Question

If S = (1 x 1!) + (2 x 2!) + (3 x 3!) + … + (11 x 11!), then (S + 5) when divided by 12! leaves a remainder of

• A. 0
• B. 1
• C. 4
• D. 6
• E. 7

Answer 1

The Correct Option is C

To solve this problem, we need to evaluate the sum \( S = (1 \times 1!) + (2 \times 2!) + (3 \times 3!) + \ldots + (11 \times 11!) \) and then determine the remainder when \( S + 5 \) is divided by \( 12! \).

Let's break down the components:

1. Understand the expression \( n \times n! = (n+1)! - n! \). This transformation simplifies our task significantly.
2. Applying this to the entire sum \( S \), we have: S = (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + (12! - 11!).
3. Observe that this represents a telescoping series where consecutive terms cancel each other, simplifying to: S = 12! - 1!.
4. Given that \( 1! = 1 \), we conclude: S = 12! - 1.

Now, compute \( S + 5 \) and find the remainder when this sum is divided by \( 12! \):

• S + 5 = (12! - 1) + 5 = 12! + 4.
• When \( 12! + 4 \) is divided by \( 12! \), the quotient is 1 and the remainder is 4.

Thus, the remainder when \( S + 5 \) is divided by \( 12! \) is 4.

Therefore, the correct answer is: 4.