Problem:
For a positive integer $c$, let $a$ and $b$ be positive integer divisors of $c$ such that $c=ab$ and call these divisors *corresponding*. If there exist a pair of *corresponding* divisors that differ by $23$ and another pair of distinct *corresponding* divisors that differ by $20$, what are the sum of the digits of $c$?
Solution:
Think of the rainbow multiplication method of factors. Consider two factors which are 20 apart and multiply to $N$. Let these be $p, q$. Now, in order to have two factors which are $23$ apart and yet are part of the rainbow multiplication, one of the factors, $m$, must be $1$ away from $p$ and the other, $n$, must be $2$ away from $q$, or the other way around. Thus, we have two cases here.
Case 1:
We have $$N = p(p+20) = (p-1)(p+22),$$ which gives us $p=22$. This first case already has a solution, so we don't need to check the other case. Just to confirm, we have $(p, q) = (22, 42)$ and $(m, n) = (21, 44)$, and both of these multiply to $N = 22 \cdot 42 = 21 \cdot 44 = 924$, which has a sum of digits of $9+2+4 = \boxed{15}$.
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