TheCATExam
Sign in

CAT 2005 — QA Question 11

Number TheoryEasy

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

Answer & solution

Correct answer: 1

  • A

    10

  • B

    0

  • C

    7

  • 1

Solution

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).

Related Number Theory questions

See all Number Theory questions →
CAT 2005 QA Q11: Let n! = 1 × 2 × 3 × ... × n for integer n ≥ 1. If p = 1! + (2 × 2!) + (3 &ti — Solution | TheCATExam