A number theory problem by Parth Bhardwaj

We need to find the first number which contains 100 as a factor i.e:- $2^{2}\times 5^{2}$ as all the next numbers in the series would contain it as a factor Since 2 is more abundant, we'll search for a number which has 5 as a factor. We see $10!$ is the first such number to have $5^{2}$ as a factor. Now we find:- $1!\ + \ 2!\ +\ 3! \ + \ 4! \ + \ 5! \ + \ 6! \ + \ 7! \ + \ 8! \ + \ 9! \ + \ 10!$ $= \ 4037913$

$4037913^{2} \ mod \ (100)$ = $13^{2} \ mod \ (100)$ = $69$

Thus $6\times 9 \ = \boxed{54}$