I am working on preparing for JEE and was working on this math problem.
We have the sum, $$\sum_{n=1}^{120}n!=1!+2!+3!+\ldots+120!$$ Now I am given the question, which says that what happens when this sum is divided by $120$. Does it divide evenly? If not, then what is its remainder?
Note that $5!=120$, and all terms after $n=5$ are also divisible by $120$. Therefore, we can conclude that $120$ does not divide evenly and our remainder is the sum of the terms before $n=5$: $$\sum_{n=1}^4{n!}=1!+2!+3!+4!=33$$