problem_id
stringlengths 32
32
| link
stringlengths 75
84
| problem
stringlengths 14
5.33k
| solution
stringlengths 15
6.63k
| letter
stringclasses 5
values | answer
stringclasses 957
values |
|---|---|---|---|---|---|
ff0d0aa7726decda609363c20fe2607f | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_4 | The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ | Let $a$ be the bigger number and $b$ be the smaller.
$a + b = 5(a - b)$
Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives
$\frac{a}{b} = \frac32$ , so the answer is $\boxed{32}$ | B | 32 |
ff0d0aa7726decda609363c20fe2607f | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_4 | The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ | Without loss of generality, let the two numbers be $3$ and $2$ , as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{32}$ | B | 32 |
77ab84124f7ac8447faf4ec39cd1263b | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_6 | Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ $1$
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$ | This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.
The first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$
To solve the system of equations:
$p=3c-4$
$p=4c-12$
$3c-4=4c-12$
$c=8$
$p=20.$
Let $x$ be the number of years until Pete is twice as old as his cousin.
$20+x=2(8+x)$
$20+x=16+2x$
$x=4$
The answer is $\boxed{4}$ | B | 4 |
4e1d4fe9b33b3c4b05e667646c323f81 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_7 | Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?
$\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.}\\ \textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}\\ \textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}$ | Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \[r_2=\frac{11r_1}{10}\] \[\pi r_1^2h_1=\pi r_2^2h_2\] Substituting the first equation into the second and dividing both sides by $\pi$ , we get \[r_1^2h_1=\frac{121r_1^2}{100}h_2\implies h_1=\frac{121h_2}{100}.\] Therefore, $\boxed{21}$ | D | 21 |
49b42b41aafea8216cb179b2b5ef684e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_10 | Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$ . What is $x$
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26$ | Use SFFT to get $(x+1)(y+1)=81$ . The terms $(x+1)$ and $(y+1)$ must be factors of $81$ , which include $1, 3, 9, 27, 81$ . Because $x > y$ $x+1$ is equal to $27$ or $81$ . But if $x+1=81$ , then $y=0$ and so $x=\boxed{26}$ | E | 26 |
d90f6fb9e0a446d6b660086dcbde2014 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_11 | On a sheet of paper, Isabella draws a circle of radius $2$ , a circle of radius $3$ , and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly $k \ge 0$ lines. How many different values of $k$ are possible?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$ | Isabella can get $0$ lines if the circles are concentric, $1$ if internally tangent, $2$ if overlapping, $3$ if externally tangent, and $4$ if non-overlapping and not externally tangent. There are $\boxed{5}$ values of $k$ | D | 5 |
36ce7e7dfd6ecbc2d67692cfdf2e914b | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_12 | The parabolas $y=ax^2 - 2$ and $y=4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12$ . What is $a+b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1.5\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2.5\qquad\textbf{(E)}\ 3$ | Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by $4 - (-2)$ (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are $(2, 0), (-2, 0).$ Then $0 = 4a - 2 \rightarrow a = 0.5$ , and $0 = 4 - 4b \rightarrow b = 1.$ Then $a + b = \boxed{1.5}$ | B | 1.5 |
36ce7e7dfd6ecbc2d67692cfdf2e914b | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_12 | The parabolas $y=ax^2 - 2$ and $y=4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12$ . What is $a+b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1.5\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2.5\qquad\textbf{(E)}\ 3$ | The parabolas must intersect the x-axis at the same two points for the kite to form. We find the x values at which they intersect by equating them and solving for x as shown below. $y = ax^2-2$ and $y = 4 - bx^2\rightarrow ax^2-2 = 4-bx^2\rightarrow (a+b)x^2 = 6 \rightarrow x = +\sqrt{\dfrac{6}{a+b}}$ or $-\sqrt{\dfrac{6}{a+b}}$ . The x-values of the y-intercepts is 0, so we plug in zero in each of them and get $-2$ and $4$ . The area of a kite is $\dfrac{d_1*d_2}{2}$ . The $d_1$ is $2+4 = 6$ . The $d_2$ is $2\sqrt{\dfrac{6}{a+b}}$ . Solving for the area $\rightarrow \dfrac{1}{2}*(6)*(2*\sqrt{\dfrac{6}{a+b}}) = 12 \rightarrow (2*\sqrt{\dfrac{6}{a+b}}) = 4 \rightarrow (\sqrt{\dfrac{6}{a+b}}) = 2 \rightarrow \dfrac{6}{a+b} = 4 \rightarrow \dfrac{6}{4} = (a+b)$ , therefore $a + b = \boxed{1.5}$ | B | 1.5 |
6010dc1308c1c5e59cba0f69a366c167 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_15 | What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$ | We can rewrite the fraction as $\frac{123456789}{2^{22} \cdot 10^4} = \frac{12345.6789}{2^{22}}$ . Since the last digit of the numerator is odd, a $5$ is added to the right if the numerator is divided by $2$ , and this will continuously happen because $5$ , itself, is odd. Indeed, this happens twenty-two times since we divide by $2$ twenty-two times, so we will need $22$ more digits. Hence, the answer is $4 + 22 = \boxed{26}$ | C | 26 |
6010dc1308c1c5e59cba0f69a366c167 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_15 | What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$ | Multiply the numerator and denominator of the fraction by $5^{22}$ (which is the same as multiplying by 1) to give $\frac{5^{22} \cdot 123456789}{10^{26}}$ . Now, instead of thinking about this as a fraction, think of it as the division calculation $(5^{22} \cdot 123456789) \div 10^{26}$ . The dividend is a huge number, but we know it doesn't have any digits to the right of the decimal point. Also, the dividend is not a multiple of 10 (it's not a multiple of 2), so these 26 divisions by 10 will each shift the entire dividend one digit to the right of the decimal point. Thus, $\boxed{26}$ is the minimum number of digits to the right of the decimal point needed. | C | 26 |
6010dc1308c1c5e59cba0f69a366c167 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_15 | What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$ | The denominator is $10^4 \cdot 2^{22}$ . Each $10$ adds one digit to the right of the decimal, and each additional $2$ adds another digit. The answer is $4 + 22 = \boxed{26}$ | C | 26 |
cec941b6aa50d270483f7e28da47e06c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$ | The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.
The quadratic formula gives the roots of the quadratic equation: $x=\frac{a\pm\sqrt{a^2-8a}}{2}$
As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$ , for some nonnegative integer $k$
$a^2-8a=k^2$
$a(a-8)=k^2$
$((a-4)+4)((a-4)-4)=k^2$
$(a-4)^2-4^2=k^2$
$(a-4)^2=k^2+4^2$
From this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k,4,|a-4|)$ must be a Pythagorean triple unless $k = 0$
In the case $k=0$ , the equation simplifies to $|a-4|=4$ . From this equation, we have $a=0,8$ . For both $a=0$ and $a=8$ $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")
If $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k,4,|a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3,4,5)$ triple. Here, $k=3$ and $|a-4|=5$ . Hence, $a=-1,9$ . Again, $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers for both $a=-1$ and $a=9$ , so these two values also satisfy the original constraints.
There are a total of four possible values for $a$ $-1,0,8,$ and $9$ . Hence, the sum of all of the possible values of $a$ is $\boxed{16}$ | C | 16 |
cec941b6aa50d270483f7e28da47e06c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | The zeros 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 the quadratic formula, the roots $r$ can be represented by \[r=\frac{a\pm\sqrt{a^2-8a}}{2}\] For $r\in\mathbb{Z}$ $a\in\mathbb{Z}$ , since $\frac{\sqrt{a^2-8a}}{2}$ and $\frac{a}{2}$ will have different mantissas (mantissae?).
Now observe the discriminant $\sqrt{a^2-8a}=\sqrt{a(a-8)}$ and have two cases.
Positive $a$
$a\geq8$ and $a\leq0$ , since $1\geq a \geq7$ gives imaginary roots. Testing positive $a$ values, quickly see that $a\leq9$ . After $16$ and $36$ , the difference between the closest nonzero factor pairs of perfect squares exceeds $8$ . For $8\geq a \geq9$ $a=8,9$ . Checking both yields an integer.
Negative $a$
We can instead test with $\sqrt{-a(8-a)}$ . If $b=8-a$ , we have our original expression. Thus, for the same reasons, $b=8,9\implies 8,9=8-a$ $a=-1$ (0 does not affect the answer).
$-1+8+9=16\implies\boxed{16}$ | C | 16 |
cec941b6aa50d270483f7e28da47e06c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | The zeros 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 $m$ and $n$ be the roots of $x^2-ax+2a$
By Vieta's Formulas, $n+m=a$ and $mn=2a$
Substituting gets us $n+m=\frac{mn}{2}$
$2n-mn+2m=0$
Using Simon's Favorite Factoring Trick:
$n(2-m)+2m=0$
$-n(2-m)-2m=0$
$-n(2-m)-2m+4=4$
$(2-n)(2-m)=4$
This means that the values for $(m,n)$ are $(0,0),(4,4),(3,6),(1,-2)$ giving us $a$ values of $-1,0,8,$ and $9$ . Adding these up gets $\boxed{16}$ | C | 16 |
cec941b6aa50d270483f7e28da47e06c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$ | The quadratic formula gives \[x = \frac{a \pm \sqrt{a(a-8)}}{2}\] . For $x$ to be an integer, it is necessary (and sufficient!) that $a(a-8)$ to be a perfect square. So we have $a(a-8) = b^2$ ; this is a quadratic in itself and the quadratic formula gives \[a = 4 \pm \sqrt{16 + b^2}\]
We want $16 + b^2$ to be a perfect square. From smartly trying small values of $b$ , we find $b = 0, b = 3$ as solutions, which correspond to $a = -1, 0, 8, 9$ . These are the only ones; if we want to make sure then we must hand check up to $b=8$ . Indeed, for $b \geq 9$ we have that the differences between consecutive squares are greater than $16$ so we can't have $b^2 + 16$ be a perfect square. So summing our values for $a$ we find $\boxed{16}$ . as the answer. | C | 16 |
cec941b6aa50d270483f7e28da47e06c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18 | The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$ | First of all, we know that $a$ is the sum of the quadratic's two roots, by Vieta's formulas. Thus, $a$ must be an integer. Then, we notice that the discriminant $a^2-8a$ must be equal to a perfect square so that the roots are integers. Thus, $a(a-8)=b^2$ where $b$ is an integer.
We can complete the square and rearrange to get $(a-4)^2-b^2=16$ . Let's define $m=a-4$ , just to make things a little easier to write, so now we have $(m+b)(m-b)=16$ . We can now list out the integer factor pairs of 16 and the resulting values of $m$ and $b$ . (Note that $m$ and $b$ must both be integers)
$(m+b)(m-b)=16$
$16*1=16$ $\Rightarrow$ Doesn't work
$8*2=16$ $\Rightarrow$ $m=5, b=3$
$4*4=16$ $\Rightarrow$ $m=4, b=0$
$2*8=16$ $\Rightarrow$ $m=5, b=-3$
$1*16=16$ $\Rightarrow$ Doesn't work
$-16*-1=16$ $\Rightarrow$ Doesn't work
$-8*-2=16$ $\Rightarrow$ $m=-5, b=-3$
$-4*-4=16$ $\Rightarrow$ $m=-4, b=0$
$-2*-8=16$ $\Rightarrow$ $m=-5, b=-3$
$-1*-16=16$ $\Rightarrow$ Doesn't work
We want the possible values of $m$ , which are $-5,-4,4,$ and $5$ . As $m+4=a$ $a$ can equal $-1,0,8,$ or $9.$ Adding all of that up gets us our answer, $\boxed{16}$ | C | 16 |
dd5e33c4b5be9193d381b911fcbe2b0c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_19 | 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?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$ | 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 $p = 2 + x + 2y = 2y + 2\sqrt{y - 1} + 2$ must be less than 2015. Simple calculations demonstrate that $y = 31^2 + 1 = 962$ is valid, but $y = 32^2 + 1 = 1025$ is not. On the lower side, $y = 1$ does not work (because $x > 0$ ), but $y = 1^2 + 1$ does work. Hence, there are 31 valid $y$ (all $y$ such that $y = n^2 + 1$ for $1 \le n \le 31$ ), and so our answer is $\boxed{31}$ | B | 31 |
dd5e33c4b5be9193d381b911fcbe2b0c | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_19 | 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?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$ | 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 down and testing, it appears that this holds for all $x$ . However, since there is a dividend of 4, the numerator must be divisible by 4 to conform to the criteria that the side lengths are positive integers.
In effect, $x$ must be a multiple of 2 to let the side lengths be integers. We test, and soon reach 62. It gives us $p = 1988$ , which is less than 2015. However, 64 gives us $2116 > 2015$ , so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get $\boxed{31}$ | B | 31 |
bf5f44f72a6c392d33e6b464d87c3f51 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$ | The area of $T$ is $\dfrac{1}{2} \cdot 8 \cdot 3 = 12$ and the perimeter is 18.
The area of $T'$ is $\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ and the perimeter is $2a + b$
Thus $2a + b = 18$ , so $2a = 18 - b$
Thus $12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ , so $48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}$
We square and divide 36 from both sides to obtain $64 = b^2 (9 - b)$ , so $b^3 - 9b^2 + 64 = 0$ . Since we know $b = 8$ is a solution, we divide by $b - 8$ to get the other solution. Thus, $b^2 - b - 8 = 0$ , so $b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.$ The answer is $\boxed{3}$ | A | 3 |
bf5f44f72a6c392d33e6b464d87c3f51 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$ | The area is $12$ , the semiperimeter is $9$ , and $a = 9 - \frac12b$ . Using Heron's formula, $\sqrt{9\left(\frac{b}{2}\right)\left(\frac{b}{2}\right)(9-b)} = 12$ . Squaring both sides and simplifying, we have $-b^3+9b^2-64=0$ . Since we know $b = 8$ is a solution, we divide by $b - 8$ to get the other solution. Thus, $b^2 - b - 8 = 0$ , so $b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.$ The answer is $\boxed{3}$ | A | 3 |
bf5f44f72a6c392d33e6b464d87c3f51 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$ | Triangle $T$ , being isosceles, has an area of $\frac{1}{2}(8)\sqrt{5^2-4^2}=12$ and a perimeter of $5+5+8=18$ .
Triangle $T'$ similarly has an area of $\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12$ and $2a+b=18$
Now we apply our computational fortitude.
\[\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12\] \[(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=24\] \[(b)\sqrt{4a^2-b^2}=48\] \[b^2(4a^2-b^2)=48^2\] \[b^2(2a+b)(2a-b)=48^2\] Plug in $2a+b=18$ to obtain \[18b^2(2a-b)=48^2\] \[b^2(2a-b)=128\] Plug in $2a=18-b$ to obtain \[b^2(18-2b)=128\] \[2b^3-18b^2+128=0\] \[b^3-9b^2+64=0\] We know that $b=8$ is a valid solution by $T$ . Factoring out $b-8$ , we obtain \[(b-8)(b^2-b-8)=0 \Rightarrow b^2-b-8=0\] Utilizing the quadratic formula gives \[b=\frac{1\pm\sqrt{33}}{2}\] We clearly must pick the positive solution. Note that $5<\sqrt{33}<6$ , and so ${3<\frac{1+\sqrt{33}}{2}<\frac{7}{2}}$ , which clearly gives an answer of $\boxed{3}$ , as desired. | A | 3 |
bf5f44f72a6c392d33e6b464d87c3f51 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$ | Triangle T has perimeter $5 + 5 + 8 = 18$ so $18 = 2a + b$
Using Heron's, we get $\sqrt{(9)(4)^2(1)} = \sqrt{(\frac{2a+b}{2})\left(\frac{b}{2}\right)^2(\frac{2a-b}{2})}$
We know that $2a + b = 18$ from above so we plug that in, and we also know that then $2a - b = 18 - 2b$
$12 = \frac{3b}{2}\sqrt{9-b}$
$64 = 9b^2 - b^3$
We plug in 3 for $b$ in the LHS, and we get 54 which is too low. We plug in 4 for $b$ in the LHS, and we get 80 which is too high. We now know that $b$ is some number between 3 and 4.
If $b \geq 3.5$ , then we would round up to 4, but if $b < 3.5$ , then we would round down to 3. So let us plug in 3.5 for $b$
We get $67.375$ which is too high, so we know that $b < 3.5$
Thus the answer is $\boxed{3}$ | A | 3 |
bf5f44f72a6c392d33e6b464d87c3f51 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$ | For this new triangle, say its legs have length $d$ and the base length $2c$ . To see why I did this, draw the triangle on a Cartesian plane where the altitude is part of the y-axis! Then, we notice that $c+d=9$ and $c*\sqrt{d^2-c^2}=12$ . It's better to let a side be some variable so we avoid having to add non-square roots and square-roots!!
Now, modify the square-root equation with $d=9-c$ ; you get $c^2*(81-18c)=144$ , so $-18c^3+81c^2=144$ . Divide by $-9$ to get $2c^3-9c^2+16=0$ . Obviously, $c=4$ is a root as established by triangle $T$ ! So, use synthetic division to obtain $2c^2-c-4=0$ , upon which $c=\frac{1+\sqrt{33}}{4}$ , which is closest to $\frac{3}{2}$ (as opposed to $2$ ). That's enough to confirm that the answer has to be $\boxed{3}$ | A | 3 |
bf5f44f72a6c392d33e6b464d87c3f51 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20 | Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$ | Since triangles $T'$ and $T$ have the same area and the same perimeter, $2a+b=18$ and $9*(9-a)^2(9-b) = 9*4^2*1$ By trying each answer choice, it is clear that the answer is $\boxed{3}$ | A | 3 |
39969399f54f55376569d104990e7b73 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_22 | For each positive integer $n$ , let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B$ , with no more than three $A$ s in a row and no more than three $B$ s in a row. What is the remainder when $S(2015)$ is divided by $12$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$ | The huge $n$ value in place, as well as the "no more than... in a row" are key phrases that indicate recursion is the right way to go.
Let's go with finding the case of $S(n)$ from previous cases.
So how can we make the words of $S(n)$ ? Do we choose 3-in-a-row of one letter, $A$ or $B$ , or do we want $2$ consecutive ones or $1$ ? Note that this covers all possible cases of ending with $A$ and $B$ with a certain number of consecutive letters. And obviously they are all distinct.
[Convince yourself that each case for $S(n)$ is considered exactly once by using these cues: does it end in $3$ $2$ , or $1$ consecutive letter(s) ( $1$ consecutive means a string like ... $BA$ , ... $AB$ , as in the letter switches) and does it $WLOG$ consider both $A$ and $B?$
From there we realize that $S(n)=S(n-1)+S(n-2)+S(n-3)$ because 3 in a row requires $S(n-3)$ , and so on. We need to find $S(2015)$ mod 12. We first compute $S(2015)$ mod $3$ and mod $4$ . By listing out the residues mod $3$ , we find that the cycle length for mod $3$ is $13$ $2015$ is $0$ mod $13$ so $S(2015) = S(13) = 2$ mod $3$ . By listing out the residues mod $4$ , we find that the cycle length for mod $4$ is $4$ . S(2015) = S(3) = mod $4$ . By Chinese Remainder Theorem, $S(2015) =\boxed{8}$ mod $12$ | null | 8 |
39969399f54f55376569d104990e7b73 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_22 | For each positive integer $n$ , let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B$ , with no more than three $A$ s in a row and no more than three $B$ s in a row. What is the remainder when $S(2015)$ is divided by $12$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$ | We can start off by finding patterns in $S(n)$ . When we calculate a few values we realize either from performing the calculation or because the calculation was performed in the exact same way that $S(n) = 2^n - 2((n_4)- (n_5) \dots (n_n))$ . Rearranging the expression we realize that the terms aside from $2^{2015}$ are congruent to $0$ mod $12$ (Just put the equation in terms of 2^{2015} and the four combinations excluded and calculate the combinations mod $12$ ). Using patterns we can see that $2^{2015}$ is congruent to $8$ mod $12$ . Therefore $\boxed{8}$ is our answer. | null | 8 |
219c15fc97a359e1068d1d662ccff137 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_23 | 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 $\frac12$ is $\frac{a-b\pi}{c}$ , where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a+b+c$
$\textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 63$ | WLOG, let the first point be on the bottom side of the square. The points where the second point could exist are outside a circle of radius 0.5 centered on the first point. The parts of the square that lie in this circle are the distance from the point to the closest side of the square $n$ , the distance from the point to the outside of the circle (the radius $0.5$ ), and any portion of the nearest side that lies within the circle as represented by the Pythagorean Theorem $\sqrt{\frac{1}{4}-n^2}$ . Thus, the total length the second point can exist in can be represented by $f(n)=4-(0.5+n+\sqrt{\frac{1}{4}-n^2})$ . Distributing, $f(n)=3.5-n-\sqrt{\frac{1}{4}-n^2}$
Then, we can find the average of this function through calculus (wow more calc?). This formula is as follows, \[\frac{1}{a_{f} -a_{i}} \int_{a_{i}}^{a_{f}} f(x)\;dx\]
For this case, the limits of integration are $0$ and $0.5$ $0\leq n\leq 0.5$ ). Then, we have,
\[2\int_{0}^{\frac{1}{2}} 3.5-n-\sqrt{\frac{1}{4}-n^2}\;dn\] \[2(\int_{0}^{\frac{1}{2}} 3.5\;dn -\int_{0}^{\frac{1}{2}}n\;dn -\int_{0}^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}\;dn)\]
Even if you don't know how to integrate, as long as you know the idea of integration, you can figure these out. Graphing the first two, you can see that the first is a rectangle of length $0.5$ and width $3.5$ . The second is an isosceles right triangle of leg length $0.5$
\[\int_{0}^{\frac{1}{2}} 3.5\;dn=\frac{7}{4}, \int_{0}^{\frac{1}{2}}n\;dn=\frac{1}{8}\]
Recognize that the third integral is a semicircle of radius $0.5$ and centered at the origin. This is where $\pi$ comes in. From $0$ to $0.5$ , the integral is simply a quarter circle. $\frac{\frac{1}{2}^2\pi}{4}=\frac{\pi}{16}$ \[\int_{0}^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}\;dn)=\frac{\pi}{16}\]
If you want me to actually integrate these, look below. Do note that this is for those that have a limited knowledge of integration or those that have little time but are being very clever.
Overall, where second point could lie to satisfy the problem is a length of $2(\frac{7}{4}-\frac{\pi}{16}-\frac{1}{8})=\frac{26-\pi}{8}$ . By contrast, the total length where it could lie is the perimeter of the square $4$ . So the possible points that the second point could be make up $\frac{\frac{26-\pi}{8}}{4}=\frac{26-\pi}{32}$ of the square's perimeter. Obviously, $\gcd(32, 26, 1)=1$ $32+26+1=59\implies\boxed{59}$ | A | 59 |
219c15fc97a359e1068d1d662ccff137 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_23 | 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 $\frac12$ is $\frac{a-b\pi}{c}$ , where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a+b+c$
$\textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 63$ | Choose a certain side for one of the points to be on. Let the distance from the point to the vertex on its left be $x$
We split this into two cases:
Case 1: $0\leq x\leq \frac12$
The total length of the segments for which the other point can be on such that the straight-line distance between the points is less than $\frac12$ is \[\sqrt{\frac14-x^2}+x+\frac12.\] We can graph this in the Cartesian plane and find the area of the region below the curve and above the line $y=0$
Case 2: $\frac12< x\leq 1$
This is basically Case 1 but flipped over the line $x=\frac12$
So our total probability is 1 minus the area of the graph over the total area (4, perimeter of square). Notice that the desired area of the region below the curve we found earlier is the sum of a quarter circle with radius $\frac12$ and centered at $(0,0)$ and a trapezoid with height $\frac12$ and bases of length $\frac12$ and $\frac32$ . Adding this all up then multiplying by 2, we have \[\frac{\pi}{8}+\frac34\] and then the probability of the desired result would be \[1-\frac{\pi+6}{32}=\frac{26-\pi}{32}\] and our answer is $26+1+32=\boxed{59}$ . ~caroline2023 | null | 59 |
32529e016b749aa764875e8adbab3517 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_3 | 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 $a$ be the number written two times, and $b$ the number written three times. Then $2a + 3b = 100$ . Plugging in $a = 28$ doesn't yield an integer for $b$ , so it must be that $b = 28$ , and we get $2a + 84 = 100$ . Solving for $a$ , we obtain $a = \boxed{8}$ | A | 8 |
5c9f194e134ce4dd8c15c20ec4f09af7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_6 | Back in 1930, Tillie had to memorize her multiplication facts from $0 \times 0$ to $12 \times 12$ . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?
$\textbf{(A)}\; 0.21 \qquad\textbf{(B)}\; 0.25 \qquad\textbf{(C)}\; 0.46 \qquad\textbf{(D)}\; 0.50 \qquad\textbf{(E)}\; 0.75$ | There are a total of $(12+1) \times (12+1) = 169$ products, and a product is odd if and only if both its factors are odd. There are $6$ odd numbers between $0$ and $12$ , namely $1, 3, 5, 7, 9, 11,$ hence the number of odd products is $6 \times 6 = 36$ . Therefore the answer is $36/169 \doteq \boxed{0.21}$ | A | 0.21 |
5c9f194e134ce4dd8c15c20ec4f09af7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_6 | Back in 1930, Tillie had to memorize her multiplication facts from $0 \times 0$ to $12 \times 12$ . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?
$\textbf{(A)}\; 0.21 \qquad\textbf{(B)}\; 0.25 \qquad\textbf{(C)}\; 0.46 \qquad\textbf{(D)}\; 0.50 \qquad\textbf{(E)}\; 0.75$ | Note that if we had an $11$ by $11$ multiplication table, the fraction of odd products becomes $0.25$ . If we add the $12$ by $12$ , the fraction of odd products decreases. Because $0.21$ is the only option less than $0.25$ , our answer is $\boxed{0.21}$ | A | 0.21 |
7fd15658523751ed5e813788bec60fc9 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_7 | A regular 15-gon has $L$ lines of symmetry, and the smallest positive angle for which it has rotational symmetry is $R$ degrees. What is $L+R$
$\textbf{(A)}\; 24 \qquad\textbf{(B)}\; 27 \qquad\textbf{(C)}\; 32 \qquad\textbf{(D)}\; 39 \qquad\textbf{(E)}\; 54$ | From consideration of a smaller regular polygon with an odd number of sides (e.g. a pentagon), we see that the lines of symmetry go through a vertex of the polygon and bisect the opposite side. Hence $L=15$ , the number of sides / vertices. The smallest angle for a rotational symmetry transforms one side into an adjacent side, hence $R = 360^\circ / 15 = 24^\circ$ , the number of degrees between adjacent sides. Therefore the answer is $L + R = 15 + 24 = \boxed{39}$ | D | 39 |
da261d48a23ade6bf54ea0a13e803d7a | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$ | $(625^{\log_5 2015})^\frac{1}{4} = ((5^4)^{\log_5 2015})^\frac{1}{4} = (5^{4 \cdot \log_5 2015})^\frac{1}{4} = (5^{\log_5 2015 \cdot 4})^\frac{1}{4} = ((5^{\log_5 2015})^4)^\frac{1}{4} = (2015^4)^\frac{1}{4} = \boxed{2015}$ | D | 2015 |
da261d48a23ade6bf54ea0a13e803d7a | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$ | We can rewrite $\log_5 2015$ as as $5^x = 2015$ . Thus, $625^{x \cdot \frac{1}{4}} = 5^x = \boxed{2015}.$ | null | 2015 |
da261d48a23ade6bf54ea0a13e803d7a | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$ | $(625^{\log_5 2015})^{\frac{1}{4}} = (625^{\frac{1}{4}})^{\log_5 2015} = 5^{\log_5 2015} = \boxed{2015}$ | D | 2015 |
da261d48a23ade6bf54ea0a13e803d7a | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8 | What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$ | We note that the year number is just $2015$ , so just guess $\boxed{2015}$ | D | 2015 |
aba17359f83e09a70f289038fd3219f7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_10 | How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?
$\textbf{(A)}\; 3 \qquad\textbf{(B)}\; 4 \qquad\textbf{(C)}\; 5 \qquad\textbf{(D)}\; 6 \qquad\textbf{(E)}\; 7$ | Since we want non-congruent triangles that are neither isosceles nor equilateral, we can just list side lengths $(a,b,c)$ with $a<b<c$ . Furthermore, "positive area" tells us that $c < a + b$ and the perimeter constraints means $a+b+c < 15$
There are no triangles when $a = 1$ because then $c$ must be less than $b+1$ , implying that $b \geq c$ , contrary to $b < c$
When $a=2$ , similar to above, $c$ must be less than $b+2$ , so this leaves the only possibility $c = b+1$ . This gives 3 triangles $(2,3,4), (2,4,5), (2,5,6)$ within our perimeter constraint.
When $a=3$ $c$ can be $b+1$ or $b+2$ , which gives triangles $(3,4,5), (3,4,6), (3,5,6)$ . Note that $(3,4,5)$ is a right triangle, so we get rid of it and we get only 2 triangles.
All in all, this gives us $3+2 = \boxed{5}$ triangles. | C | 5 |
0453a7987c7fd70b7d11b69b012dbe86 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_13 | Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,$ and $BC=6$ . What is $AC$
$\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7$ | $\angle ADB$ and $\angle ACB$ are both subtended by segment $AB$ , hence $\angle ACB = \angle ADB = 40^\circ$ . By considering $\triangle ABC$ , it follows that $\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ$ . Hence $\triangle ABC$ is isosceles, and $AC = BC = \boxed{6}.$ | B | 6 |
0c611776aa8eb8c021cef9622f3342d2 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_18 | For every composite positive integer $n$ , define $r(n)$ to be the sum of the factors in the prime factorization of $n$ . For example, $r(50) = 12$ because the prime factorization of $50$ is $2 \times 5^{2}$ , and $2 + 5 + 5 = 12$ . What is the range of the function $r$ $\{r(n): n \text{ is a composite positive integer}\}$
$\textbf{(A)}\; \text{the set of positive integers} \\ \textbf{(B)}\; \text{the set of composite positive integers} \\ \textbf{(C)}\; \text{the set of even positive integers} \\ \textbf{(D)}\; \text{the set of integers greater than 3} \\ \textbf{(E)}\; \text{the set of integers greater than 4}$ | This problem becomes simple once we recognize that the domain of the function is $\{4, 6, 8, 9, 10, 12, 14, 15, \dots\}$ . By evaluating $r(4)$ to be $4$ , we can see that $\textbf{(E)}$ is incorrect. Evaluating $r(6)$ to be $5$ , we see that both $\textbf{(B)}$ and $\textbf{(C)}$ are incorrect. Since our domain consists of composite numbers, which, by definition, are a product of at least two positive primes, the minimum value of $r(n)$ is $4$ , so $\textbf{(A)}$ is incorrect. That leaves us with $\boxed{3}$ | D | 3 |
0c611776aa8eb8c021cef9622f3342d2 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_18 | For every composite positive integer $n$ , define $r(n)$ to be the sum of the factors in the prime factorization of $n$ . For example, $r(50) = 12$ because the prime factorization of $50$ is $2 \times 5^{2}$ , and $2 + 5 + 5 = 12$ . What is the range of the function $r$ $\{r(n): n \text{ is a composite positive integer}\}$
$\textbf{(A)}\; \text{the set of positive integers} \\ \textbf{(B)}\; \text{the set of composite positive integers} \\ \textbf{(C)}\; \text{the set of even positive integers} \\ \textbf{(D)}\; \text{the set of integers greater than 3} \\ \textbf{(E)}\; \text{the set of integers greater than 4}$ | Think backwards. The range is the same as the numbers $y$ that can be expressed as the sum of two or more prime positive integers.
The lowest number we can get is $y = 2+2 = 4$ . For any number greater than 4, we can get to it by adding some amount of 2's and then possibly a 3 if that number is odd. For example, 23 can be obtained by adding 2 ten times and adding a 3; this corresponds to the argument $n = 2^{10} \times 3$ . Thus our answer is $\boxed{3}$ | D | 3 |
879f415f2f62a1f9bfc66e7c11c75113 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_20 | For every positive integer $n$ , let $\text{mod}_5 (n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}$ recursively as follows:
\[f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text\\ f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\ f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4. \end{cases}\]
What is $f(2015,2)$
$\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4$ | Simply take some time to draw a table of values of $f(i,j)$ for the first few values of $i$
\[\begin{array}{|c || c | c | c | c | c |} \hline i \text{\ \textbackslash\ } j & 0 & 1 & 2 & 3 & 4\\ \hline\hline 0 & 1 & 2 & 3 & 4 & 0\\ \hline 1 & 2 & 3 & 4 & 0 & 1\\ \hline 2 & 3 & 0 & 2 & 4 & 1\\ \hline 3 & 0 & 3 & 4 & 1 & 0\\ \hline 4 & 3 & 1 & 3 & 1 & 3\\ \hline 5 & 1 & 1 & 1 & 1 & 1\\ \hline \end{array}\]
Now we claim that for $i \ge 5$ $f(i,j) = 1$ for all values $0 \le j \le 4$ . We will prove this by induction on $i$ and $j$ . The base cases for $i = 5$ , have already been proven.
For our inductive step, we must show that for all valid values of $j$ $f(i, j) = 1$ if for all valid values of $j$ $f(i - 1, j) = 1$
We prove this itself by induction on $j$ . For the base case, $j=0$ $f(i, 0) = f(i-1, 1) = 1$ . For the inductive step, we need $f(i, j) = 1$ if $f(i, j-1) = 1$ . Then, $f(i, j) = f(i-1, f(i, j-1)).$ $f(i, j-1) = 1$ by our inductive hypothesis from our inner induction and $f(i-1, 1) = 1$ from our outer inductive hypothesis. Thus, $f(i, j) = 1$ , completing the proof.
It is now clear that for $i \ge 5$ $f(i,j) = 1$ for all values $0 \le j \le 4$
Thus, $f(2015,2) = \boxed{1}$ | B | 1 |
879f415f2f62a1f9bfc66e7c11c75113 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_20 | For every positive integer $n$ , let $\text{mod}_5 (n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}$ recursively as follows:
\[f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text\\ f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\ f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4. \end{cases}\]
What is $f(2015,2)$
$\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4$ | We are given that \[f(0,n) \equiv n+1\pmod{5} .\] Then, $f(1,n) = f(0, f(1, n-1)) \equiv f(1,n-1) + 1\pmod{5}$ . Thus $f(1,n) \equiv n+f(1,0)\pmod{5}$ . Since $f(1,0)=f(0,1)=2$ , we get \[f(1,n) \equiv n+2\pmod{5} .\] Then, $f(2,n) = f(1, f(2, n-1)) \equiv f(2,n-1) + 2\pmod{5}$ . Thus $f(2,n) \equiv 2n+f(2,0)\pmod{5}$ . Since $f(2,0)=f(1,1)=3$ , we get \[f(2,n) \equiv 2n+3\pmod{5} .\] Now $f(3,n) = f(2, f(3, n-1)) \equiv 2f(3,n-1) + 3\pmod{5}$ . Thus \begin{align*} f(3,n) &\equiv 2f(3,n-1) + 3 &\pmod{5} \\ 2f(3,n-1) &\equiv 2^2f(3,n-2) + 3\cdot 2 &\pmod{5} \\ \vdots \qquad &\quad\vdots \quad\qquad \vdots \qquad\qquad \vdots &\\ 2^{n-1}f(3,1) &\equiv 2^nf(3,0) + 3\cdot 2^{n-1} &\pmod{5} \end{align*} Adding them all up we get \[f(3,n) \equiv 3(2^n-1)\pmod{5} .\] This means that $f(3,0)=0$ $f(3,1)=3$ $f(3,2)=4$ $f(3,3)=1$ , and $f(3,4)=0$ . Thus $f(3,n)$ never takes the value 2.
Since $f(4,n)=f(3,f(4,n-1))$ , this implies that $f(4,n) \neq 2$ for any $n$ . By induction, $f(3,n) \neq f(3,2) = 4$ for any $n$ . It follows that $f(3,n) \neq f(3,4) = 0$ for any $n$ . Thus $f(4,n)$ only takes values in $\{1,3\}$ . In fact, it alternates between 1 and 3: $f(4,0)=f(3,1)=3$ , then $f(4,1)=f(3,f(4,0))=f(3,3)=1$ , then $f(4,2)=f(3,f(4,1))=f(3,1)=3$ , and so on.
Repeating the argument above, we see that $f(5,n) = f(4, f(5,n-1))$ can only take values in $\{1,3\}$ . However, $f(5,n-1)\neq 0$ for any $n$ implies that $f(5,n)\neq f(4,0)=3$ for any $n$ . Thus $f(5,n)=1$ for all $n$ . We can easily verify this: $f(5,0)=f(4,1)=1$ , then $f(5,1)=f(4,f(5,0))=f(4,1)=1$ , then $f(5,2)=f(4,f(5,1))=f(4,1)=1$ , and so on.
Then $f(6,0)=f(5,1)=1$ . Moreover, $f(6,n) = f(5,f(6,n-1)) = 1$ for all $n$ . Continuing in this manner we see that $f(m,n)=1$ for all $m\ge 5$
In particular, $f(2015,2) = \boxed{1}$ | B | 1 |
aa2e2d2448a5ca2de318d97c9b62f9b9 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21 | 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 necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$ | We can translate this wordy problem into this simple equation:
\[\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil\]
We will proceed to solve this equation via casework.
Case 1: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}$
Our equation becomes $\frac{s}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=1$ and $j=4$ yield $s=64$ and $s=66$ , respectively.
Case 2: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}+\frac{1}{2}$
Our equation becomes $\frac{s}{2} +\frac{1}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=2$ yields $s=63$
Summing up we get $63+64+66=193$ . The sum of the digits is $\boxed{13}$ | D | 13 |
aa2e2d2448a5ca2de318d97c9b62f9b9 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21 | 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 necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$ | It can easily be seen that the problem can be expressed by the equation: \[\left\lceil \frac{s}{2} \right\rceil - \left\lceil \frac{s}{5} \right\rceil = 19\]
However, because the ceiling function is difficult to work with, we can rewrite the previous equation as:
\[\frac{s+a}{2} - \frac{s+b}{5} = 19\] Where $a \in \{0,1\}$ and $b \in \{0,1,2,3,4\}$ Multiplying both sides by ten and simplifying, we get: \[5s+5a-2s-2b=190\] \[3s = 190+2b-5a\] \[s = 63 + \frac{1+2b-5a}{3}\]
Because s must be an integer, we need to find the values of $a$ and $b$ such that $2b-5a \equiv 2 \mod 3$ . We solve using casework.
Case 1: $a = 0$
If $a = 0$ , we have $2b \equiv 2 \mod 3$ . We can easily see that $b = 1$ or $b = 4$ , which when plugged into our original equation lead to $s = 64$ and $s=66$ respectively.
Case 2: $a = 1$
If $a = 1$ , we have $2b-5 \equiv 2 \mod 3$ , which can be rewritten as $2b \equiv 1 \mod 3$ . We can again easily see that $b = 2$ is the only solution, which when plugged into our original equation lead to $s = 63$
Adding these together we get $64+66+63=193$ . The sum of the digits is $\boxed{13}$ | D | 13 |
aa2e2d2448a5ca2de318d97c9b62f9b9 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21 | 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 necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$ | As before, we write the equation:
\[\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil.\]
To get a ballpark estimate of where $s$ might lie, we remove the ceiling functions to find:
\[\frac{s}{2} - 19 = \frac{s}{5}.\]
This gives $\frac{3s}{10} = 19$ , and thus values for $s$ will be around $\frac{190}{3} = 63.\overline3$
Now, to establish some bounds around this estimated working value, we note that if $s=60$ , Cozy takes 30 steps while Dash takes 12, a difference of 18. If $s=70$ , Cozy takes 35 steps while Dash takes 14, a difference of 21. When $s$ increases from a multiple of ten, the difference will never decrease beyond what it is at the multiple of ten, and likewise, when it decreases, it never becomes greater than at the multiple of ten, so any working values of $s$ will be between $60$ and $70$
Then, by inspection, $s=63, 64,$ or $66$ , so $\sum s = 193 \implies \boxed{13}.$ | D | 13 |
aa2e2d2448a5ca2de318d97c9b62f9b9 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21 | 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 necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$ | Notice that the possible number of steps in the staircase is around 60 to 70. By testing all of the values between 60 and 70, we see that 63, 64 and 66 work. Adding those up gives 193, so the answer is $1+9+3=\boxed{13}.$ | D | 13 |
aa2e2d2448a5ca2de318d97c9b62f9b9 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21 | 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 necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$ | We represent C's steps with $2c + a = n$ and D's steps with $5d + b = n$ , where $a \in \{1,2\}$ and $b \in \{1,2,3,4,5\}$ , where $n$ is the number of steps, $c$ is the number of jumps C takes bar the last one, and $d$ is the number of jumps D takes bar the last.
The reason for starting at 1 and ending at 5 instead of 0 through 4 is that the last step can be quite problematic to deal with, especially if it is possible to make it in one go, so we treat it as a different jump that can take all possible jump values. We know that D makes it in 19 fewer than C, so $c+1-(d+1) = 19 \implies c = 19+d \implies 5d + b = 2(19+d) + a \implies 3d = 38 + a - b$
Now that we have this nice equivalence, we can do the thing and it works.
If we take both sides mod 3 and rearrange, we get $b \equiv 2+a \pmod{3}$ This gives us the following satisfactory $(a,b)$ relational pairs: $\{(1,3),(2,1),(2,4)\}$ . We can now just find the corresponding $d$ value for each pair, sum it all up using $5d + b = n$ and sum the digits to reveal $\boxed{13}$ as the answer. | D | 13 |
0ed85e4b626d8fb1cffdd36cbc5c7f4e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_22 | Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\; 14 \qquad\textbf{(B)}\; 16 \qquad\textbf{(C)}\; 18 \qquad\textbf{(D)}\; 20 \qquad\textbf{(E)}\; 24$ | Consider shifting every person over three seats left after each person has gotten up and sat back down again. Now, instead of each person being seated not in the same chair and not in an adjacent chair, each person will be seated either in the same chair or an adjacent chair. The problem now becomes the number of ways in which six people can sit down in a chair that is either the same chair or an adjacent chair in a circle.
Consider the similar problem of $n$ people sitting in a chair that is either the same chair or an adjacent chair in a row. Call the number of possibilities for this $F_n$ . Then if the leftmost person stays put, the problem is reduced to a row of $n-1$ chairs, and if the leftmost person shifts one seat to the right, the new person sitting in the leftmost seat must be the person originally second from the left, reducing the problem to a row of $n-2$ chairs. Thus, $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3$ . Clearly $F_1 = 1$ and $F_2 = 2$ , so $F_3 = 3$ $F_4 = 5$ , and $F_5 = 8$
Now consider the six people in a circle and focus on one person. If that person stays put, the problem is reduced to a row of five chairs, for which there are $F_5 = 8$ possibilities. If that person moves one seat to the left, then the person who replaces him in his original seat will either be the person originally to the right of him, which will force everyone to simply shift over one seat to the left, or the person originally to the left of him, which reduces the problem to a row of four chairs, for which there are $F_4 = 5$ possibilities, giving $1 + 5 = 6$ possibilities in all. By symmetry, if that person moves one seat to the right, there are another $6$ possibilities, so we have a total of $8+6+6 = \boxed{20}$ possibilities. | D | 20 |
0ed85e4b626d8fb1cffdd36cbc5c7f4e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_22 | Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\; 14 \qquad\textbf{(B)}\; 16 \qquad\textbf{(C)}\; 18 \qquad\textbf{(D)}\; 20 \qquad\textbf{(E)}\; 24$ | Label the people sitting at the table $A, B, C, D, E, F,$ and assume that they are initially seated in the order $ABCDEF$ . The possible new positions for $A, B, C, D, E,$ and $F$ are respectively (a dash indicates a non-allowed position):
\[\begin{tabular}{| c | c | c | c | c | c |} \hline - & - & A & A & A & - \\ \hline - & - & - & B & B & B \\ \hline C & - & - & - & C & C \\ \hline D & D & - & - & - & D \\ \hline E & E & E & - & - & - \\ \hline - & F & F & F & - & - \\ \hline \end{tabular}\]
The permutations we are looking for should use one letter from each column, and there should not be any repeated letters:
$\begin{tabular}{c} CDEFAB \\ CEAFBD \\ CEFABD \\ CEFBAD \\ CFEABD \\ CFEBAD \\ DEAFBC \\ DEAFCB \\ DEFABC \\ DEFACB \\ DEFBAC \\ DFEABC \\ DFEACB \\ DFEBAC \\ EDAFBC \\ EDAFCB \\ EDFABC \\ EDFACB \\ EDFBAC \\ EFABCD \end{tabular}$
There are $\boxed{20}$ such permutations. | D | 20 |
0ed85e4b626d8fb1cffdd36cbc5c7f4e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_22 | Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\; 14 \qquad\textbf{(B)}\; 16 \qquad\textbf{(C)}\; 18 \qquad\textbf{(D)}\; 20 \qquad\textbf{(E)}\; 24$ | We can represent each rearrangement as a permutation of the six elements $\{1,2,3,4,5,6\}$ in cycle notation. Note that any such permutation cannot have a 1-cycle, so the only possible types of permutations are 2,2,2-cycles, 4,2-cycles, 3,3-cycles, and 6-cycles. We deal with each case separately.
For 2,2,2-cycles, suppose that one of the 2-cycles switches the people across from each other, i.e. $(14)$ $(25)$ , or $(36)$ . WLOG, we may assume it to be $(14)$ . Then we could either have both of the other 2-cycles be across from each other, giving the permutation $(14)(25)(36)$ or else neither of the other 2-cycles is across from each other, in which case the only possible permutation is $(14)(26)(35)$ . This can happen for $(25)$ and $(36)$ as well. So since the first permutation is not counted twice, we find a total of $1+3=4$ permutations that are 2,2,2-cycles where at least one of the 2-cycles switches people diametrically opposite from each other. Otherwise, since the elements in a 2-cycle cannot differ by 1, 3, or 5 mod 6, they must differ by 2 or 4 mod 6, i.e. they must be of the same parity. But since we have three odd and three even elements, this is impossible. Hence there are exactly 4 such permutations that are 2,2,2-cycles.
For 4,2-cycles, we assume for the moment that 1 is part of the 2-cycle. Then the 2-cycle can be $(13)$ $(15)$ , or $(14)$ . The first two are essentially the same by symmetry, and we must arrange the elements 2, 4, 5, 6 into a 4-cycle. However, 5 must have two neighbors that are not next to it, which is impossible, hence the first two cases yield no permutations. If the 2-cycle is $(14)$ , then we must arrange the elements 2, 3, 5, 6 into a 4-cycle. Then 2 must have the neighbors 5 and 6. We find that the 4-cycles $(2536)$ and $(2635)$ satisfy the desired properties, yielding the permutations $(14)(2536)$ and $(14)(2635)$ . This can be done for the 2-cycles $(25)$ and $(36)$ as well, so we find a total of 6 such permutations that are 4,2-cycles.
For 3,3-cycles, note that if 1 neighbor 4, then the third element in the cycle will neighbor one of 1 and 4, so this is impossible. Therefore, the 3-cycle containing 1 must consist of elements 1, 3, and 5. Therefore, we obtain the four 3,3-cycles $(135)(246)$ $(153)(246)$ $(135)(264)$ , and $(153)(264)$
For 6-cycles, note that the neighbors of 1 can be 3 and 4, 3 and 5, or 4 and 5. In the first case, we may assume that it looks like $(314\dots)$ -- the form $(413\dots)$ is also possible but equivalent to this case. Then we must place the elements 2, 5, and 6. Note that 5 and 6 cannot go together, so 2 must go in between them. Also, 5 cannot neighbor 4, so we are left with one possibility, namely $(314625)$ , which has an analogous possibility $(413526)$ . In the second case, we assume that it looks like $(315\dots)$ . The 2 must go next to the 5, and the 6 must go last (to neighbor the 3), so the only possibility here is $(315246)$ , with the analogous possibility $(513642)$ . In the final case, we may assume that it looks like $(415\dots)$ . Then the 2 and 3 cannot go together, so the 6 must go in between them. Therefore, the only possibility is $(415362)$ , with the analogous possibility $(514263)$ . We have covered all possibilities for 6-cycles, and we have found 6 of them.
Therefore, there are $4+6+4+6 = \boxed{20}$ such permutations. | D | 20 |
a77ae6fc7467da59b7eed28399a53979 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_23 | 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?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$ | 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. | B | 10 |
a77ae6fc7467da59b7eed28399a53979 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_23 | 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?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$ | 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 have $a>2$ , or $a\geqslant3$
Also note that $c \geq b \geq a > 0$ , hence $\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}$ .
Thus, $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}$ , so $a \leq 6$
So we have $a=3, 4, 5$ or $6$
Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$ . From $\frac{1}{b}<k$ , we have $b>\frac{1}{k}$ . From $\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k$ , we have $b \leq \frac{2}{k}$ . Thus $\frac{1}{k}<b \leq \frac{2}{k}$
When $a=3$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$ , so $b=7, 8, 9, 10, 11, 12$ . We find the solutions $(a, b, c)=(3, 7, 42)$ $(3, 8, 24)$ $(3, 9, 18)$ $(3, 10, 15)$ $(3, 12, 12)$ , for a total of $5$ solutions.
When $a=4$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$ , so $b=5, 6, 7, 8$ . We find the solutions $(a, b, c)=(4, 5, 20)$ $(4, 6, 12)$ $(4, 8, 8)$ , for a total of $3$ solutions.
When $a=5$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$ , so $b=5, 6$ . The only solution in this case is $(a, b, c)=(5, 5, 10)$
When $a=6$ $b$ is forced to be $6$ , and thus $(a, b, c)=(6, 6, 6)$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions. | B | 10 |
d4c5cd6e0df2570f7ef63483756498ca | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24 | Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$ | First, note that $PQ$ lies on the radical axis of any of the pairs of circles. Suppose that $O_1$ and $O_2$ are the centers of two circles $C_1$ and $C_2$ that intersect exactly at $P$ and $Q$ , with $O_1$ and $O_2$ lying on the same side of $PQ$ , and $O_1 O_2=39$ . Let $x=O_1 R$ $y=O_2 R$ , and suppose that the radius of circle $C_1$ is $r$ and the radius of circle $C_2$ is $\tfrac{5}{8}r$
Then the power of point $R$ with respect to $C_1$ is
\[(r+x)(r-x) = r^2 - x^2 = 24^2\]
and the power of point $R$ with respect to $C_2$ is
\[\left(\frac{5}{8}r + y\right) \left(\frac{5}{8}r - y\right) = \frac{25}{64}r^2 - y^2 = 24^2.\]
Also, note that $x-y=39$
Subtract the above two equations to find that $\tfrac{39}{64}r^2 - x^2 + y^2 = 0$ or $39 r^2 = 64(x^2-y^2)$ . As $x-y=39$ , we find that $r^2=64(x+y) = 64(2y+39)$ . Plug this into an earlier equation to find that $25(2y+39)-y^2=24^2$ . This is a quadratic equation with solutions $y=\tfrac{50 \pm 64}{2}$ , and as $y$ is a length, it is positive, hence $y=57$ , and $x=y+39=96$ . This is the only possibility if the two centers lie on the same side of their radical axis.
On the other hand, if they lie on opposite sides, then it is clear that there is only one possibility, and then it is clear that $O_1 R + O_2 R = O_1 O_2 = 39$ . Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is $57+96+39 = \boxed{192}$ | D | 192 |
d4c5cd6e0df2570f7ef63483756498ca | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24 | Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$ | Note that if a circle passes through a pair of points, the center of the circle is on the perpendicular bisector of the line segment between the pair of points. This means that $A$ $B$ $C$ , and $D$ are all on the perpendicular bisector of $PQ$ . Let us say the distance from $A$ to the line $PQ$ is some $a$ . Therefore, the distance from $B$ to the line $PQ$ is $\sqrt{(\frac{8}{5}\sqrt{a^2 + 24^2})^2 - 24^2}$ , which comes out to be $\sqrt{\frac{64}{25}a^2 + \frac{39}{25}\cdot 576}$ . Since $AB = 39$ , we have one of $\sqrt{\frac{64}{25}a^2 + \frac{39}{25}\cdot 576} \pm a$ to be equal to $39$ . We can solve both equations to get that out of the four possible solutions, and only two are positive: $7$ and $57$ . Note that since no two circles can be congruent, we need the radius of one of $A$ or $C$ to be $7$ and the other to be $57$ . Plugging in to find the corresponding radii of $B$ and $D$ gives $32$ and $96$ , and adding everything up gives $\boxed{192}$ | D | 192 |
d4c5cd6e0df2570f7ef63483756498ca | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24 | Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$ | Let's start by drawing $PQ$ . Because all circles contain $P$ and $Q$ , all the centers lie on the perpendicular bisector of $PQ$ , and point $R$ is on this bisector.
For all the circle radii to be different (there can't be two congruent circles), two centers are on the same side of $PQ$ , and two are on the opposite side of $PQ$ . For the latter two circles--call them $A$ and $B$ -- $AR+BR=39$
Let's consider the next case, where $C$ and $D$ lie on the same side. Construct right triangles from the picture, and use the Pythagorean Theorem (divide by 3 to negate big numbers). You will get that the distance from $R$ to the closest circle center is $57$ . Therefore, the answer is $39+2\cdot57+39=\boxed{192}$ | D | 192 |
d4c5cd6e0df2570f7ef63483756498ca | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24 | Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$ | Since the radical axis $PQ$ is perpendicular to the line connecting the center of the circles, we have that $A$ $B$ $C$ $D$ , and $R$ are collinear. WLOG, assume that $A$ and $B$ are on the same side of $R$ and let $AR=y$ and let $BP=x$ so that $AP=\frac{5}{8}x$
Then, using the Pythagorean Theorem on right triangles $PBR$ and $PAR$ \[(39+y)^2+24^2=x^2\qquad(1)\] \[y^2+24^2=\frac{25}{64}x^2\qquad(2)\] Subtracting the $(2)$ from $(1)$ gives \[x^2=64(2y+39)\qquad(3)\] Substituting $(3)$ into $(2)$ gives $y^2-50y-399=0=(y-57)(y+7).$ Taking the positive solution ( $y>0$ ), $y=AR=57$ and $BR=(57)+39=96$
Since none of the circles are congruent, $C$ and $D$ must be on the opposite side of $R$ so $CR+DR=CD=39$ . Hence, $AR+BR+CR+DR=57+96+39=\boxed{192}$ | D | 192 |
d4c5cd6e0df2570f7ef63483756498ca | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24 | Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$ | Note that the four circles are coaxial, meaning $A,B,C,D,R$ are all collinear. Let $AR=x$ . By Pythagorean Theorem , the radius of circle $A$ squared would be $r_A^2 = x^2+24^2$ and the radius of circle $B$ squared would be $r_B^2 = (x+39)^2+24^2.$ Since $r_A^2 = \dfrac{25}{64}r_B^2$ \[x^2+24^2 = \dfrac{25}{64}((x+39)^2+24^2)\] Solving this will give $x^2-50x-399 = 0$ , or $x=57, -7$ $AR = 57$ . The same equation will apply to $CR$ $CR$ would be the other root: $CR=7$ (the negative $7$ signals that $C$ is the negative, or opposite, direction of $D$ , about $R$ ). Thus, $AR = 57$ $BR = 57+39 = 96$ $CR = 7$ $DR = 39-7 = 32$ , implying $AR+BR+CR+DR = \boxed{192}$ | D | 192 |
d5bd7b2c4f883ed89c5c69057a9e66dd | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$ | Let $x=e^{i\pi/6}$ , a $30^\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is:
\[1+2x+3x^2+\cdots+(k+1)x^k\]
We need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetico-geometric series
\begin{align*} S &=1+2x+3x^2+\cdots+2015x^{2014} \\ xS &=x+2x^2+3x^3+\cdots+2015x^{2015} \\ (1-x)S &=1+x+x^2+\cdots+x^{2014}-2015x^{2015} \\ S &= \frac{1-x^{2015}}{(1-x)^2}-\frac{2015x^{2015}}{1-x} \end{align*}
We want to find $|S|$ . First, note that $x^{2015}=x^{11}=x^{-1}$ because $x^{12}=1$ . Therefore
\[S =\frac{1-\frac{1}{x}}{(1-x)^2}-\frac{2015}{x(1-x)}=-\frac{1}{x(1-x)}-\frac{2015}{x(1-x)}=-\frac{2016}{x(1-x)}.\]
Hence, since $|x|=1$ , we have $|S| = \frac{2016}{|1-x|}.$
Now we just have to find $|1-x|$ . This can just be computed directly:
\[1-x=1-\frac{\sqrt{3}}{2}-\frac{1}{2}i\]
\[|1-x|^2=\left(1-\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=2-\sqrt{3}={\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)}^2\]
\[|1-x|=\frac{\sqrt{6}-\sqrt{2}}{2}\]
Therefore $|S|=2016\cdot\frac{2}{\sqrt{6}-\sqrt{2}}=2016\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)=1008\sqrt{2}+1008 \sqrt{6}$
Thus the answer is $1008+2+1008+6=\boxed{2024}$ | B | 2024 |
d5bd7b2c4f883ed89c5c69057a9e66dd | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$ | Here is an alternate solution that does not use complex numbers:
The distance from $P_{2015}$ to $P_0$ can be evaluated using the Pythagorean theorem . Assuming $P_0$ lies at the origin, we can calculate the distance the bee traveled to $P_{2015}$ by evaluating the distance the bee traveled in the x-direction and the y-direction. Let's start by summing each movement:
$x=1\cos{0}+2\cos{30}+3\cos{60}+\cdots+2014\cos{270}+2015\cos{300}$
A movement of $p$ units at $q$ degrees is the same thing as a movement of $-p$ units at $q-180$ degrees, so we can adjust all the cosines with arguments greater than $180$ as follows:
$x=1\cos{0}+2\cos{30}+3\cos{60}+4\cos{90}+5\cos{120}+6\cos{150}-7\cos{0}-8\cos{30}-\cdots-2015\cos{120}$
Grouping terms with like-cosines and factoring out the cosines:
$x=(1-7+13-\cdots+2005-2011)\cos{0}+\cdots+(6-12+18-\cdots-2004+2010)\cos{150}$
Each sum in the parentheses has $336$ terms (except the very last one, which has $335$ ). By pairing each term, we see there are $\frac{336}{2}$ pairs of $-6$ . Therefore, each sum equals $168\cdot-6=-1008$ except the very last sum, which has $167$ pairs of $-6$ plus an extra 2010 and equals $167\cdot-6+2010=1008$ . Plugging in these values:
$x=-1008\cos{0}-1008\cos{30}-1008\cos{60}-1008\cos{90}-1008\cos{120}+1008\cos{150}$ $x=1008(-1-\frac{\sqrt{3}}{2}-\frac{1}{2}-0+\frac{1}{2}-\frac{\sqrt{3}}{2})=-1008(1+\sqrt{3})$
We can find how far the bee traveled in the y-direction using the same logic as above, we arrive at the sum:
$y=-1008\sin{0}-1008\sin{30}-1008\sin{60}-1008\sin{90}-1008\sin{120}+1008\sin{150}$
$y=1008(0-\frac{1}{2}-\frac{\sqrt{3}}{2}-1-\frac{\sqrt{3}}{2}+\frac{1}{2})=-1008(1+\sqrt{3})$
Finally, we use the Pythagorean to find the distance from $P_0$ . This distance is given by:
$\sqrt{x^2+y^2}=\sqrt{(-1008(1+\sqrt{3}))^2+(-1008(1+\sqrt{3}))^2}=\sqrt{2\cdot1008^2\cdot(1+\sqrt{3})^2}=1008(1+\sqrt{3})\sqrt{2}=1008\sqrt{2}+1008\sqrt{6}$ , so the answer is $1008+2+1008+6=\boxed{2024}$ | B | 2024 |
d5bd7b2c4f883ed89c5c69057a9e66dd | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$ | We first notice that if the bee is turning 30 degrees each turn, it will take 12 turns to be looking in the same direction when the bee initially left. This means we simply need to answer the question; how far will the bee be when the bee is facing in the same direction?
First we use the fact that after 3 turns, the bee will be facing in a direction perpendicular to the the initial direction. From here we can draw a perpendicular from $P_2$ to the line $\overline{P_0P_1}$ intersecting a point $C_0$ . We will also place the point $C_1$ at the intersection of $\overline{P_0P_1}$ and $\overline{P_3P_4}$ . In addition, the point $C_2$ is placed at the perpendicular dropped from $P_2$ to the line $\overline{P_3C_1}$ . We will also set the distance $\overline{P_0P_1} = n$ and thus $\overline{P_1P_2} = n+1$ . With this perpendicular we see that the triangle $\triangle{P_1P_2C_0}$ is a 30-60-90 triangle. This means that the length $\overline{P_1C_0} = \frac{(n+1)\sqrt{3}}{2}$ and the length $\overline{C_1C_2} = \frac{n+1}{2}$ . We can also see that the triangle $\triangle{P_2C_1P_3}$ is a 30-60-90 triangle and thus $\overline{C_0C_1} = \frac{n+2}{2}$ and $\overline{C_2P_3} = \frac{(n+2)\sqrt{3}}{2}$ . Now if we continue this across all $P_i$ and set the point $P_0$ to the coordinates $(0, 0)$ . As you can see, we are inherently putting a “box” around the figure. Doing similar calculations for all four “sides” of this spiral we get that the length
\[\overline{P_0C_1} = n + \frac{(n+1)\sqrt{3}}{2} + \frac{n+2}{2}\] \[\overline{C_1C_4} = \frac{(n+1)}{2} + \frac{(n+2)\sqrt{3}}{2} + (n+3) + \frac{(n+4)\sqrt{3}}{2} + \frac{n+5}{2}\] \[\overline{C_4C_7} = \frac{(n+4)}{2} + \frac{(n+5)\sqrt{3}}{2} + (n+6) + \frac{(n+7)\sqrt{3}}{2} + \frac{n+8}{2}\] \[\overline{C_7C_{10}} = \frac{(n+7)}{2} + \frac{(n+8)\sqrt{3}}{2} + (n+9) + \frac{(n+10)\sqrt{3}}{2} + \frac{n+11}{2}\] , and finally \[\overline{C_{10}P_{12}} = \frac{(n+10)}{2} + \frac{(n+11)\sqrt{3}}{2}\]
Here the point $C_4$ is defined as the intersection of lines $\overline{P_3P_4}$ and $\overline{P_6P_7}$ . The point $C_7$ is defined as the intersection of lines $\overline{P_6P_7}$ and $\overline{P_9P_{10}}$ . Finally, the point $C_10$ is defined as the intersection of lines $\overline{P_{9}P_{10}}$ and $\overline{P_{12}P_{13}}$ . Note that our spiral stops at $P_{12}$ before the next spiral starts. Calculating the offset from the x and the y direction, we see that the offset, or the new point $P_{12}$ , is $({-6}, {-6}-12 \sqrt{3})$ . This is an interesting property that the points’ coordinate changes by a constant offset no matter what $n$ is. Since the new point’s subscript changes by 12 each time and we see that 2016 is divisible by 12, the point $P_{2016} = ({-168} \cdot {6}, {168} \cdot ({-6} \sqrt{3} {-12}))$ . Using similar 30-60-90 triangle properties, we see that $P_{2015} = ({-6} \cdot 168-1008 \sqrt{3}, 168({-6} \sqrt{3} - 12) + 1008)$ . Using the distance formula, the numbers cancel out nicely (1008 is divisible by 168, so take 168 when using the distance formula) and we see that the final answer is $(1008)(1+\sqrt{3})(\sqrt{2})$ which gives us a final answer of $\boxed{2024}$ | null | 2024 |
d5bd7b2c4f883ed89c5c69057a9e66dd | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$ | Suppose that the bee makes a move of distance $i$ . After $6$ turns it will be facing the opposite direction and move $i+6$ units. Combining these opposite movements gives a total movement of $-6$ units in the original direction. This means that every $12$ moves, the bee will move $-6$ units in each direction of $0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ$
We want to find the displacement vector for every $12$ moves. Factoring out the $-6$ for now (which flips the direction), we draw a quick diagram of one unit in each direction. [asy] draw((0,0)--(1,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2+sqrt(3)/2)--(3/2+sqrt(3)/2, 3/2+sqrt(3)/2)--(1+sqrt(3)/2, 3/2+sqrt(3))--(1, 2+sqrt(3)), EndArrow); draw((1,0)--(3/2+sqrt(3)/2, 0), dashed); draw((1,2+sqrt(3))--(3/2+sqrt(3)/2, 2+sqrt(3)), dashed); draw((3/2+sqrt(3)/2, 2+sqrt(3))--(3/2+sqrt(3)/2, 0), dashed); draw((1+sqrt(3)/2,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2), dashed); draw((1+sqrt(3)/2,2+sqrt(3))--(1+sqrt(3)/2, 3/2+sqrt(3))--(3/2+sqrt(3)/2, 3/2+sqrt(3)), dashed); [/asy] Using the 30-60-90 triangles, it is clear that the displacement vector (factoring back in the $-6$ ) is $-6\left\langle 1, 2+\sqrt{3}\right\rangle$
To compute the distance to $P_{2015}$ , we can compute the position of $P_{2016}$ (a multiple of $12$ moves) and then subtract the vector from $P_{2015}$ to $P_{2016}$
The bee reaches $P_{2016}$ after $\frac{2016}{12} = 168$ sets of $12$ moves, so the total displacement vector to $P_{2016}$ is $168(-6)\left\langle 1, 2+\sqrt{3}\right\rangle = \left\langle -1008, -2006-1008\sqrt{3}\right\rangle$
The bee moves at an angle of $-30^\circ$ from $P_{2015}$ to $P_{2016}$ , so subtracting it means moving an angle of $150^\circ$ . Since the vector is $2016$ units long, by a 30-60-90 triangle, it is $\left\langle -1008\sqrt{3}, 1008\right\rangle$
Therefore the total displacement vector to $P_{2015}$ is $\left\langle -1008, -2006-1008\sqrt{3}\right\rangle + \left\langle -1008\sqrt{3}, 1008\right\rangle = \left\langle -1008-1008\sqrt{3}, -1008-1008\sqrt{3}\right\rangle$ . The displacement is thus $\sqrt{2}\left|-1008-1008\sqrt{3}\right| = 1008\sqrt{2}+1008\sqrt{6} \implies 1008+2+1008+6 = \boxed{2024}$ | B | 2024 |
d5bd7b2c4f883ed89c5c69057a9e66dd | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$ | Let $P_0$ be the origin. East would be the real axis in the positive direction. Then we can assign each $P_n$ a complex value. The displacement would then be the magnitude of the complex number.
Notice that after the $n$ th move the value of $P_n$ is $P_{n-1}+ne^{\frac{i(n-1)\pi}{6}}$ . Also notice that after six moves the bee is facing in the opposite direction. And because we have found a recursion, we can add these up.
Then we have \[P_n=e^{0}+2e^{\frac{i\pi}{6}}+\cdots+2015e^{\frac{2014i\pi}{6}},\] and this becomes \[(1-7+13-\cdots-2011)e^0+(2-8+\cdots-2012)e^{\frac{i\pi}{6}}+\cdots+(6-12+\cdots+2010)e^{\frac{5i\pi}{6}}.\]
Simplifying, we have \[-6\cdot168-6\cdot168e^{\frac{i\pi}{6}}-6\cdot168e^{\frac{2i\pi}{6}}-6\cdot168e^{\frac{3i\pi}{6}}-6\cdot168e^{\frac{4i\pi}{6}}-(2010-6\cdot167)e^{\frac{5i\pi}{6}},\] which eventually simplifies to \[-1008-(2+\sqrt3)1008i,\] and this is a $15-75-90$ triangle which has ratios of $1:(2+\sqrt3):(\sqrt2+\sqrt6)$ so the magnitude is $1008\sqrt2+1008\sqrt6$ and the answer is $1008+2+1008+6=\boxed{2024}$ | B | 2024 |
d5bd7b2c4f883ed89c5c69057a9e66dd | https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25 | A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$ | After each 12 moves, the bee will be facing the same direction as it started. Let $P_0$ be the origin and let $P_n$ (with $n$ divisible by $12$ ) be $(x,y)$ . We now notice that each of the move pairs with lengths $n+1$ $n+7$ $n+2$ $n+8$ $n+3$ $n+9$ $n+4$ $n+10$ $n+5$ $n+11$ $n+6$ $n+12$ will move the bee 6 units in the directions corresponding to the moves with lengths $n+7$ $n+8$ $n+9$ $n+10$ $n+11$ , and $n+12$ . This equates to moving the bee from $(x,y)$ to $(x-6,y-12-6\sqrt{3})$ , a move that repeats every 12 moves. Since $\frac{2016}{12} = 168$ , we have that $P_(2016) = (-1008, -2016-1008\sqrt{3})$ . It follows that $P_(2015) = (-1008-1008\sqrt{3}, -1008-1008\sqrt{3})$ so the distance to $P_0$ is $1008(\sqrt{2} + \sqrt{6})$ so the answer is $1008 + 2 + 1008 + 6 = \boxed{2024}$ | B | 2024 |
8ced0fa4e62d2571a1f4c024fc2c4f23 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_2 | At the theater children get in for half price. The price for $5$ adult tickets and $4$ child tickets is $$24.50$ . How much would $8$ adult tickets and $6$ child tickets cost?
$\textbf{(A) }$35\qquad \textbf{(B) }$38.50\qquad \textbf{(C) }$40\qquad \textbf{(D) }$42\qquad \textbf{(E) }$42.50$ | Suppose $x$ is the price of an adult ticket. The price of a child ticket would be $\frac{x}{2}$
\begin{eqnarray*} 5x + 4(x/2) = 7x &=& 24.50\\ x &=& 3.50\\ \end{eqnarray*}
Plug in for 8 adult tickets and 6 child tickets.
\begin{eqnarray*} 8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\ &=&\boxed{38.50} | B | 38.50 |
9f71a48715f97d1d12960b17799f7db2 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | 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:
$\text{B}-\text{O}-\text{R}-\text{Y}$
$\text{O}-\text{B}-\text{R}-\text{Y}$
The answer is $1+2=\boxed{3}$ | B | 3 |
9f71a48715f97d1d12960b17799f7db2 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | 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 symmetry, exactly half of the $12$ arrangements have the blue house before the yellow house, and exactly half of those $6$ arrangements have the orange house before the red house, so our answer is $12\cdot\frac{1}{2}\cdot\frac{1}{2}= \boxed{3}$ | B | 3 |
9f71a48715f97d1d12960b17799f7db2 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4 | 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: $\text{O-B-R-Y}$ .
Therefore, our answer is $\boxed{3}$ | B | 3 |
396f2f976638a27e600105ba37c3a839 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | 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 \[90-87=3\implies\boxed{3}\] | C | 3 |
396f2f976638a27e600105ba37c3a839 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | 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 $70$ or $80$ ).The mean is $10\%\cdot70+35\%\cdot80+30\%\cdot90+25\%\cdot100=7+28+27+25=87$ .
So, our solution is $90-87=3\Rightarrow\boxed{3}$ ~sosiaops | C | 3 |
396f2f976638a27e600105ba37c3a839 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | 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 $\le 80$ -point scores make up $10\%+35\% = 45\% < 50\%$ of the scores, but the $\le 90$ -point scores make up $45\%+30\% = 75\% > 50\%$ of the scores, so the median is $90$
$10\%$ of scores were $70-90 = -20$ more than the median, $35\%$ were $-10$ more, $100\%-75\% = 25\%$ were $10$ more, and the rest were equal. This means that the mean score is $10\%\cdot(-20)+35\%\cdot(-10)+25\%\cdot10 = -2 + (-3.5) + 2.5 = -3$ more than the median, so their difference is $\left|-3\right| = \boxed{3}$ .
~ emerald_block | C | 3 |
4ef2a255c0275ab2581bf02895b7f68a | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_6 | The difference between a two-digit number and the number obtained by reversing its digits is $5$ times the sum of the digits of either number. What is the sum of the two digit number and its reverse?
$\textbf{(A) }44\qquad \textbf{(B) }55\qquad \textbf{(C) }77\qquad \textbf{(D) }99\qquad \textbf{(E) }110$ | Let the two digits be $a$ and $b$ . Then, $5a + 5b = 10a + b - 10b - a = 9a - 9b$ , or $2a = 7b$ . This yields $a = 7$ and $b = 2$ because $a, b < 10$ . Then, $72 + 27 = \boxed{99}.$ | D | 99 |
4ef2a255c0275ab2581bf02895b7f68a | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_6 | The difference between a two-digit number and the number obtained by reversing its digits is $5$ times the sum of the digits of either number. What is the sum of the two digit number and its reverse?
$\textbf{(A) }44\qquad \textbf{(B) }55\qquad \textbf{(C) }77\qquad \textbf{(D) }99\qquad \textbf{(E) }110$ | We start like above. Let the two digits be $a$ and $b$ . Therefore, $5(a+b) = 10a+b-10b-a=9(a-b)$ . Since we are looking for $10a+b+10b+a=11(a+b)$ and we know that $a+b$ must be a multiple of $9$ , the only answer choice that works is $\boxed{99}.$ | D | 99 |
bdf1157d287c23856113e8dc3c9bdb32 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_11 | problem_id
bdf1157d287c23856113e8dc3c9bdb32 A customer who intends to purchase an applianc...
bdf1157d287c23856113e8dc3c9bdb32 A customer who intends to purchase an applianc...
Name: Text, dtype: object | Let the listed price be $x$ . Since all the answer choices are above $\textdollar100$ , we can assume $x > 100$ . Thus the discounts after the coupons are used will be as follows:
Coupon 1: $x\times10\%=.1x$
Coupon 2: $20$
Coupon 3: $18\%\times(x-100)=.18x-18$
For coupon $1$ to give a greater price reduction than the other coupons, we must have $.1x>20\implies x>200$ and $.1x>.18x-18\implies.08x<18\implies x<225$
The only choice that satisfies such conditions is $\boxed{219.95}$ | C | 219.95 |
bdf1157d287c23856113e8dc3c9bdb32 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_11 | problem_id
bdf1157d287c23856113e8dc3c9bdb32 A customer who intends to purchase an applianc...
bdf1157d287c23856113e8dc3c9bdb32 A customer who intends to purchase an applianc...
Name: Text, dtype: object | For coupon $1$ to be the most effective, we want 10% of the price to be greater than 20. This clearly occurs if the value is over 200. For coupon 1 to be more effective than coupon 3, we want to minimize the value over 200, so $\boxed{219.95}$ is the smallest number over 200. | C | 219.95 |
319997b31b053c975a9c9d5ff9b8fbcc | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 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 $\boxed{4}$ | B | 4 |
319997b31b053c975a9c9d5ff9b8fbcc | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 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 $\boxed{4}$ | B | 4 |
319997b31b053c975a9c9d5ff9b8fbcc | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 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 $b=a+2$ $b+2=(a+2)+2=\boxed{4}$ | null | 4 |
319997b31b053c975a9c9d5ff9b8fbcc | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 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 $a+4 \Longrightarrow \boxed{4}$ | B | 4 |
d52fac580a189f0980006e4b2e9b254b | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | 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 $7d+525=10d$ , or $d=175$
Now, we must add an extra $35$ miles traveled in the first hour, giving a total of $\boxed{210}$ miles. | C | 210 |
d52fac580a189f0980006e4b2e9b254b | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | 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 $35 \text{ mph}$ , David has $175$ miles left. This then takes him $3.5$ hours at $50 \text{ mph}$ . But $210/35 = 6 \text{ hours}$ . Since $1+3.5 = 4.5 \text{ hours}$ is $1.5$ hours less than $6$ , our answer is $\boxed{210}$ | C | 210 |
d52fac580a189f0980006e4b2e9b254b | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | 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 $d=35(5+1)=35\cdot 6=210 \Longrightarrow \boxed{210}$ | C | 210 |
105d383b26e9011174560fc6b6c71914 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_13 | A fancy bed and breakfast inn has $5$ rooms, each with a distinctive color-coded decor. One day $5$ friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no more than $2$ friends per room. In how many ways can the innkeeper assign the guests to the rooms?
$\textbf{(A) }2100\qquad \textbf{(B) }2220\qquad \textbf{(C) }3000\qquad \textbf{(D) }3120\qquad \textbf{(E) }3125\qquad$ | We can work in reverse by first determining the number of combinations in which there are more than $2$ friends in at least one room. There are three cases:
Case 1: Three friends are in one room. Since there are $5$ possible rooms in which this can occur, we are choosing three friends from the five, and the other two friends can each be in any of the four remaining rooms,
there are $5\cdot\binom{5}{3}\cdot4\cdot4 = 800$ possibilities.
Case 2: Four friends are in one room. Again, there are $5$ possible rooms, we are choosing four of the five friends, and the other one can be in any of the other four rooms, so there are $5\cdot\binom{5}{4}\cdot4= 100$ possibilities.
Case 3: Five friends are in one room. There are $5$ possible rooms in which this can occur, so there are $5$ possibilities.
Since there are $5^5 = 3125$ possible combinations of the friends, the number fitting the given criteria is $3125 - (800+100+5) = \boxed{2220}$ | null | 2220 |
c4a83a2ae23b8860abab288e824c030b | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_15 | A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$ | For each digit $a=1,2,\ldots,9$ there are $10\cdot10$ (ways of choosing $b$ and $c$ ) palindromes. So the $a$ s contribute $(1+2+\cdots+9)(100)(10^4+1)$ to the sum.
For each digit $b=0,1,2,\ldots,9$ there are $9\cdot10$ (since $a \neq 0$ ) palindromes. So the $b$ s contribute $(0+1+2+\cdots+9)(90)(10^3+10)$ to the sum.
Similarly, for each $c=0,1,2,\ldots,9$ there are $9\cdot10$ palindromes, so the $c$ contributes $(0+1+2+\cdots+9)(90)(10^2)$ to the sum.
It just so happens that \[(1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000\] so the sum of the digits of the sum is $\boxed{18}$ | B | 18 |
c4a83a2ae23b8860abab288e824c030b | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_15 | A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$ | Notice that $10001+ 99999 = 110000.$ In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is $110000.$ We have $9\cdot 10\cdot 10$ palindromes, or $450$ pairs of palindromes summing to $110000.$ Performing the multiplication gives $49500000$ , so the sum $\boxed{18}$ | B | 18 |
c4a83a2ae23b8860abab288e824c030b | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_15 | A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$ | As shown above, there are a total of $900$ five-digit palindromes. We can calculate their sum by finding the expected value of a randomly selected palindrome satisfying the conditions given, then multiplying it by $900$ to get our sum. The expected value for the ten-thousands and the units digit is $\frac{1+2+3+\cdots+9}{9}=5$ , and the expected value for the thousands, hundreds, and tens digit is $\frac{0+1+2+\cdots+9}{10}=4.5$ . Therefore our expected value is $5\times10^4+4.5\times10^3+4.5\times10^2+4.5\times10^1+5\times10^0=55,\!000$ . Since the question asks for the sum of the digits of the resulting sum, we do not need to keep the trailing zeros of either $55,\!000$ or $900$ . Thus we only need to calculate $55\times9=495$ , and the desired sum is $\boxed{18}$ | B | 18 |
c4a83a2ae23b8860abab288e824c030b | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_15 | A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$ | First, allow $a$ to be zero, and then subtract by how much we overcount. We'll also sum each palindrome with its $\textit{complement}$ . If $\overline{abcba}$ (the line means a, b, and c are digits and $abcba\ne a\cdot b\cdot c\cdot b\cdot a$ ) is a palindrome, then its complement is $\overline{defed}$ where $d=9-a$ $e=9-b$ $f=9-c$ . Notice how every palindrome has a unique compliment, and that the sum of a palindrome and its complement is $99999$ . Therefore, the sum of our palindromes is $99999\times (10^3/2)$ . (There are $10^3/2$ pairs.)
However, we have overcounted, as something like $05350$ $\textit{isn't}$ a palindrome by the problem's definition, but we've still included it. So we must subtract the sum of numbers in the form $\overline{0nmn0}$ . By the same argument as before, these sum to $9990\times (10^2/2)$ . Therefore, the sum that the problem asks for is:
\[500\times99999-50\times 9990\] \[=500\times99999-500\times 999\] \[=500(99999-999)\] \[=500\times 99000\]
Since all we care about is the sum of the digits, we can drop the $0$ 's.
\[5\times99\] \[=5\times(100-1)\] \[=495\]
And finally, $4+9+5=\boxed{18}$ | B | 18 |
6a2cc5541e1749d3d0bb739aacabadf4 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_20 | 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 $7$ $k-2$ $1$ 's, and $04$ . We want to make the numbers sum to 1000, so $7+4+(k-2) = 1000$ . Solving, we get $k = 991$ , meaning the answer is $\fbox{(D)}$
Another way to proceed is that we know the difference between the sum of the digits of each product and $k$ is always $9$ , so we just do $1000-9=\boxed{991}$ | D | 991 |
a6fe97e6abf371a042d7cdc12829da49 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_18 | The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A) }19\qquad \textbf{(B) }31\qquad \textbf{(C) }271\qquad \textbf{(D) }319\qquad \textbf{(E) }511\qquad$ | For all real numbers $a,b,$ and $c$ such that $b>0$ and $b\neq1,$ note that:
Therefore, we have \begin{align*} \log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))) \text{ is defined} &\implies \log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))>0 \\ &\implies \log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))>1 \\ &\implies 0<\log_{16}(\log_{\frac1{16}}x)<\frac14 \\ &\implies 1<\log_{\frac1{16}}x<2 \\ &\implies \frac{1}{256}<x<\frac{1}{16}. \end{align*} The domain of $f(x)$ is an interval of length $\frac{1}{16}-\frac{1}{256}=\frac{15}{256},$ from which the answer is $15+256=\boxed{271}.$ | C | 271 |
a6fe97e6abf371a042d7cdc12829da49 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_18 | The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A) }19\qquad \textbf{(B) }31\qquad \textbf{(C) }271\qquad \textbf{(D) }319\qquad \textbf{(E) }511\qquad$ | For simplicity, let $a=\log_{\frac{1}{16}}{x},b=\log_{16}a,c=\log_{\frac{1}{4}}b$ , and $d=\log_4c$
The domain of $\log_{\frac{1}{2}}x$ is $x \in (0, \infty)$ , so $d \in (0, \infty)$ .
Thus, $\log_4{c} \in (0, \infty) \Rightarrow c \in (1, \infty)$ .
Since $c=\log_{\frac{1}{4}}b$ we have $b \in \left(0, \left(\frac{1}{4}\right)^1\right)=\left(0, \frac{1}{4}\right)$ .
Since $b=\log_{16}{a}$ , we have $a \in (16^0,16^{1/4})=(1,2)$ .
Finally, since $a=\log_{\frac{1}{16}}{x}$ $x \in \left(\left(\frac{1}{16}\right)^2,\left(\frac{1}{16}\right)^1\right)=\left(\frac{1}{256},\frac{1}{16}\right)$
The length of the $x$ interval is $\frac{1}{16}-\frac{1}{256}=\frac{15}{256}$ and the answer is $\boxed{271}$ | C | 271 |
a6fe97e6abf371a042d7cdc12829da49 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_18 | The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A) }19\qquad \textbf{(B) }31\qquad \textbf{(C) }271\qquad \textbf{(D) }319\qquad \textbf{(E) }511\qquad$ | The domain of $f(x)$ is the range of the inverse function $f^{-1}(x)=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}$ . Now $f^{-1}(x)$ can be seen to be strictly decreasing, since $\left(\frac12\right)^x$ is decreasing, so $4^{\left(\frac12\right)^x}$ is decreasing, so $\left(\frac14\right)^{4^{\left(\frac12\right)^x}}$ is increasing, so $16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}$ is increasing, therefore $\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}$ is decreasing.
Therefore, the range of $f^{-1}(x)$ is the open interval $\left(\lim_{x\to\infty}f^{-1}(x), \lim_{x\to-\infty}f^{-1}(x)\right)$ . We find: \begin{align*} \lim_{x\to-\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}&= \lim_{a\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^a}}}\\ &= \lim_{b\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^b}}\\ &= \left(\frac1{16}\right)^{16^0}\\ &= \frac{1}{16}. \end{align*} Similarly, \begin{align*} \lim_{x\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}&=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^0}}}\\ &= \left(\frac1{16}\right)^{16^{\frac14}}\\ &= \left(\frac1{16}\right)^2\\ &= \frac{1}{256}. \end{align*} Hence the range of $f^{-1}(x)$ (which is then the domain of $f(x)$ ) is $\left(\frac{1}{256},\frac{1}{16}\right)$ and the answer is $\boxed{271}$ | C | 271 |
28df386612800e3289364bd41bf31d8d | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_19 | There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and \[5x^2+kx+12=0\] has at least one integer solution for $x$ . What is $N$
$\textbf{(A) }6\qquad \textbf{(B) }12\qquad \textbf{(C) }24\qquad \textbf{(D) }48\qquad \textbf{(E) }78\qquad$ | Factor the quadratic into \[\left(5x + \frac{12}{n}\right)\left(x + n\right) = 0\] where $-n$ is our integer solution. Then, \[k = \frac{12}{n} + 5n,\] which takes rational values between $-200$ and $200$ when $|n| \leq 39$ , excluding $n = 0$ . This leads to an answer of $2 \cdot 39 = \boxed{78}$ | E | 78 |
28df386612800e3289364bd41bf31d8d | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_19 | There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and \[5x^2+kx+12=0\] has at least one integer solution for $x$ . What is $N$
$\textbf{(A) }6\qquad \textbf{(B) }12\qquad \textbf{(C) }24\qquad \textbf{(D) }48\qquad \textbf{(E) }78\qquad$ | Solve for $k$ so \[k=-\frac{12}{x}-5x.\] Note that $x$ can be any integer in the range $[-39,0)\cup(0,39]$ so $k$ is rational with $\lvert k\rvert<200$ . Hence, there are $39+39=\boxed{78}.$ | E | 78 |
28df386612800e3289364bd41bf31d8d | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_19 | There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and \[5x^2+kx+12=0\] has at least one integer solution for $x$ . What is $N$
$\textbf{(A) }6\qquad \textbf{(B) }12\qquad \textbf{(C) }24\qquad \textbf{(D) }48\qquad \textbf{(E) }78\qquad$ | Plug in $k=200$ to find the upper limit. You will find the limit to be a number from $0<x<-1$ and one that is just below $-39.$ All the integer values from $-1$ to $-39$ can be attainable through some value of $k$ . Since the question asks for the absolute value of $k$ , we see that the answer is $39\cdot2 = \boxed{78.}$ | E | 78. |
ebffaf2feb1d3baba43941777480fdf2 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_20 | In $\triangle BAC$ $\angle BAC=40^\circ$ $AB=10$ , and $AC=6$ . Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$ | Let $C_1$ be the reflection of $C$ across $\overline{AB}$ , and let $C_2$ be the reflection of $C_1$ across $\overline{AC}$ . Then it is well-known that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$ . (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.) As $A$ lies on both $AB$ and $AC$ , we have $C_2A=C_1A=CA=6$ . Furthermore, $\angle CAC_1=2\angle CAB=80^\circ$ by the nature of the reflection, so $\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ$ . Therefore by the Law of Cosines \[BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{14}.\] | D | 14 |
ebffaf2feb1d3baba43941777480fdf2 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_20 | In $\triangle BAC$ $\angle BAC=40^\circ$ $AB=10$ , and $AC=6$ . Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$ | In $\triangle BAC$ , the three lines look like the Chinese character 又. Let $\triangle DEA$ $\triangle CDA$ , and $\triangle BEA$ have bases $DE$ $CD$ , and $BE$ respectively. Then, $\triangle DEA$ has the same side $DA$ as $\triangle CDA$ and the same side $EA$ as $\triangle BEA$ . Connect all three triangles with $\triangle DEA$ in the center and the two triangles sharing one of its sides. Then, $\pentagon BACDE$ is formed with $BE+DE+CD$ forming the base.
Intuitively, the pentagon's base is minimized when all three bottom sides are collinear. This is simply the original $\triangle BAC$ except that $\angle BAC =120^\circ$ . (In $\triangle DEA$ $\triangle CDA$ , and $\triangle BEA$ $\angle A = 40^\circ$ , and the three triangles connect at $A$ to form the pentagon). Thus, $m\angle BAC = 40 \cdot 3$ ).
$BC$ in this new triangle is then the minimum of $BE+DE+CD$ . Applying law of cosines, $BC=\sqrt{6^2+10^2-2(6)(10)\cos (120^\circ)}=\sqrt{196}=14 \implies \boxed{14}$ | D | 14 |
ebffaf2feb1d3baba43941777480fdf2 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_20 | In $\triangle BAC$ $\angle BAC=40^\circ$ $AB=10$ , and $AC=6$ . Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$ | [asy] size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,Ep,Bp,Cp; A = (0,0); B = 10*dir(-110);C = 6*dir(-70); Bp = 10*dir(-30);Cp = 6*dir(-150); D = IP(Cp--Bp, A--B); Ep = IP(Cp--Bp, A--C); draw(A--B--C--A--Cp--Bp--A); draw(Cp--B); draw(C--Bp); draw(C--D); draw(B--Ep); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$B'$", Bp, E); dot("$C'$", Cp, W); dot("$D$", D, dir(-70)); dot("$E$", Ep, dir(60)); MA("40^\circ",Cp,A,D, 1); MA("40^\circ",D,A,Ep, 1); MA("40^\circ",Ep,A,Bp, 1); label("$6$", A--Cp); label("$10$", Bp--A); [/asy]
(Diagram by shihan)
Reflect $C$ across $AB$ to $C'$ . Similarly, reflect $B$ across $AC$ to $B'$ . Clearly, $BE = B'E$ and $CD = C'D$ . Thus, the sum $BE + DE + CD = B'E + DE + C'D$ . This value is minimized when $B'$ $C'$ $D$ and $E$ are collinear. To finish, we use the law of cosines on the triangle $AB'C'$ $B'C' = \sqrt{6^2 + 10^2 - 2(6)(10)\cos 120^{\circ}} = \boxed{14}.$ | D | 14 |
d51bffd65de9e0c252b2ec0e65d7f979 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_21 | For every real number $x$ , let $\lfloor x\rfloor$ denote the greatest integer not exceeding $x$ , and let \[f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).\] The set of all numbers $x$ such that $1\leq x<2014$ and $f(x)\leq 1$ is a union of disjoint intervals. What is the sum of the lengths of those intervals?
$\textbf{(A) }1\qquad \textbf{(B) }\dfrac{\log 2015}{\log 2014}\qquad \textbf{(C) }\dfrac{\log 2014}{\log 2013}\qquad \textbf{(D) }\dfrac{2014}{2013}\qquad \textbf{(E) }2014^{\frac1{2014}}\qquad$ | Let $\lfloor x\rfloor=k$ for some integer $1\leq k\leq 2013$ . Then we can rewrite $f(x)$ as $k(2014^{x-k}-1)$ . In order for this to be less than or equal to $1$ , we need $2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)$ . Combining this with the fact that $\lfloor x\rfloor =k$ gives that $x\in\left[k,k+\log_{2014}\left(\dfrac{k+1}k\right)\right]$ , and so the length of the interval is $\log_{2014}\left(\dfrac{k+1}k\right)$ . We want the sum of all possible intervals such that the inequality holds true; since all of these intervals must be disjoint, we can sum from $k=1$ to $k=2013$ to get that the desired sum is \[\sum_{i=1}^{2013}\log_{2014}\left(\dfrac{i+1}i\right)=\log_{2014}\left(\prod_{i=1}^{2013}\dfrac{i+1}i\right)=\log_{2014}\left(\dfrac{2014}1\right)=\boxed{1}.\] | A | 1 |
8dd58d71d622cdca7859918e42e278d8 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_25 | 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 $4 \cdot 5^n$
The third power of $2$ is between $4 \cdot 5^n$ and $8 \cdot 5^n$ , meaning that it can be between $5^n$ and $5^{n+1}$ or not.
If there are only $2$ power of $2$ s between every consecutive powers of $5$ up to $5^{867}$ , there would be $867\cdot 2 = 1734$ power of $2$ s. However, there are $2013$ powers of $2$ before $5^{867}$ , meaning the answer is $2013 - 1734 = \boxed{279}$ | B | 279 |
aa18879eaf88a69e9b2edb86f2045609 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_23 | The fraction
\[\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},\]
where $n$ is the length of the period of the repeating decimal expansion. What is the sum $b_0+b_1+\cdots+b_{n-1}$
$\textbf{(A) }874\qquad \textbf{(B) }883\qquad \textbf{(C) }887\qquad \textbf{(D) }891\qquad \textbf{(E) }892\qquad$ | $\frac{1}{99^2}\\\\ =\frac{1}{99} \cdot \frac{1}{99}\\\\ =\frac{0.\overline{01}}{99}\\\\ =0.\overline{00010203...9799}$
So, the answer is $0+0+0+1+0+2+0+3+...+9+7+9+9=2\cdot10\cdot\frac{9\cdot10}{2}-(9+8)$ or $\boxed{883}$ | B | 883 |
cb010b64769a59dd885495b2be9d59f6 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_24 | Let $f_0(x)=x+|x-100|-|x+100|$ , and for $n\geq 1$ , let $f_n(x)=|f_{n-1}(x)|-1$ . For how many values of $x$ is $f_{100}(x)=0$
$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$ | 1. Draw the graph of $f_0(x)$ by dividing the domain into three parts. [asy] unitsize(0); int w = 250; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); yaxis(-h,h,Ticks(100.0),Arrows); draw((-100,-h)--(-100,h),dashed); draw((100,-h)--(100,h),dashed); real f0(real x) { return x + abs(x-100) - abs(x+100); } draw(graph(f0,-w,w),Arrows); label("$f_0$",(-w,f0(-w)),W); [/asy]
2. Apply the recursive rule a few times to find the pattern.
Note: $f_n(x) = |f_{n-1}(x)| - 10$ is used to enlarge the difference, but the reasoning is the same. [asy] unitsize(0); int w = 350; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); yaxis(-h,h,Ticks(100.0),Arrows); draw((-100,-h)--(-100,h),dashed); draw((100,-h)--(100,h),dashed); int s = 10; real f0(real x) { return x + abs(x-100) - abs(x+100); } real f(real x, int k) { real a = abs(f0(x))-s*k; return a >= -s ? a : k%2 == 0 ? abs(x%(2*s)-s)-s : -abs(x%(2*s)-s); } real f1(real x) { return f(x,1); } real f2(real x) { return f(x,2); } real f3(real x) { return f(x,3); } real f4(real x) { return f(x,4); } draw(graph(f0,-w,w,w*2#s),Arrows); draw(graph(f1,-w,w,w*2#s),red,Arrows); draw(graph(f2,-w,w,w*2#s),orange,Arrows); draw(graph(f3,-w,w,w*2#s),lightolive,Arrows); label("$f_0$",(-w,f0(-w)),W); label("$f_1$",(-w,f1(-w)),NW,red); label("$f_2$",(-w,f2(-w)),W,orange); label("$f_3$",(-w,f3(-w)),SW,lightolive); [/asy]
3. Extrapolate to $f_{100}$ . Notice that the summits start $100$ away from $0$ and get $1$ closer each iteration, so they reach $0$ exactly at $f_{100}$ [asy] unitsize(0); int w = 350; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); yaxis(-h,h,Ticks(100.0),Arrows); draw((-100,-h)--(-100,h),dashed); draw((100,-h)--(100,h),dashed); int s = 10; real f0(real x) { return x + abs(x-100) - abs(x+100); } real f(real x, int k) { real a = abs(f0(x))-s*k; return a >= -s ? a : k%2 == 0 ? abs(x%(2*s)-s)-s : -abs(x%(2*s)-s); } real f1(real x) { return f(x,1); } real f2(real x) { return f(x,2); } real f3(real x) { return f(x,3); } real f4(real x) { return f(x,4); } real f98(real x) { return f(x,100#s-2); } real f99(real x) { return f(x,100#s-1); } real f100(real x) { return f(x,100#s); } draw(graph(f0,-w,w,w*2#s),Arrows); draw(graph(f1,-w,w,w*2#s),red,Arrows); draw(graph(f2,-w,w,w*2#s),orange,Arrows); draw(graph(f3,-w,w,w*2#s),lightolive,Arrows); draw(graph(f98,-w,w,w*2#s),heavygreen,Arrows); draw(graph(f99,-w,w,w*2#s),blue,Arrows); draw(graph(f100,-w,w,w*2#s),purple,Arrows); label("$f_0$",(-w,f0(-w)),W); label("$f_1$",(-w,f1(-w)),NW,red); label("$f_2$",(-w,f2(-w)),W,orange); label("$f_3$",(-w,f3(-w)),SW,lightolive); label("$f_{98}$",(-w,f98(-w)),NW,heavygreen); label("$f_{99}$",(-w,f99(-w)),W,blue); label("$f_{100}$",(-w,f100(-w)),SW,purple); [/asy]
$f_{100}(x)$ reaches $0$ at $x = -300$ , then zigzags between $0$ and $-1$ , hitting $0$ at every even $x$ , before leaving $0$ at $x = 300$
This means that $f_{100}(x) = 0$ at all even $x$ where $-300 \le x \le 300$ . This is a $601$ -integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or $\frac{601+1}{2} = \boxed{301}$ .
(Revised by Flamedragon & Jason,C & emerald_block | C | 301 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.