2023 AMC 10A Problem 23

Complementary divisors with product $N$ differing by $20$ are $b$ and $b + 20,$ so $N + 100 = (b + 10)^2$

Then $4N + 529$ is also a perfect square; combine the two to get a difference of squares equal to $129$

Complementary divisors differing by $20$ are $b$ and $b + 20$ with product $N$, so $N = b^2 + 20b$ and $N + 100 = (b + 10)^2$. A pair differing by $23$ gives $4N + 529 = (2d + 23)^2$. Set $N + 100 = k^2$. Then $4k^2 + 129 = m^2$, so $(m - 2k)(m + 2k)$ $= 129$ $= 3 \cdot 43$. The positive factor pair $3,43$ gives $k=10$, hence the inadmissible value $N=0$. The pair $1,129$ gives $m = 65$, $k = 32$, hence $N = 32^2 - 100 = 924$. Check it: $924 = 22 \cdot 42 = 21 \cdot 44$, and the digit sum is $9 + 2 + 4 = 15$. Thus, C is the correct answer.