$\bmod 120$

Observe that $5! = 120 \equiv 0 \pmod{120}$ and that n! has 5! as its factor for any n >= 5. Then we have

$\begin{aligned} 1!+2!+3!+ \cdots + 120! & \equiv 1!+2!+3!+4! \pmod{120} \\ & \equiv 33 \pmod{120}. \end{aligned}$ Similar problem by brilliant staff