# Summation of factorials modulo ten

I have read that$$\sum\limits_{i=1}^n i!\equiv3\;(\text{mod }10),\quad n> 3.$$

Why is the sum constant, and why is it $3$?

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Hint

For $i\ge 5$ we have $$i!\equiv 0\mod 10$$

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Think about what you are summing:

$$1+2+6+24+120+720+\dots = 33 + 120 + 720 + \dots$$

Taking mod $10$ of the sum, you can see that $33$ gives $3$, can you see that all other sumands are divisible by $10$?

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