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Functions

Let n !=1 × 2...

Question

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

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Solution

The correct option is C 1
Option (D) is the correct answer. Check the video for the approach.

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Similar questions

n! = 1 \times 2 \times 3 \dots \dots n

p = 1.1! + (2.2!) + (3.3!) + \dots \dots + (10.10!)

n! = 1 \times 2 \times 3 \times . . . . . . . . . . . \times n

n > 1

p = 1! + (2 \times 2!) + (3 \times 3!) + . . . . . . . . . . . + (10 \times 10!),

n! = 1 \times 2 \times 3 \times . . . . . . . . . . . \times n

n > 1

p = 1! + (2 + 2!) + (3 + 3!) + . . . . . . . . . . . + (10 + 10!),

p + 2

p = 1 \times 1! + 2 \times 2! + 3 \times 3! + . . . + 10 \times 10!

p + 2

11!

p = 1 \times 1! + 2 \times 2! + 3 \times 3! + . . . + 10 \times 10!

p + 2

11!

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