# What is the remainder when the infinite sum (1!)

What is the remainder when the infinite sum (1!)² + (2!)² + (3!)² + ··· is divided by 1152?

## Answer/Solution

41

### Steps/Work

Explanation :

We have to find out the remainder when (1!)² + (2!)² + (3!)² + ··· is divided by 1152

1152 = 27 * 32

=> (6!)2 is divisble by 1152

=> All (n!)2 are divisible by 1152 as long as n > 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 : B

We have to find out the remainder when (1!)² + (2!)² + (3!)² + ··· is divided by 1152

1152 = 27 * 32

=> (6!)2 is divisble by 1152

=> All (n!)2 are divisible by 1152 as long as n > 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 : B