What are the last three digits of the sum $1!+2!+3!+4!+.....2020! \;$?
I got $313$.
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 $233+080=313$.
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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 $233+080=313$.