The remainder when (1!)² + (2!)² + (3!)² + ..... + (100!)² is divided by 10² is
A
28
B
27
C
14
D
17
Correct Option:
D
Explanation
Work modulo 100. For every n ≥ 5, (n!)² is divisible by 100, so only the first four terms matter: 1² + 2² + 6² + 24² = 1 + 4 + 36 + 576 = 617. The remainder on division by 100 is 17.
