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Updated on: 16 Oct 2019, 19:08
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.
Last edited by CareerGeek on 16 Oct 2019, 19:08, edited 1 time in total.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
42% (02:41) correct
58%
(02:25)
wrong
based on 133
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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
A. 0
B. 1
C. 2
D. 19
E. 20
21 Oct 2019, 17:55
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Dillesh4096
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
General Discussion
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
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
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