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# If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem

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VP
Joined: 20 Jul 2017
Posts: 1076
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem  [#permalink]

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Updated on: 16 Oct 2019, 19:08
2
8
00:00

Difficulty:

95% (hard)

Question Stats:

33% (02:42) correct 67% (02:14) wrong based on 58 sessions

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If $$N = 1! + 2! + 3! + . . . . . + 11! + 12!$$, what is the remainder when $$5^N$$ is divided by $$21$$ ?

A. 0
B. 1
C. 2
D. 19
E. 20

Originally posted by Dillesh4096 on 16 Oct 2019, 12:09.
Last edited by Dillesh4096 on 16 Oct 2019, 19:08, edited 1 time in total.
Math Expert
Joined: 02 Aug 2009
Posts: 8182
If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem  [#permalink]

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21 Oct 2019, 17:55
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1
Dillesh4096 wrote:
If $$N = 1! + 2! + 3! + . . . . . + 11! + 12!$$, what is the remainder when $$5^N$$ is divided by $$21$$ ?

A. 0
B. 1
C. 2
D. 19
E. 20

ManjariMishra

$$5^n=5^{1!+2!+3!+.......11!+12!}=5^1*5^2+5^{3!}.......5^{12!}$$

Now the entire number will give the same remainder as the product of remainders of each term in the product.
Remainder when 5^1 or 5 is divided by 21 is 5
5^2 or 25 will give 4
5^6=125^2=(126-1)^2.. On expansion all terms except (-1)^2 will be divisible by 21 as 126=21*6. Thus remainder =1.
All the other power contain 3! As 4!=4*3! and so on. Therefore each term thereafter will leave a remainder of 1..
Remainder therefore =5*4*1*1*1......*1=20
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VP
Joined: 19 Oct 2018
Posts: 1077
Location: India
Re: If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem  [#permalink]

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16 Oct 2019, 13:48
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2
We can notice 2 things about N

1. N is a multiple of 3.

N= 1!+2!+3!......+12!
= 1+2+3(X)
=3+3X
=3(1+X)

2. N is odd

N = 1!+2(Y)
N= 2Y+1

Hence we can write N=3k, where k is odd.

$$5^3$$= (21*6)-1

$$5^3$$= -1 mod 21
$$5^{3k}$$= $$(-1)^{odd number}$$ mod 21
$$5^N$$= -1 mod 21

$$5^N$$= (21-1) mod 21
$$5^N$$= 20 mod 21

E

Dillesh4096 wrote:
If $$N = 1! + 2! + 3! + . . . . . + 11! + 12!$$, what is the remainder when $$5^N$$ is divided by $$21$$ ?

A. 0
B. 1
C. 2
D. 19
E. 20

Posted from my mobile device
[/quote]
Manager
Joined: 10 May 2018
Posts: 50
Re: If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem  [#permalink]

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21 Oct 2019, 11:14
Manager
Joined: 16 Oct 2011
Posts: 109
GMAT 1: 570 Q39 V41
GMAT 2: 640 Q38 V31
GMAT 3: 650 Q42 V38
GMAT 4: 650 Q44 V36
GMAT 5: 570 Q31 V38
GPA: 3.75
Re: If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem  [#permalink]

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21 Oct 2019, 18:26
Notice 1 + 2! + ...... is odd. Its also a mutilple of 3 since 1 + 2! + .... = 3 + 3! + .... is a multiple of 3. so N is an odd multiple of 3. Odd multiples of 3 are 3, 9 , 27, ...
So 5^n = 5^3k, therefore 5^n is a multiple of 125. Notice 21 divides 126, so the remainder when 21 divides 125 is 20. Since 5^n = 125^x the remainder is 20. E

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Re: If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem   [#permalink] 21 Oct 2019, 18:26
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