses non-adjacent.
By symme... | 3 |
2,737 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | 3 | problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object | To start with, the blue house is either the first or second house.
If the blue house is the first, then the orange must follow, leading to $2$ cases: $\text{B-O-R-Y}$ and $\text{B-O-Y-R}$
If the blue house is second, then the orange house must be first and the yellow house last, leading to $1$ case: $\text{O-B-R-Y}$ .
... | 3 |
2,738 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | 1 | problem_id
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object | Without loss of generality, let there be $20$ students (the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points. Therefore, the mean
is $87$ and the median is $90$
Thus, the solution is... | 3 |
2,739 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | 2 | problem_id
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object | The percentage who scored $100$ points is $100\%-(10\%+35\%+30\%)=100\%-75\%=25\%$ . Now, we need to find the median, which is the score that splits the upper and lower $50\%$ .The lower $10\%+35\%=45\%$ scored $70$ or $80$ points, so the median is $90$ (since the upper $25\%$ is $100$ points and the lower $45\%$ is $7... | 3 |
2,740 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_7 | 1 | Nonzero real numbers $x$ $y$ $a$ , and $b$ satisfy $x < a$ and $y < b$ . How many of the following inequalities must be true?
$\textbf{(I)}\ x+y < a+b\qquad$
$\textbf{(II)}\ x-y < a-b\qquad$
$\textbf{(III)}\ xy < ab\qquad$
$\textbf{(IV)}\ \frac{x}{y} < \frac{a}{b}$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)... | Let us denote $a = x + k$ where $k > 0$ and $b = y + l$ where $l > 0$ . We can write that $x + y < x + y + k + l \implies x + y < a + b$
It is important to note that $1$ counterexample fully disproves a claim. Let's try substituting $x=-3,y=-4,a=1,b=4$
$\textbf{(II)}$ states that $x-y<a-b \implies -3 - (-4) < 1 - 4 \i... | 1 |
2,741 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_9 | 1 | The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We find that the area of the triangle is $\frac{6\times 2\sqrt{3}}{2} =6\sqrt{3}$ . By the Pythagorean Theorem , we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$ . Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.
Let $h$ be the third... | 3 |
2,742 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_9 | 2 | The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | By the Pythagorean Theorem, we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$ . Notice that we now have a 30-60-90 triangle, with the angle between sides $2\sqrt{3}$ and $4\sqrt{3}$ equal to $60^{\circ}$ . Dropping an altitude from the right angle to the hypotenuse, we see that our desir... | 3 |
2,743 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 1 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | Let $a=1$ . Our list is $\{1,2,3,4,5\}$ with an average of $15\div 5=3$ . Our next set starting with $3$ is $\{3,4,5,6,7\}$ . Our average is $25\div 5=5$
Therefore, we notice that $5=1+4$ which means that the answer is $\boxed{4}$ | 4 |
2,744 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 2 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | We are given that \[b=\frac{a+a+1+a+2+a+3+a+4}{5}\] \[\implies b =a+2\]
We are asked to find the average of the 5 consecutive integers starting from $b$ in terms of $a$ . By substitution, this is \[\frac{a+2+a+3+a+4+a+5+a+6}5=a+4\]
Thus, the answer is $\boxed{4}$ | 4 |
2,745 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 3 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | We know from experience that the average of $5$ consecutive numbers is the $3^\text{rd}$ one or the $1^\text{st} + 2$ . With the logic, we find that $b=a+2$ $b+2=(a+2)+2=\boxed{4}$ | 4 |
2,746 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 4 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | The list of numbers is $\left\{a,\ a+1,\ b,\ a+3,\ a+4\right\}$ so $b=a+2$ . The new list is $\left\{a+2,\ a+3,\ a+4,\ a+5,\ a+6\right\}$ and the average is $a+4 \Longrightarrow \boxed{4}$ | 4 |
2,747 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_14 | 1 | The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | [asy]//Needs refining (hmm I think it's fine --bestwillcui1) size(12cm); fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7)); for(int i=-2;i<=8;i+=1) draw((i,-12)--(i,12),grey); for(int j=-12;j<=12;j+=1) draw((-2,j)--(8,j),grey); draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis draw((0,-13)--(0,13),linewidth(1),Arro... | 60 |
2,748 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_14 | 2 | The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | We can let the two lines be \[y=mx+b\] \[y=-\frac{1}{m}x-b\] This is because the lines are perpendicular, hence the $m$ and $-\frac{1}{m}$ , and the sum of the y-intercepts is equal to 0, hence the $b, -b$
Since both lines contain the point $(6,8)$ , we can plug this into the two equations to obtain \[8=6m+b\] and \[8=... | 60 |
2,749 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_14 | 3 | The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | Since their sum is $0$ , let the y intercepts be P $(0,a)$ and Q $(0,-a)$ . The slope of $AP$ is $\frac{8-a}{6}$ . The slope of AQ is $\frac{8+a}{6}$ . Since multiplying the slopes of perpendicular lines yields a product of $-1$ , we have $\frac{64-a^2}{36}=-1$ , which results in $a^2=100$ . We can use either the posit... | 60 |
2,750 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_14 | 4 | The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | Using point-slope form, the first line has the equation \[y-8=m\left(x-6\right) \longrightarrow y=mx-6m+8\] The second line has the equation \[y-8=-\frac{1}{m}\left(x-6\right) \longrightarrow y=-\frac{x}{m}+\frac{6}{m}+8\] At the y-intercept, the value of the x-coordinate is $0$ , hence: the first equation is $y=-6m+8$... | 60 |
2,751 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | 1 | problem_id
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
Name: Text, dtype: object | Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives.
Let $d$ be the distance still needed to travel after $1$ hour. We have that $\dfrac{d}{50}+1.5=\dfrac{d}{35}$ , where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early.
Simplifying gives $7d+525=10d$... | 210 |
2,752 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | 2 | problem_id
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
Name: Text, dtype: object | Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices.
Quickly checking, we know that neither choice $\textbf{(A)}$ or choice $\textbf{(B)}$ work, but $\textbf{(C)}$ does. We can verify as follows. After $1$ hour at $35 \text{ mph}$... | 210 |
2,753 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | 3 | problem_id
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
Name: Text, dtype: object | Let the total distance be $d$ . Then $d=35(t+1)$ . Since 1 hour has passed, and he increased his speed by $15$ miles per hour, and he had already traveled $35$ miles, the new equation is $d-35=50(t-1-\frac{1}{2})=50(t-\frac{3}{2})$ . Solving, $d=35t+35=50t-40$ $15t=75$ $t=5$ , and $d=35(5+1)=35\cdot 6=210 \Longrightarr... | 210 |
2,754 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_16 | 1 | In rectangle $ABCD$ $AB=1$ $BC=2$ , and points $E$ $F$ , and $G$ are midpoints of $\overline{BC}$ $\overline{CD}$ , and $\overline{AD}$ , respectively. Point $H$ is the midpoint of $\overline{GE}$ . What is the area of the shaded region?
[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (... | Denote $D=(0,0)$ . Then $A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)$ . Let the intersection of $AF$ and $DH$ be $X$ , and the intersection of $BF$ and $CH$ be $Y$ . Then we want to find the coordinates of $X$ so we can find $XY$ . From our points, the slope of $AF$ is $\bigg(\dfrac{-2}{\tfrac12}\b... | 16 |
2,755 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_16 | 2 | In rectangle $ABCD$ $AB=1$ $BC=2$ , and points $E$ $F$ , and $G$ are midpoints of $\overline{BC}$ $\overline{CD}$ , and $\overline{AD}$ , respectively. Point $H$ is the midpoint of $\overline{GE}$ . What is the area of the shaded region?
[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (... | From the diagram in Solution 1, let $e$ be the height of $XHY$ and $f$ be the height of $XFY$ . It is clear that their sum is $1$ as they are parallel to $GD$ . Let $k$ be the ratio of the sides of the similar triangles $XFY$ and $AFB$ , which are similar because $XY$ is parallel to $AB$ and the triangles share angle $... | 16 |
2,756 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18 | 1 | A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ $1$ $4$ , and $5$ . What is the area of the square?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$ | Let the points be $A=(x_1,0)$ $B=(x_2,1)$ $C=(x_3,5)$ , and $D=(x_4,4)$
Note that the difference in $y$ value of $B$ and $C$ is $4$ . By rotational symmetry of the square, the difference in $x$ value of $A$ and $B$ is also $4$ . Note that the difference in $y$ value of $A$ and $B$ is $1$ . We now know that $AB$ , the s... | 17 |
2,757 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18 | 2 | A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ $1$ $4$ , and $5$ . What is the area of the square?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$ | By translation, we can move the square with point $A$ at the origin. Then, $A=(0,0), B=(x_1,1), C=(x_2,5), D=(x_3,4)$ . We will use the relationship among the 4 sides of being perpendicular and equal.
The slope of $AB$ is $\frac{1-0}{x_1-0}=\frac{1}{x_1}$
Because $BC$ is perpendicular to $AB$ , the slope of $BC=-x_1$ .... | 17 |
2,758 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18 | 3 | A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ $1$ $4$ , and $5$ . What is the area of the square?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$ | In this solution, we will use the fact that the diagonals of a square bisect each other, they are perpendicular to each other, and they are equal in length.
Using the fact that the diagonals bisect each other, we get the equation:
Now we use the fact that the diagonals are perpendicular to each other:
Using the fact th... | 17 |
2,759 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_18 | 4 | A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ $1$ $4$ , and $5$ . What is the area of the square?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$ | Draw it out, by inspection the coordinates are $(-1, 4)$ $(0, 0)$ $(4, 1)$ , and $(3, 5)$ . The side length is $\sqrt{17}, [ABCD]=\boxed{17}$ ~JH. L | 17 |
2,760 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_20 | 1 | problem_id
6a2cc5541e1749d3d0bb739aacabadf4 The product $(8)(888\dots8)$ , where the secon...
6a2cc5541e1749d3d0bb739aacabadf4 The product $(8)(888\dots8)$ , where the secon...
Name: Text, dtype: object | We can list the first few numbers in the form $8 \cdot (8....8)$
(Hard problem to do without the multiplication, but you can see the pattern early on)
$8 \cdot 8 = 64$
$8 \cdot 88 = 704$
$8 \cdot 888 = 7104$
$8 \cdot 8888 = 71104$
$8 \cdot 88888 = 711104$
By now it's clear that the numbers will be in the form $7$ $k-2$... | 991 |
2,761 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_21 | 1 | Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$ -axis at the same point. What is the sum of all possible $x$ -coordinates of these points of intersection?
$\textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ ... | Note that when $y=0$ , the $x$ values of the equations should be equal by the problem statement. We have that \[0 = ax + 5 \implies x = -\dfrac{5}{a}\] \[0 = 3x+b \implies x= -\dfrac{b}{3}\] Which means that \[-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15\] The only possible pairs $(a,b)$ then are $(a,b) = (1,15), (3,... | 8 |
2,762 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_21 | 2 | Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$ -axis at the same point. What is the sum of all possible $x$ -coordinates of these points of intersection?
$\textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ ... | Going off of Solution 1, for the first equation, notice that the value of $x$ cannot be less than $-5$ . We also know for the first equation that the values of $x$ have to be $5$ divided by something. Also, for the second equation, the values of $x$ can only be $-\frac13,-\frac23,-\frac33, \dots$ . Therefore, we see th... | 8 |
2,763 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | 1 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | Note that $\tan 15^\circ=2-\sqrt{3}=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3$ . (It is important to memorize the sin, cos, and tan values of $15^\circ$ and $75^\circ$ .) Therefore, we have $DE=10\sqrt 3$ . Since $ADE$ is a $30-60-90$ triangle, $AE=2 \cdot AD=2 \cdot 10=\boxed{20}$ | 20 |
2,764 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | 2 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | Let $F$ be a point on line $\overline{CD}$ such that points $C$ and $F$ are distinct and that $\angle EBF = 15^\circ$ . By the angle bisector theorem, $\frac{BC}{BF} = \frac{CE}{EF}$ . Since $\triangle BFC$ is a $30-60-90$ right triangle, $CF = \frac{10\sqrt{3}}{3}$ and $BF = \frac{20\sqrt{3}}{3}$ . Additionally, \[CE ... | 20 |
2,765 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | 3 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | Reflect $\triangle{ECB}$ over line segment $\overline{CD}$ . Let the point $F$ be the point where the right angle is of our newly reflected triangle. By subtracting $90 - (15+15) = 60$ to find $\angle ABF$ , we see that $\triangle{ABC}$ is a $30-60-90$ right triangle. By using complementary angles once more, we can see... | 20 |
2,766 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | 4 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | Square.PNG
We are going to use pure Euclidian geometry to prove $AE=AB$
Reflect rectangle $ABCD$ along line $CD$ . Let the square be $ABFG$ as shown. Construct equilateral triangle $\triangle EFH$
Because $HF=EF$ $GF=BF=20$ , and $\angle GFH=\angle BFE=15^{\circ}$ $\triangle GFH\cong \triangle BFE$ by $SAS$
So, $GH=BE$... | 20 |
2,767 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | 5 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | Rectangle.PNG
We are going to use pure Euclidian geometry to prove $AE=AB$
Construct equilateral triangle $\triangle BEF$ , and let $GF$ be the height of $\triangle ABF$
$\angle GBF=90^{\circ}-15^{\circ}-60^{\circ}=15^{\circ}$ $\angle GBF=\angle CBE$ $\angle BGF=\angle BCE=90^{\circ}$ $BF=BE$
$\triangle BGF \cong \tria... | 20 |
2,768 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_22 | 6 | In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$ | All trigonometric functions in this solution are in degrees. We know \[\sin\left(a+b\right)=\sin\left(a\right)\cos\left(b\right)+\sin\left(b\right)\cos\left(a\right)\] so \[\sin\left(15\right)=\sin\left(45-30\right)=\sin\left(45\right)\cos\left(-30\right)+\sin\left(-30\right)\cos\left(45\right)\] \[=\frac{\sqrt{2}}{2}\... | 20 |
2,769 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_25 | 1 | problem_id
8dd58d71d622cdca7859918e42e278d8 The number $5^{867}$ is between $2^{2013}$ and...
8dd58d71d622cdca7859918e42e278d8 The number $5^{867}$ is between $2^{2013}$ and...
Name: Text, dtype: object | The problem is asking for between how many consecutive powers of $5$ are there $3$ power of $2$
There can be either $2$ or $3$ powers of $2$ between any two consecutive powers of $5$ $5^n$ and $5^{n+1}$
The first power of $2$ is between $5^n$ and $2 \cdot 5^n$
The second power of $2$ is between $2 \cdot 5^n$ and $4 \cd... | 279 |
2,770 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_1 | 1 | Leah has $13$ coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?
$\textbf {(A) } 33 \qquad \textbf {(B) } 35 \qquad \textbf {(C) } 37 \qquad \textbf {(D) } 39 \qquad \textbf {(E)... | If Leah has $1$ more nickel, she has $14$ total coins. Because she has the same number of nickels and pennies, she has $7$ nickels and $7$ pennies. This is after the nickel has been added, so we must subtract $1$ nickel to get $6$ nickels and $7$ pennies. Therefore, Leah has $6\cdot5+7=\boxed{37}$ cents. | 37 |
2,771 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_2 | 1 | What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$
$\textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64$ | We can synchronously multiply ${2^3}$ to the expresions both above and below the fraction bar. Thus, \[\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.\] Hence, the fraction equals to $\boxed{64}$ | 64 |
2,772 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_2 | 2 | What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$
$\textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64$ | We have \[\frac{2^3+2^3}{2^{-3}+2^{-3}} = \frac{8 + 8}{\frac{1}{8} + \frac{1}{8}} = \frac{16}{\frac{1}{4}} = 16 \cdot 4 = 64,\] so our answer is $\boxed{64}$ | 64 |
2,773 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_5 | 1 | Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2$ , and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window?
[asy] fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); fill... | We note that the total length must be the same as the total height, as it is given in the problem. Calling the width of each small rectangle $2x$ , and the height $5x$ , we can see that the length is composed of $4$ widths and $5$ bars of length $2$ . This is equal to two heights of the small rectangles as well as $3$ ... | 26 |
2,774 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_7 | 1 | Suppose $A>B>0$ and A is $x$ % greater than $B$ . What is $x$
$\textbf {(A) } 100\left(\frac{A-B}{B}\right) \qquad \textbf {(B) } 100\left(\frac{A+B}{B}\right) \qquad \textbf {(C) } 100\left(\frac{A+B}{A}\right)\qquad \textbf {(D) } 100\left(\frac{A-B}{A}\right) \qquad \textbf {(E) } 100\left(\frac{A}{B}\right)$ | We have that A is $x\%$ greater than B, so $A=\frac{100+x}{100}(B)$ . We solve for $x$ . We get
$\frac{A}{B}=\frac{100+x}{100}$
$100\frac{A}{B}=100+x$
$100\left(\frac{A}{B}-1\right)=x$
$100\left(\frac{A-B}{B}\right)=x$ $\boxed{100}$ | 100 |
2,775 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_7 | 2 | Suppose $A>B>0$ and A is $x$ % greater than $B$ . What is $x$
$\textbf {(A) } 100\left(\frac{A-B}{B}\right) \qquad \textbf {(B) } 100\left(\frac{A+B}{B}\right) \qquad \textbf {(C) } 100\left(\frac{A+B}{A}\right)\qquad \textbf {(D) } 100\left(\frac{A-B}{A}\right) \qquad \textbf {(E) } 100\left(\frac{A}{B}\right)$ | The question is basically asking the percentage increase from $B$ to $A$ . We know the formula for percentage increase is $\frac{\text{New-Original}}{\text{Original}}$ . We know the new is $A$ and the original is $B$ . We also must multiple by $100$ to get $x$ out of it's fractional/decimal form. Therefore, the answer ... | 100 |
2,776 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_8 | 1 | A truck travels $\dfrac{b}{6}$ feet every $t$ seconds. There are $3$ feet in a yard. How many yards does the truck travel in $3$ minutes?
$\textbf {(A) } \frac{b}{1080t} \qquad \textbf {(B) } \frac{30t}{b} \qquad \textbf {(C) } \frac{30b}{t}\qquad \textbf {(D) } \frac{10t}{b} \qquad \textbf {(E) } \frac{10b}{t}$ | Converting feet to yards and minutes to second, we see that the truck travels $\dfrac{b}{18}$ yards every $t$ seconds for $180$ seconds. We see that he does $\dfrac{180}{t}$ cycles of $\dfrac{b}{18}$ yards. Multiplying, we get $\dfrac{180b}{18t}$ , or $\dfrac{10b}{t}$ , or $\boxed{10}$ | 10 |
2,777 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_8 | 2 | A truck travels $\dfrac{b}{6}$ feet every $t$ seconds. There are $3$ feet in a yard. How many yards does the truck travel in $3$ minutes?
$\textbf {(A) } \frac{b}{1080t} \qquad \textbf {(B) } \frac{30t}{b} \qquad \textbf {(C) } \frac{30b}{t}\qquad \textbf {(D) } \frac{10t}{b} \qquad \textbf {(E) } \frac{10b}{t}$ | We set a proportion by letting the $x$ being the number of feet the truck travels in $3$ minutes.
$\frac{\frac{b}{6}}{t}=\frac{x}{180}$
$\frac{b}{6t}=\frac{x}{180}$
$\frac{180b}{6t}=x$
$x=\frac{30b}{t}$
Remember $x$ is the number of feet the truck travels, so we divide by $3$ to convert to yards.
$\frac{x}{3}=\frac{10b... | 10 |
2,778 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_10 | 1 | In the addition shown below $A$ $B$ $C$ , and $D$ are distinct digits. How many different values are possible for $D$
\[\begin{array}[t]{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}\]
$\textbf {(A) } 2 \qquad \textbf {(B) } 4 \qquad \textbf {(C) } 7 \qquad \textbf {(D) } 8 \qquad \textbf {(E) } 9$ | Note from the addition of the last digits that $A+B=D\text{ or }A+B=D+10$ .
From the addition of the frontmost digits, $A+B$ cannot have a carry, since the answer is still a five-digit number. Also $A + B$ cant have a carry since then for the second column, $C + 1 + D$ cant equal $D$ .
Therefore $A+B=D$
Using the sec... | 7 |
2,779 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_11 | 1 | For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:
(1) two successive $15\%$ discounts
(2) three successive $10\%$ discounts
(3) a $25\%$ discount followed by a $5\%$ discount
What is the smallest possible positive integer value of $n$
$\textbf{(A)}\ \ 27\qquad\textbf... | Let the original price be $x$ . Then, for option $1$ , the discounted price is $(1-.15)(1-.15)x = .7225x$ . For option $2$ , the discounted price is $(1-.1)(1-.1)(1-.1)x = .729x$ . Finally, for option $3$ , the discounted price is $(1-.25)(1-.05) = .7125x$ . Therefore, $n$ must be greater than $\max(x - .7225x, x-.729x... | 29 |
2,780 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_11 | 2 | For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:
(1) two successive $15\%$ discounts
(2) three successive $10\%$ discounts
(3) a $25\%$ discount followed by a $5\%$ discount
What is the smallest possible positive integer value of $n$
$\textbf{(A)}\ \ 27\qquad\textbf... | Assume the original price was $100$ dollars. Thus, after a discount of $n\%$ , the price will be $100-n$ dollars. Using basic calculations, find out the value of the other discounts. This leaves us with the prices: $100-n$ $\frac{289}{4}$ $\frac{9^3}{10}$ , and $\frac{15\cdot19}{4}$ . Simplify these to get $100-n$ $72$... | 29 |
2,781 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_11 | 3 | For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:
(1) two successive $15\%$ discounts
(2) three successive $10\%$ discounts
(3) a $25\%$ discount followed by a $5\%$ discount
What is the smallest possible positive integer value of $n$
$\textbf{(A)}\ \ 27\qquad\textbf... | Assume WLOG that the original price was $$100$ . Then, option 1 would cost $100 \cdot \frac{17}{20} \cdot \frac{17}{20} = $ 72.25$ . Option 2 would cost $100 \cdot \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} = $72.90$ . Option 3 would cost $100 \cdot \frac{3}{4} \cdot \frac{19}{20} = $71.25$ . Therefore, the lar... | 29 |
2,782 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_14 | 1 | Danica drove her new car on a trip for a whole number of hours, averaging $55$ miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a $3$ -digit number with $a\ge1$ and $a+b+c\le7$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2$
$\textb... | We can set up an algebraic equation for this problem.
From what's given, we have that $100c+10b+a=55x+100a+10b+c$
This simplifies to be $0=55x+99a-99c\implies -55x=99a-99c$
Factoring, we get that $-55x=99(a-c)\implies x=-\frac{9(a-c)}{5}$
Hence, notice that we want $a-c=-5$ so that $x=9$
The only pair that works for th... | 37 |
2,783 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_18 | 1 | A list of $11$ positive integers has a mean of $10$ , a median of $9$ , and a unique mode of $8$ . What is the largest possible value of an integer in the list?
$\textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35$ | We start off with the fact that the median is $9$ , so we must have $a, b, c, d, e, 9, f, g, h, i, j$ , listed in ascending order. Note that the integers do not have to be distinct.
Since the mode is $8$ , we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we w... | 35 |
2,784 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_18 | 2 | A list of $11$ positive integers has a mean of $10$ , a median of $9$ , and a unique mode of $8$ . What is the largest possible value of an integer in the list?
$\textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35$ | Note that $x_1 + \ldots + x_{11} = 110$ let $x_6 = 9$ so $x_1 + \ldots + x_5 + x_7 + \ldots + x_{11} = 101$ . To maximize the value of $x_i$ where $i$ ranges from $1$ to $11$ , we let any $7$ elements be $1,2,\ldots,7$ so $x_1 + x_2 + x_3 = 57$ . Now we have to let one of above $3$ values = $8$ hence $x_1 + x_2 = 49$ n... | 35 |
2,785 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_19 | 1 | Two concentric circles have radii $1$ and $2$ . Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{4}\qquad \textbf{(C)}\ \frac{2-\sqrt{2}}{2}\... | Let the center of the two circles be $O$ . Now pick an arbitrary point $A$ on the boundary of the circle with radius $2$ . We want to find the range of possible places for the second point, $A'$ , such that $AA'$ passes through the circle of radius $1$ . To do this, first draw the tangents from $A$ to the circle of rad... | 13 |
2,786 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_20 | 1 | For how many integers $x$ is the number $x^4-51x^2+50$ negative?
$\textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16$ | First, note that $50+1=51$ , which motivates us to factor the polynomial as $(x^2-50)(x^2-1)$ . Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^2-50<0<x^2-1$ . Solving this inequality, we find $1<x^2<50$ . There are e... | 12 |
2,787 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_20 | 2 | For how many integers $x$ is the number $x^4-51x^2+50$ negative?
$\textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16$ | Since the $x^4-51x^2$ part of $x^4-51x^2+50$ has to be less than $-50$ (because we want $x^4-51x^2+50$ to be negative), we have the inequality $x^4-51x^2<-50 \rightarrow x^2(x^2-51) <-50$ $x^2$ has to be positive, so $(x^2-51)$ is negative. Then we have $x^2<51$ . We know that if we find a positive number that works, i... | 12 |
2,788 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_21 | 1 | Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$ . The other two sides are of lengths $10$ and $14$ . The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$
$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qq... | [asy] size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(B+C)/2,E); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",DD,N... | 25 |
2,789 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_21 | 2 | Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$ . The other two sides are of lengths $10$ and $14$ . The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$
$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qq... | [asy] size(7cm); pair A,B,C,D,E; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(12,1),E); label("14",(2,1),E); label("12",(A+E)/2,N); label("21",(E+B)/2,N); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); ... | 25 |
2,790 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_24 | 1 | problem_id
77e0fabb4f8f99b4274e980292ddff5d The numbers $1, 2, 3, 4, 5$ are to be arranged...
77e0fabb4f8f99b4274e980292ddff5d The numbers $1, 2, 3, 4, 5$ are to be arranged...
Name: Text, dtype: object | We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$ , or $12$ cases, which is ... | 2 |
2,791 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_2 | 1 | Alice is making a batch of cookies and needs $2\frac{1}{2}$ cups of sugar. Unfortunately, her measuring cup holds only $\frac{1}{4}$ cup of sugar. How many times must she fill that cup to get the correct amount of sugar?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 16 \qquad\tex... | To get how many cups we need, we realize that we simply need to divide the number of cups needed by the number of cups collected in her measuring cup each time. Thus, we need to evaluate the fraction $\frac{2\frac{1}{2}}{\frac{1}{4}}$ . Simplifying, this is equal to $\frac{5}{2}(4) = \boxed{10}$ | 10 |
2,792 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_4 | 1 | A softball team played ten games, scoring $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , and $10$ runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?
$\textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\text... | We know that, for the games where they scored an odd number of runs, they cannot have scored twice as many runs as their opponents, as odd numbers are not divisible by $2$ . Thus, from this, we know that the five games where they lost by one run were when they scored $1$ $3$ $5$ $7$ , and $9$ runs, and the others are ... | 45 |
2,793 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_5 | 1 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175. In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf... | The total amount paid is $105 + 125 + 175 = 405$ . To get how much each should have paid, we do $405/3 = 135$
Thus, we know that Tom needs to give Sammy 30 dollars, and Dorothy 10 dollars. This means that $t-d = 30 - 10 = \boxed{20}$ | 20 |
2,794 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_5 | 2 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175. In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf... | The difference in the money that Tom paid and Dorothy paid is $20$ . In order for them both to have paid the same amount, Tom must pay $20$ more than Dorothy. The answer is $\boxed{20}$ | 20 |
2,795 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_5 | 3 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175. In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf... | The meaning of sharing costs equally is meaning that, after the vacation, they are equally dividing the money in a way such that, each person would have the same amount left. As each person spends an amount of money, greater than 100, let it be that they all had $200$ dollars to spend. This means that after the vacatio... | 20 |
2,796 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_6 | 1 | Joey and his five brothers are ages $3$ $5$ $7$ $9$ $11$ , and $13$ . One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home. How old is Joey?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C... | Because the $5$ -year-old stayed home, we know that the $11$ -year-old did not go to the movies, as the $5$ -year-old did not and $11+5=16$ . Also, the $11$ -year-old could not have gone to play baseball, as he is older than $10$ . Thus, the $11$ -year-old must have stayed home, so Joey is $\boxed{11}$ | 11 |
2,797 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_6 | 2 | Joey and his five brothers are ages $3$ $5$ $7$ $9$ $11$ , and $13$ . One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home. How old is Joey?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C... | There are only $4$ kids who are under $10$ but since the $5$ -year old stayed home, the only possible ages who went to play baseball are the brothers who are $3,7,9$ , either $13+3$ or $7+9$ is $16$ but since we need $2$ kids to go to baseball who are under $10$ $13,3$ must have been the pair to go to the movies and $9... | 11 |
2,798 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_7 | 1 | A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\t... | Let us split this up into two cases.
Case $1$ : The student chooses both algebra and geometry.
This means that $3$ courses have already been chosen. We have $3$ more options for the last course, so there are $3$ possibilities here.
Case $2$ : The student chooses one or the other.
Here, we simply count how many ways we... | 9 |
2,799 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_7 | 2 | A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\t... | We can use complementary counting. Since there must be an English class, we will add that to our list of classes leaving $3$ remaining spots for rest of classes. We are also told that there needs to be at least one math class. This calls for complementary counting. The total number of ways of choosing $3$ classes out o... | 9 |
2,800 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_7 | 3 | A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\t... | Similar to Solution 1, note that for Case 1 of solution the answer is simply $\binom{4}2$ and for the second case it is $\binom{3}2$ hence $\boxed{9}$ ways | 9 |
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