imc

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 the other 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 $(a)(a-20)=(a+2)(a-21)$ (under the requirement that one of the variables in the second case must be smaller than $a-20$). 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 $9+2+4=\boxed{15}$.