Found this question on a previous exam:
What is the answer of the following expression?
$$(1!+2!+3!...+100!)\pmod{18} =$$
The answer $0$ isn't right!
If you know the answer please explain me it step by step.
Any help will be appreciated!
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1Note that you only need to add the first $5$ terms. – Tobias Kildetoft Sep 24 '13 at 13:16
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1Which factorials are multiples of $18$? Only the others count. – Daniel Fischer Sep 24 '13 at 13:16
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The answer is 9? – Gil Sep 24 '13 at 13:19
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Basically what we are doing is (1!)mod18 + (2!)mod18 + (3!) mod18 ... + (100!)mod18? – Gil Sep 24 '13 at 13:24
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And then take the remainder of the sum. – Daniel Fischer Sep 24 '13 at 13:26
1 Answers
6
As noted in the comments, if $18$ divides $n!$, then you can ignore that particular term. Now, $18$ divides $6!$ and therefore any factorial greater than or equal to $6!$. So we are left with $1!+2!+3!+4!+5!=1+2+6+24+120 \equiv 9\pmod {18}$.
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