2011 AMC 12B Problems/Problem 21

Problem

The arithmetic mean of two distinct positive integers $x$ and $y$ is a two-digit integer. The geometric mean of $x$ and $y$ is obtained by reversing the digits of the arithmetic mean. What is $|x - y|$ ?

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 54 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ 70$

Solution 1

Answer: (D)

$\frac{x + y}{2} = 10 a+b$ for some $1\le a\le 9$ , $0\le b\le 9$ .

$\sqrt{xy} = 10 b+a$

Squaring the first and second equations, $\frac{x^2 + 2xy + y^2}{4}=100 a^2 + 20 ab + b^2$

$xy = 100b^2 + 20ab + a^2$

Subtracting the previous two equations, $\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)$

$|x-y| = 2\sqrt{99(a^2 - b^2)}=6\sqrt{11(a^2 - b^2)}$

Note that for x-y to be an integer, $(a^2 - b^2)$ has to be $11n$ for some perfect square $n$ . Since $a$ is at most $9$ , $n = 1$ or $4$

If $n = 1$ , $|x-y| = 66$ , if $n = 4$ , $|x-y| = 132$ . In AMC, we are done. Otherwise, we need to show that $n=4$ , or $a^2 -b^2 = 44$ is impossible.

$(a-b)(a+b) = 44$ -> $a-b = 1$ , or $2$ or $4$ and $a+b = 44$ , $22$ , $11$ respectively. And since $a+b \le 18$ , $a+b = 11$ , $a-b = 4$ , but there is no integer solution for $a$ , $b$ .

Short Cut

We can arrive at $|x-y| = 6\sqrt{11(a^2 - b^2)}$ using the method above. Because we know that $|x-y|$ is an integer, it must be a multiple of 6 and 11. Hence the answer is $66.$

In addition: Note that $11n$ with $n = 1$ may be obtained with $a = 6$ and $b = 5$ as $a^2 - b^2 = 36 - 25 = 11$ .

Sidenote

It is easy to see that $(a,b)=(6,5)$ is the only solution. This yields $(x,y)=(98,32)$ . Their arithmetic mean is $65$ and their geometric mean is $56$ .

Solution 2

Let $(x+y)/2 = 10a + b$ and $\sqrt{xy} = 10b + a$ . By AM-GM we know that $a \ge b$ . Squaring and multiplying by 4 on the first equation we get $x^2 + y^2 + 2xy = 400a^2 + 4b^2 + 80ab$ . Squaring and multiplying the second equation by 4 we get $4xy = 400b^2 + 4a^2 + 80ab$ . Subtracting we get $(x-y)^2 = 396(a^2 - b^2)$ . Note that $396 = 2^2 \cdot 3^2 \cdot 11$ . So to make it a perfect square $a^2 - b^2 = 11$ . From difference of squares, we see that $a = 6$ and $b = 5$ . So the answer is $3 \cdot 2 \cdot 11 = 66$ . ~coolmath_2018

Solution 3 (whee!)

Let the 2 digit number be $10a+b$ . We know $x+y=20a+2b$ and $\sqrt{xy} = 10b+a$ .

Thus. $x^2+2xy+y^2 = 400a^2+80ab+4b^2$ and $xy = 100b^2+20ab+a^2$ . Notice that $|x-y| = \sqrt{(x-y)^2}$ . We know that $(x-y)^2 = (x+y)^2-4xy = 396a^2-396b^2$ . Then, $|x-y| = 6\sqrt{11}\cdot \sqrt{b^2-a^2}$ . Clearly, $|x-y|$ must be a multiple of 11, so the only possible answer is $\boxed{\textbf{(D) }66}$ .

-skibbysiggy

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AMC 12 Problems and Solutions

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