Question

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 $x + y + m?$

$88$

$112$

$116$

$144$

$154$

Solution:

Let $x = 10a + b$ and $y = 10b + a$ with $a \gt b.$ Then $x^2 - y^2 = (10a+b)^2 - (10b+a)^2 = 99(a^2 - b^2) = 99(a+b)(a-b).$

Since $99 = 9 \cdot 11,$ for this to be a perfect square we need $11 \mid (a+b)(a-b).$ As $a + b \le 17$ and $a - b \le 8,$ the only multiple of $11$ available is $a + b = 11.$

Then $x^2 - y^2 = 9 \cdot 11^2 (a - b),$ which is a perfect square exactly when $a - b$ is a perfect square. Taking $a - b = 1$ with $a + b = 11$ gives $(a, b) = (6, 5).$

So $x = 65,$ $y = 56,$ and $m = \sqrt{65^2 - 56^2} = \sqrt{1089} = 33.$ Thus $x + y + m = 65 + 56 + 33 = 154.$

Thus, the correct answer is E.

2005 AMC 12B Problem 19

Solution:

Problem 19 in Other Years