Question

If $\left(1 !\right)^{2}+\left(2 !\right)^{2}+\left(3 !\right)^{2}+\ldots +\left(99 !\right)^{2}+\left(100 !\right)^{2}$ is divided by $100,$ then the remainder obtained is

• A. $27$
• B. $28$
• C. $17$
• D. $14$

Answer 1

Answer: (C)

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