2005 AMC 10B Problem 24

Reversed digits suggest writing the numbers with their digits. Let $x = 10a + b$ and $y = 10b + a$, where $a$ and $b$ are digits. Since $m$ is positive, $x > y$, so $a > b$.

1. Factor. $x^2 - y^2 = (x - y)(x + y)$ (difference of squares). Now
- $x - y = 9a - 9b = 9(a - b)$,
- $x + y = 11a + 11b = 11(a + b)$.

So $m^2 = 9 \cdot 11 \cdot (a - b)(a + b)$.
2. What makes a perfect square? In a perfect square, every prime appears an even number of times. The $9 = 3 \cdot 3$ is fine, but $11$ appears only once so far. So $(a - b)(a + b)$ must supply another $11$.
3. Where can the $11$ come from? $a - b$ is between $1$ and $8$, so it cannot be a multiple of $11$. So $a + b$ must be. Since $a + b$ is at most $9 + 8 = 17$, we need $a + b = 11$.
4. Now $m^2 = 9 \cdot 11 \cdot 11 \cdot (a - b) = 33^2 \cdot (a - b)$. For this to be a perfect square, $a - b$ must be a perfect square: $1$ or $4$.
5. Find the digits.
- $a + b = 11$, $a - b = 1$: adding gives $2a = 12$, so $a = 6$, $b = 5$.
- $a + b = 11$, $a - b = 4$: adding gives $2a = 15$, not a whole number. Rejected.
6. Answer. $x = 65$, $y = 56$, and $m^2 = 33^2 \cdot 1$, so $m = 33$. Check: $65^2 - 56^2 = 4225 - 3136 = 1089 = 33^2$.
$$x + y + m = 65 + 56 + 33 = 154.$$

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

For problems about reversing digits, write the numbers as $10a + b$ and $10b + a$; their sum is always a multiple of $11$ and their difference a multiple of $9$. A perfect square needs each prime an even number of times, and the small size of digits leaves only a couple of options to check.

From step 2, $m^2$ is a multiple of $99$, so $m$ must be a multiple of both $3$ and $11$, that is, a multiple of $33$. Try $m = 33$: we need $x^2 = y^2 + 1089$ with $x$, $y$ reversed two-digit numbers. Checking $65$ and $56$: $4225 - 3136 = 1089$. It works, giving $65 + 56 + 33 = 154$.

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 $\boxed{\textbf{(E)}\ 154}$.