# What is the remainder when $1! + 2! + 3! +\cdots+ 1000!$ is divided by $12$?

What is the remainder when $$1! + 2! + 3! +\cdots+ 1000!$$ is divided by $12$.

I have tried to find the answer using the Binomial Theorem but that doesn't help. How will we do this?

• I'd love to see how you used the Binomial Theorem. Oct 28, 2014 at 11:41

If $n\ge 4$, then $4!=24$ divides $n!$ $-$ in particular $12$ divides $n!$ when $\ge 4$.
Thus $$1!+2!+\cdots+1000!=1!+2!+3! \!\!\!\!\pmod{12}=9\!\!\!\!\pmod{12}.$$
Hint: Every term from $12!$ onward is divisible by $12$, so they don't matter.
• actually, every term from $4!$ onward is divisible by $12$, too Dec 26, 2013 at 18:35
• Every element from $4!$ onwards is divisible by 12 Dec 26, 2013 at 18:36