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What are the last three digits of the sum $1!+2!+3!+4！+.....2020! \;$?

I got $313$.

J. W. Tanner
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Selwyn Liu
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    Welcome to Mathematics Stack Exchange. Note that the last $3$ digits of $n!$ for $n\ge15$ are $000$, so it's the same as the last three digits of $1!+2!+3!+4!+\cdots+14!$ – J. W. Tanner Sep 21 '20 at 02:54
  • So this means I just have to work up to 15! ? – Selwyn Liu Sep 21 '20 at 02:57
  • Is there a specific answer to this? so I can check mine, Thanks！ – Selwyn Liu Sep 21 '20 at 03:01
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    Edit your question to include your attempt – J. W. Tanner Sep 21 '20 at 03:02
  • See https://math.stackexchange.com/questions/1314219/find-the-last-digit-of-the-sum-123-49 and https://math.stackexchange.com/questions/997439/find-the-least-positive-residue-of-123-100-modulo-each-of-the-follow and https://math.stackexchange.com/questions/1531634/what-is-the-remainder-when-12345-50-is-divided-by-5 and https://math.stackexchange.com/questions/1509183/find-the-remainder-when-123-49-is-divided-by-7 – Gerry Myerson Sep 21 '20 at 03:05
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    Quick computer confirmation: Indeed $313$. – David G. Stork Sep 21 '20 at 03:17

1 Answers1

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Your answer is correct.

For $n\ge15$, $5^3$ and $2^3$ divide $n!$, so the last three digits of $n!$ are $000$.

$1!+2!=3$; $3!+4!=30$; $5!+6!+7!+8!$ ends with $120+720+040+320\equiv200$;

$9!+10!+11!+12!$ ends with $880+800+800+600\equiv080$;

and $13!+14!$ ends with $800+200\equiv000$.

Therefore, the answer is $233+080=313$.

J. W. Tanner
  • 60,406