2005 AMC 10B Problems/Problem 24

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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 $x + y + m$ ?

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

Solution

Let $x = 10a+b, y = 10b+a$ , without loss of generality with $a>b$ . Then $x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$ . It follows that $11|(a-b)(a+b)$ , but $a-b < 10$ so $11|a+b \Longrightarrow a+b=11$ . Then we have $33^2(a-b) = m^2$ . Thus $a-b$ is a perfect square. Also, since $a-b$ and $a+b$ have the same parity, so $a-b$ is a one-digit odd perfect square, namely $1$ or $9$ . The latter case gives $(a,b) = (10,1)$ , which does not work. The former case gives $(a,b) = (6,5)$ , which works, and we have $x+y+m = 65 + 56 + 33 = 154\ \mathbf{(E)}$ .

2005 AMC 10B (Problems • Resources)
Preceded by [[2005 AMC 10B Problems/Problem {{{num-b}}}\|Problem {{{num-b}}}]]	Followed by [[2005 AMC 10B Problems/Problem {{{num-a}}}\|Problem {{{num-a}}}]]
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2005 AMC 10B Problems/Problem 24

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Solution

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