Question

Let $n! = 1 \times 2 \times 3 \times \dots \times n$ for integer $n \geq 1$. If $p = 11! + (2 \times 2!) + (3 \times 3!) + \dots + (10 \times 10!)$, then $p+2$ when divided by $11!$ leaves a remainder of:

• A. 10
• B. 0
• C. 7
• D. 1

Hint:

Telescoping factorial sums collapse to a few terms.

Answer 1

The Correct Option is B

Using identity $k \times k! = (k+1)! - k!$, the sum telescopes. $p = 11! + [(3! - 2!) + (4! - 3!) + \dots + (11! - 10!)] = 11! + (11! - 2!)$. Then $p+2 = 2 \times 11!$, divisible by $11!$. Remainder = 0. \[ \boxed{0} \]