Question

Digit at the unit place of the sum of $\left(1 !\right)^{2}+\left(2 !\right)^{2}+\left(3 !\right)^{2}\ldots \ldots \ldots +\left(2008 !\right)^{2}$ is

• A. 5
• B. 7
• C. 9
• D. 6

Answer 1

Answer: (B)

Given the sum $S = \left(\right. 1 ! \left.\right)^{2} + \left(\right. 2 ! \left.\right)^{2} + \left(\right. 3 ! \left.\right)^{2} + ....... + \left(\right. 2008 ! \left.\right)^{2}$
Since the digit at units place is zero in $n !$ for $n \geq 5$
Hence, $S = \left(\right. 1 \left.\right)^{2} + \left(\right. 2 \left.\right)^{2} + \left(\right. 6 \left.\right)^{2} + \left(\right. 24 \left.\right)^{2} +$ numbers having zero at units place
$= 617 +$ all other numbers having zero at units place