Terms greater than 5!
i.e.,
(5!)2,(6!)2,...,(100!)2 is divisible
by 100
∴ For terms
(5!)2,(6!)2, ...,
(100!)2 remainder is 0.
Now consider
(1!)2+(2!)2+(3!)2+(4!)2
= 1 + 4+ 36+ 576
= 617
When 617 is divided by 100, its remainder is 17.
∴ Required remainder is 17.