# Last 5 digits of factorial sum

Find the last 5 digits of $$1!+2\cdot2!+3\cdot3!+\cdots+23\cdot23!+24\cdot24!$$

I have no idea how to find an elegant solution to this since the "last 5 digits" part makes it much harder. Usually the number of fives in each factorial reduces the computation here, but here it doesn't work.

• That sum is $25!-1$ – J. W. Tanner Dec 3 at 0:36

$$n\times n!=(n+1)!-n!$$, so the sum telescopes to $$25!-1$$.
$$25!$$ has $$6$$ factors of $$5$$ and more than that many of $$2$$.
• You mean $24$? Still, that would be painful – J. W. Tanner Dec 3 at 9:35