2023 AMC 10A #23

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 $9+2+4=\boxed{{\displaystyle 15}}$.