How to find 10th digit of $\sum_{k=1}^{49} k!$

Question

How to find the tenth digit (from the right) of $\sum_{k=1}^{49} (k!)$ analytically.

The worst possible method would be to actually sum each individual number which would yield a number of order $10^{62}$ . So this method is almost impossible.

Also I am new user so please point out if I am wrong somewhere.

Note: The tenth digit and not the 10's digit.

Answer 1

NOTE: The answer was posted before the edit.

The factorials starting from $10!$ onwards to $49!$ end with 2 zeroes. So they have no contribution to tens' digit.

Hence we can sum easily $$1!+2!+3!+4!+5!+6!+7!+8!+9!$$ $$=1+2+6+24+120+720+5040+40320+362880$$

The last $2$ digits of the sum add up to $13$.

Hence tens' digit is $1$.

Answer 2

You can avoid big numbers by doing the multiplies $\bmod {10^{10}}$ and just add them up.