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# If P = 1! + 2x2! + 3x3! + 4x4! + ... + 10x10!

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NUS School Moderator
Joined: 18 Jul 2018
Posts: 1020
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
If P = 1! + 2x2! + 3x3! + 4x4! + ... + 10x10!  [#permalink]

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07 Aug 2018, 20:48
2
2
00:00

Difficulty:

65% (hard)

Question Stats:

29% (01:30) correct 71% (02:05) wrong based on 18 sessions

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If P = 1! + 2x2! + 3x3! + 4x4! + ... + 10x10!
Find the remainder when $$\frac{(P+2)}{11!}$$

1) 0
2) 1
3) 2
4) 7
5) 9

Kudos if you liked the question

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VP
Status: Learning stage
Joined: 01 Oct 2017
Posts: 1014
WE: Supply Chain Management (Energy and Utilities)
Re: If P = 1! + 2x2! + 3x3! + 4x4! + ... + 10x10!  [#permalink]

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07 Aug 2018, 21:12
3
1
Afc0892 wrote:
If P = 1! + 2x2! + 3x3! + 4x4! + ... + 10x10!
Find the remainder when $$\frac{(P+2)}{11!}$$

1) 0
2) 1
3) 2
4) 7
5) 9

Kudos if you liked the question

Given P= 1! + 2x2! + 3x3! + 4x4! + ... + 10x10!

We have to find out the pattern of each term since each term is denoted in similar format (1*1!, 2*2! etc)

we can say, terms are in the format n*n!.

So, n*n! can be written as (n+1-1)*n!=(n+1)n!-n!=(n+1)!-n!
Now it is easy to find out P (Or sum of the factorial series given).

So, P=(2!-1!) + (3!-2!) + (4!-3!) + .................+ (11!-10!)=11!-1

So, P+2=11!-1+2=11!+1

So, the remainder when P+2 is divided by 11! is 1.

Ans. (B)
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Re: If P = 1! + 2x2! + 3x3! + 4x4! + ... + 10x10!   [#permalink] 07 Aug 2018, 21:12
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# If P = 1! + 2x2! + 3x3! + 4x4! + ... + 10x10!

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