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 $x + y + m?$

$88$

$112$

$116$

$144$

$154$

Answer: E

Concepts:digits difference of squares perfect square

Difficulty rating: 1880

Solution:

Write $x = 10a + b$ and $y = 10b + a$ with $a \gt b.$ Then $\begin{aligned} m^2 &= x^2 - y^2 \\ &= 99(a^2 - b^2) \\ &= 99(a+b)(a-b). \end{aligned}$

Since $99 = 9 \cdot 11,$ for $m^2$ to be a perfect square we need $11 \mid (a+b)(a-b).$ As $a + b \le 17,$ this forces $a + b = 11,$ and then $a - b$ must itself be a perfect square.

With $a - b \le 8,$ the only workable case is $a - b = 1,$ giving $(a, b) = (6, 5).$ Then $x = 65,$ $y = 56,$ and $m^2 = 99 \cdot 11 = 33^2,$ so $m = 33.$

Therefore $x + y + m$ $= 65 + 56 + 33$ $= 154.$

Thus, E is the correct answer.

2005 AMC 10B Problem 24

Answer: E

Difficulty rating: 1880

Solution:

Problem 24 in Other Years