1!+2!+...+1000! \equiv x (mod 10)

**dwsmith** · June 26th 2010, 07:33 PM

Solve for x

$1!+2!+\dots +1000! \equiv x \ \mbox{(mod 10)}$

The book say when $k\geq 5, k!\equiv 0 \ \mbox{(mod 10)} \ \mbox{and} \ \mbox{(mod 15)}$ .

Why is this? And how am I supposed to use this to solve this problem?

**undefined** · June 26th 2010, 07:42 PM

Originally Posted by dwsmith

Solve for x

$1!+2!+\dots +1000! \equiv x \ \mbox{(mod 10)}$

The book say when $k\geq 5, k!\equiv 0 \ \mbox{(mod 10)} \ \mbox{and} \ \mbox{(mod 15)}$ .

Why is this? And how am I supposed to use this to solve this problem?

5 = 5 * 4 * 3 * 2, you can see why it's divisible by 10 and 15? Also 6! = 6 * 5! so it's also divisible by 10 and 15, etc. But the mod 15 part really doesn't matter here.

So 1! + 2! + ... + 1000! looks like 1 + 2 + 6 + 4 + 0 + 0 + 0 + ... + 0 (mod 10).

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