Question: 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?

Answer: 
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 $n=5$: $$\sum\_{n=1}^4{n!}=1!+2!+3!+4!=33$$
Since every term after the first one is even, the whole sum is odd. But $120$ is even, hence it does not divide the sum.