Let $n! = 1 \times 2 \times 3 \times . . . . . . . . . . . \times n$ for integer $n > 1$ . If $p = 1! + (2 + 2!) + (3 + 3!) + . . . . . . . . . . . + (10 + 10!),$ then $p + 2$ when divided by 11! leaves a remainder of (CAT 2005)

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Solution

The correct option is C 1
option (d)

If $P = 1! = 1$

Then $P + 2 = 3,$ when divided by $2!$ Remainder will be 1.

If $P = 1! + 2 \times 2! = 5$

Then, $P + 2 = 7$ when divided by $3!$ Remainder is still 1.

Hence, $P = 1! + (2 + 2!) + (3 \times 3!) + . . . . . + (10 \times 10!)$

When divided by $11!$ Leaves remainder 1.

Alternative method :
$P = 1 + 2.2! + 3.3! + . . . .10 .10!$
$= (2 - 1) 1! + (3 - 1) 2! + (4 - 1) 3! + . . . . . + (11 - 1) 10!$
$= 2! - 1! + 3! - 2! + . . . + 11! - 10!$
$= 1 + 11!$
Hence, the remainder is 1.

Suggest Corrections

Similar questions

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 \dots \dots \times n

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

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

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

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

p + 2

11!

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