# Difference between revisions of "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 $\frac{x + y}{2} = 10 a+b$ for some $1\le a\le 9$, $0\le b\le 9$. $\sqrt{xy} = 10 b+a$ $100 a^2 + 20 ab + b^2 = \frac{x^2 + 2xy + y^2}{4}$ $xy = 100b^2 + 20ab + a^2$ $\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)$ $|x-y| = 2\sqrt{99(a^2 - b^2)}$

Note that in order for x-y to be 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 $a^2 -b^2 = 44$ 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 $a$, $b$.

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 $56$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 