Consider positive integers $a, b$ with a difference of $20$. Suppose $b = a-20$. Then, we have $(a)(a-20) = n$. If there is another pair of two integers that multiply to $n$ but have a difference of 23, one integer must be greater than $a$, and one must be smaller than $a-20$. We can create two cases and set both equal. We have $(a)(a-20) = (a+1)(a-22)$, and $(a)(a-20) = (a+2)(a-21)$. Starting with the first case, we have $a^2-20a = a^2-21a-22$,or $0=-a-22$, which gives $a=-22$, which is not possible. The other case is $a^2-20a = a^2-19a-42$, so $a=42$. Thus, our product is $(42)(22) = (44)(21)$, so $c = 924$. Adding the digits, we have $9+2+4 = \boxed{\textbf{(C) } 15}$.
-Sepehr2010
-minor edits by aidanb
-extremely minor edits by cc21
If the positive integer $n$ has positive integer divisors $a$ and $b$ with $n = ab$, then $a$ and $b$ are said to be $\textit{complementary}$ divisors of $n$. Suppose that $N$ is a positive integer that has one complementary pair of divisors that differ by $20$ and another pair of complementary divisors that differ by $23$. What is the sum of the digits of $N$?
$\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$