2005 AMC 12B Problems/Problem 19


Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits of $x$. The integers $x$ and $y$ satisfy $x^{2}-y^{2}=m^{2}$ for some positive integer $m$. What is $x+y+m$?

$\mathrm{(A)}\ 88    \qquad \mathrm{(B)}\ 112   \qquad \mathrm{(C)}\ 116   \qquad \mathrm{(D)}\ 144   \qquad \mathrm{(E)}\ 154   \qquad$


let $x=10a+b$, then $y=10b+a$ where $a$ and $b$ are nonzero digits.

By difference of squares, \[x^2-y^2=(x+y)(x-y)\] \[=(10a+b+10b+a)(10a+b-10b-a)\] \[=(11(a+b))(9(a-b))\]

For this product to be a square, the factor of $11$ must be repeated in either $(a+b)$ or $(a-b)$, and given the constraints it has to be $(a+b)=11$. The factor of $9$ is already a square and can be ignored. Now $(a-b)$ must be another square, and since $a$ cannot be $10$ or greater then $(a-b)$ must equal $4$ or $1$. If $a-b=4$ then $(a+b)+(a-b)=11+4$, $2a=15$, $a=15/2$, which is not a digit. Hence the only possible value for $a-b$ is $1$. Now we have $(a+b)+(a-b)=11+1$, $2a=12$, $a=6$, then $b=5$, $x=65$, $y=56$, $m=33$, and $x+y+m=154\Rightarrow\boxed{\mathrm{E}}$

See also

2005 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 12 Problems and Solutions

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