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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.

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    $\begingroup$ I'd love to see how you used the Binomial Theorem. $\endgroup$ – gnasher729 Oct 28 '14 at 11:41
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Hint: Every term from $12!$ onward is divisible by $12$, so they don't matter.

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    $\begingroup$ actually, every term from $4!$ onward is divisible by $12$, too $\endgroup$ – John Dvorak Dec 26 '13 at 18:35
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    $\begingroup$ Every element from $4!$ onwards is divisible by 12 $\endgroup$ – user2369284 Dec 26 '13 at 18:36
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    $\begingroup$ 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
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If $n\ge 4$, then $4!=24$ divides $n!$ $-$ in particular $12$ divides $n!$ when $\ge 4$.

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

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