Determine the sum of all single-digit replacements for $n$ such that the number ${42{,}789{,}n37}$ is divisible by 3.
I know that a number is divisible by 3 if and only if the sum of its digits is divisible by 3. So, I want to find the sum of the digits of ${42{,}789{,}n37}$ and see what values of $n$ make it a multiple of 3. The sum of the digits of ${42{,}789{,}n37}$ is $4 + 2 + 7 + 8 + 9 + n + 3 + 7 = 40 + n$. For this to be a multiple of 3, $n$ must be either 2, 5, or 8, because those are the only single-digit numbers that make $40 + n$ divisible by 3. Therefore, the sum of all single-digit replacements for $n$ is $2 + 5 + 8 = 15$. 