2023 AMC 10A Problem 23

1. Name the pairs. Call the smaller divisor in the first pair $a$, so the pair is $a$ and $a + 20$. Call the smaller divisor in the second pair $b$, so that pair is $b$ and $b + 23$. Then
$$N = a(a + 20) = b(b + 23).$$
2. Compare $a$ and $b$. If $a$ were $b$ or smaller, then $a + 20$ would also be smaller than $b + 23$, and $a(a+20)$ would be less than $b(b+23)$. So $a$ is bigger than $b$.
3. But not by much. Suppose $a$ is at least $b + 2$. Then
$$a(a + 20) \ge (b + 2)(b + 22) = b^2 + 24b + 44,$$
which is more than $b(b + 23) = b^2 + 23b$. Too big. So $a$ must be exactly $b + 1$.
4. Solve. Put $a = b + 1$:
$$(b + 1)(b + 21) = b(b + 23)$$
$$b^2 + 22b + 21 = b^2 + 23b.$$
The $b^2$ cancels, leaving $b = 21$. Then $a = 22$.
5. Find $N$ and check. $N = 22 \times 42 = 924$, and also $21 \times 44 = 924$. ✓ The pairs differ by $20$ and $23$.
6. The digit sum is $9 + 2 + 4 = 15$.

The answer is $\boxed{\textbf{(C)}\ 15}$.

Two pairs of factors with the same product must be close to each other: a pair with a bigger gap needs a smaller small factor, but only a little smaller. Squeezing the small factors between two inequalities left one possibility, $a = b + 1$, and then the equation became simple. Pinning an unknown between "too small" and "too big" is a strong tool when the answer must be a whole number.

Make perfect squares. Multiply $N = a(a+20)$ by $4$ and add $400$: $4N + 400 = (2a + 20)^2$. In the same way, $4N + 529 = (2b + 23)^2$. These two squares differ by $529 - 400 = 129$. Call the square roots $u = 2b + 23$ and $v = 2a + 20$. Then
$$(u - v)(u + v) = 129 = 3 \times 43 = 1 \times 129.$$
The pair $3, 43$ gives $u = 23$, $v = 20$, so $a = b = 0$ and $N = 0$, which is not allowed. The pair $1, 129$ gives $u = 65$, $v = 64$, so $b = 21$, $a = 22$, and $N = 924$ again.