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
Explanation:
p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!)
Now, n × n! = [(n + 1) – 1] × n! = (n + 1)! – n!
∴ p = 2! – 1! + 3! – 2! + 4! – 3! + 5! – 4! +… + 11! – 10!
∴ p = 11! – 1!
∴ p + 2 = 11! + 1
∴ p + 2 when divided by 11! leaves a remainder of 1.
Hence, option (d).
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