What is the remainder when the infinite sum (1!)² + (2!)² + (3!)² + ··· is divided by 1152?
Choose one
A) 12
B) 41
C) 152
D) 802
E) None
Explanation :   We have to find out the remainder when (1!)² + (2!)² + (3!)² + ··· is divided by 1152   1152 = 27 * 32   =&gt; (6!)2 is divisble by 1152   =&gt; All (n!)2 are divisible by 1152 as long as n &gt; 5   So, our problem is now reduced to   Rem [((1!)² + (2!)² + (3!)² + (4!)² + (5!)²)/1152]   = Rem[(1 + 4 + 36 +576 + 14400) / 1152]   = Rem [15017/1152]   = 41   Answer : 41
<result>B</result>
What is the remainder when the infinite sum (1!)² + (2!)² + (3!)² + ··· is divided by 1152?