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