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

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If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem [#permalink] New post   Updated on: 16 Oct 2019, 19:08
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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
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E

Originally posted by CareerGeek on 16 Oct 2019, 12:09.
Last edited by CareerGeek on 16 Oct 2019, 19:08, edited 1 time in total.
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chetan2u
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If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem [#permalink] New post   21 Oct 2019, 17:55
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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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Re: If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem [#permalink] New post   16 Oct 2019, 13:48
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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

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Re: If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem [#permalink] New post   21 Oct 2019, 11:14
chetan2u, Bunuel, VeritasKarishma Could you please explain this
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Re: If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem [#permalink] New post   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] New post   24 Mar 2022, 21:51
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Re: If N = 1! + 2! + 3! + . . . . . + 11! + 12!, what is the rem [#permalink]
24 Mar 2022, 21:51
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