Difference between revisions of "2005 AMC 12B Problems/Problem 19"

Revision as of 09:41, 4 July 2013

Problem

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 $x+y+m$ ?

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

Solution

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 $x+y+m=154\Rightarrow\boxed{E}$

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

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

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 == See also ==
 {{AMC12 box|year=2005|ab=B|num-b=18|num-a=20}}
+{{MAA Notice}}

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Difference between revisions of "2005 AMC 12B Problems/Problem 19"

Revision as of 09:41, 4 July 2013

Problem

Solution

See also