COMEDK Exam 2016 Solved Paper

Here, 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 .
So, required remainder is 17 .