If the sum of $1! + 2! + 3! + \cdots + 49! + 50!$ is divided by $15$, what is the remainder?
So, I want to find out how many terms in the sum are multiples of $15$. I notice that any factorial greater than or equal to $5!$ is a multiple of $15$, because it contains factors of $3$ and $5$. Therefore, the sum of $5! + 6! + \cdots + 50!$ is a multiple of $15$, and I can ignore it when computing the remainder. That leaves me with the sum of $1! + 2! + 3! + 4!$, which is $1 + 2 + 6 + 24 = 33$. If I divide $33$ by $15$, I get a quotient of $2$ and a remainder of $3$. So, the final answer is $3$. 