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: 1840

Solution:

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

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. Because $a+b=11$ is odd, $a-b$ is odd; and because $1 \le a-b \le 8,$ its only possible square value is $1.$ Hence $(a, b) = (6, 5).$

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

Answer: E

Difficulty rating: 1840

Solution:

Problem 19 in Other Years