2023 AMC 10A Problem 23

Write $N = a(a + 20) = b(b + 23)$ for positive integers $a, b$. Complete the square in each:
$$4N + 400 = (2a + 20)^2, \qquad 4N + 529 = (2b + 23)^2.$$

Set $u = 2b + 23$ and $v = 2a + 20$. Subtracting,
$$u^2 - v^2 = 129, \qquad (u - v)(u + v) = 129 = 3 \cdot 43.$$

Both factors have the same parity (they sum to $2u$), and $129$ is odd, so the only options are $(u-v, u+v) = (1, 129)$ or $(3, 43)$.

• $(3, 43)$: $u = 23$, $v = 20$, so $b = 0$ and $a = 0$, giving $N = 0$, not positive.
• $(1, 129)$: $u = 65$, $v = 64$, so $b = 21$ and $a = 22$.

Then $N = 22 \cdot 42 = 924$, and indeed $21 \cdot 44 = 924$. The digit sum is $9 + 2 + 4 = 15$.

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

A product of two numbers with a fixed difference is a square minus a constant: $a(a + k) = \left(a + \frac k2\right)^2 - \frac{k^2}{4}$. Two such descriptions of the same $N$ therefore give two squares with a known difference, and a difference of squares with a small factorization has only a handful of integer solutions. Multiplying by $4$ first keeps everything in integers.

$a^2 + 20a = b^2 + 23b$ gives $(a - b)(a + b) = 23b - 20a$. Since $a > b$ and the right side is small, $a - b$ must be tiny; $a - b = 1$ gives $2b + 1 = 3b - 20$, so $b = 21$, $a = 22$.

• Trying to guess N by testing small products, or accepting the factorization 3 times 43 that leads to N = 0.
• Answering the digit sum of $a$, $b$, or of $4N + 400$ rather than of $N = 924$.