Question: Find all nonnegative integer solutions to $2^a +3^b +5^c =n!$.
My work so far: I realize that if $n$ is greater than $2, 3$, or $5$, then $3^b+5^c$ has to be divisible by $2$, $2^a+5^c$ has to be divisible by $3$, and $2^a+3^b$ has to be divisible by $5$. I'm not sure how to prove that the sum of these powers equals $n!$ exactly.