The correct answer is option (C):
4
Let's analyze the given sum S = (1 x 1!) + (2 x 2!) + (3 x 3!) + … + (11 x 11!). We can rewrite each term in the sum using the identity n x n! = (n+1)! - n!.
So, 1 x 1! = 2! - 1!
2 x 2! = 3! - 2!
3 x 3! = 4! - 3!
...
11 x 11! = 12! - 11!
Now, substitute these back into the expression for S:
S = (2! - 1!) + (3! - 2!) + (4! - 3!) + ... + (12! - 11!)
Notice that this is a telescoping sum, meaning most terms cancel out. We are left with:
S = 12! - 1!
S = 12! - 1
We are asked to find the remainder when (S + 5) is divided by 12!.
So, consider S + 5 = (12! - 1) + 5
S + 5 = 12! + 4
Now divide (S + 5) by 12!:
(S + 5) / 12! = (12! + 4) / 12!
(S + 5) / 12! = 1 + 4/12!
Since 12! is an integer, the quotient is 1 and the remainder is 4.
Alternatively, we can notice that 12! is a multiple of 12!, so 12! divided by 12! leaves a remainder of 0. Since S + 5 = 12! + 4, and 12! contributes a 0 remainder when divided by 12!, the remainder is simply the remainder when 4 is divided by 12!, which is 4 because 4 is less than 12!.
Therefore, the remainder when (S + 5) is divided by 12! is 4.