CAT Question 116
Let n! = 1 × 2 × 3 × … × n for integer n ≥ 1. If p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!), then p + 2 when divided by 11! leaves a remainder of:
(1) 10
(2) 0
(3) 7
(4) 1
Powered by abhisays.com
Let n! = 1 × 2 × 3 × … × n for integer n ≥ 1. If p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!), then p + 2 when divided by 11! leaves a remainder of:
(1) 10
(2) 0
(3) 7
(4) 1