2005 AMC 10B Problem 24

Let $x = 10a + b$ and $y = 10b + a$ with digits $a, b$. Then $x - y = 9(a - b)$ and $x + y = 11(a + b)$, so
$$x^2 - y^2 = (x - y)(x + y) = 99\,(a - b)(a + b) = 3^2 \cdot 11 \cdot (a - b)(a + b).$$
For this to be a perfect square, the prime $11$ must appear to an even power, so $11$ divides $(a - b)(a + b)$. Since $m > 0$ we need $a > b$, so $1 \le a - b \le 8$ and $11 \nmid a - b$. Hence $11 \mid a + b$, and as $a + b \le 17$, we get $a + b = 11$.

Now $x^2 - y^2 = 3^2 \cdot 11^2 \cdot (a - b)$, so $a - b$ must be a perfect square. With $a + b = 11$ odd, $a - b$ is odd, so $a - b = 1$ (the only odd square at most $8$). Solving $a + b = 11$, $a - b = 1$ gives $a = 6$, $b = 5$.

Then $x = 65$, $y = 56$, and $m^2 = 99 \cdot 1 \cdot 11 = 1089 = 33^2$, so $m = 33$.

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

The answer is $\boxed{\textbf{(E)}\ 154}$.

Digit-reversal problems almost always reward writing the numbers as $10a + b$ and $10b + a$: sums and differences become multiples of $11$ and $9$. The perfect-square condition is then a statement about prime exponents, and the digit bounds ($a + b \le 18$, $a - b \le 9$) leave only one or two options for each factor. Bounding after factoring is the recurring pattern.

Recognize that $m^2$ must be a multiple of $99$, so $m$ is a multiple of $33$; try $m = 33$ and look for a Pythagorean triple $33, y, x$ with $x$ and $y$ digit reversals. The primitive triple $33$-$56$-$65$ fits ($1089 + 3136 = 4225 = 65^2$), and $65 + 56 + 33 = 154$.

• Trying to make a - b supply the factor 11, which is impossible for digits, or forgetting that a - b must itself be a perfect square after a + b = 11.
• Testing $a - b = 4$ with $a + b = 11$ and not noticing it yields non-integer digits, or reporting $x + y = 121$ without adding $m$.