A number theory problem by SRIJAN Singh

Since $n! \bmod 21 = 0$ for $n \ge 7$ ,

$\begin{aligned} \sum_{n=1}^{2000} n! & \equiv \sum_{n=1}^6 n! \pmod {21} \\ & \equiv 1! + 2! + 3! + 4! + 5! + 6! \pmod {21} \\ & \equiv 1 + 2 + 6 + 3 + 15 + 6 \pmod{21} \\ & \equiv \boxed{12} \pmod {21} \end{aligned}$

For any $X \geq 7$ ,

$3 | X!, 7 | X! \therefore 21| X!$

So $\sum\limits_{i=1}^{2000} {i!} \equiv \sum\limits_{i=1}^{6} {i!} = 873 \equiv 12 \mod 21$