If (1 !)2+(2 !)2+(3 !)2+ ldots +(99 !)2+(100 !)2 is divided by 100, then the remainder obtained is

Question

If (1 !)2+(2 !)2+(3 !)2+ ldots +(99 !)2+(100 !)2 is divided by 100, then the remainder obtained is (A) 27 (B) 28 (C) 17 (D) 14

Tardigrade Admin · Accepted Answer

5!=120⇒ (5 !)2=(12)2× 100 ⇒ (5 !)2,(6 !)2,(7 !)2, ldots ldots ..,(99 !)2 all are divisible by 100. and (1 !)2+(2 !)2+(3 !)2+(4 !)2 =1+4+36+576=617 So, when divided by 100 , the remainder obtained will be 17.

Q. If $(1!)^{2} + (2!)^{2} + (3!)^{2} + \dots + (99!)^{2} + (100!)^{2}$ is divided by $100,$ then the remainder obtained is

$27$

$28$

$17$

$14$

$5! = 120 \Rightarrow (5!)^{2} = (12)^{2} \times 100$
$\Rightarrow (5!)^{2}, (6!)^{2}, (7!)^{2}, \dots\dots .., (99!)^{2}$ all are divisible by $100.$
and $(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2}$
$= 1 + 4 + 36 + 576 = 617$
So, when divided by $100$ , the remainder obtained will be $17.$

Q. If (1!)2+(2!)2+(3!)2+…+(99!)2+(100!)2 is divided by 100, then the remainder obtained is

27

28

17

14