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

Question

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?

Please help.

$\begingroup$ I'd love to see how you used the Binomial Theorem. $\endgroup$ – gnasher729 Oct 28 '14 at 11:41 — gnasher729, Oct 28 '14 at 11:41

user61527 · Accepted Answer · 2013-12-26 18:34:29Z

33

Hint: Every term from $12!$ onward is divisible by $12$, so they don't matter.

answered Dec 26 '13 at 18:34

user61527

39

$\begingroup$ actually, every term from $4!$ onward is divisible by $12$, too $\endgroup$ – John Dvorak Dec 26 '13 at 18:35
1

$\begingroup$ Every element from $4!$ onwards is divisible by 12 $\endgroup$ – user2369284 Dec 26 '13 at 18:36
4

$\begingroup$ I think the answer is 9. Right! $\endgroup$ – user2369284 Dec 26 '13 at 18:37
$\begingroup$ So nice :-) Beatiful. $\endgroup$ – Umberto Sep 20 '14 at 13:07

add a comment |

Yiorgos S. Smyrlis · Accepted Answer · 2014-02-27 12:13:30Z

44

If $n\ge 4$, then $4!=24$ divides $n!$ $-$ in particular $12$ divides $n!$ when $\ge 4$.

Thus $$ 1!+2!+\cdots+1000!=1!+2!+3! \!\!\!\!\pmod{12}=9\!\!\!\!\pmod{12}. $$

answered Dec 26 '13 at 18:37

64k1386167

add a comment |

2 Answers 2