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What is the remainder when 3^123 is divided by 5?

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What is the remainder when 3^123 is divided by 5?  [#permalink]

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New post 27 Apr 2016, 11:23
1
5
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A
B
C
D
E

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  5% (low)

Question Stats:

77% (00:44) correct 23% (00:42) wrong based on 196 sessions

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Re: What is the remainder when 3^123 is divided by 5?  [#permalink]

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New post 27 Apr 2016, 20:18
1
Ok lets try this one.

\(\ 3^{123}= \ (5-2)^{123}= \ 5^{123}+(-2)^{123}\)

\(\ 5^{123}\) is always divisible by 5.

So need to look at the \(\ -2^{123}\)

Units digit of -2 to the power of n is cyclical and consists of 4 possibilities {-2, 4, -8, 6}

Hence \(\ (-2)^{120}\) will have last digit 6 and \(\ (-2)^{123}\) = -8.

Dividing -8 by 5 we have remainder

-8=5n+R
For n=-2
-8=5*-2+R

R=2
Hence Answer is (C)
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Re: What is the remainder when 3^123 is divided by 5?  [#permalink]

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New post 27 Apr 2016, 20:33
maipenrai wrote:
Ok lets try this one.

\(\ 3^{123}= \ (5-2)^{123}= \ 5^{123}+(-2)^{123}\)

\(\ 5^{123}\) is always divisible by 5.

So need to look at the \(\ -2^{123}\)

Units digit of -2 to the power of n is cyclical and consists of 4 possibilities (-2, 4, -8, 6)

Hence \(\ (-2)^{120}\) will have last digit 6 and \(\ (-2)^{123}\) = -8.

Dividing 8 by -5 we have remainder

8=-5n+R
For n=2
8=-5*2+R

R=2
Hence Answer is (C)


hi,
another approach after the highlighted portion
So need to look at the \(\ -2^{123}\)
\(\ -2^{123}= (-2^4)^{\frac{123}{4}} = 16^{30}*-2^3=(15+1)^{30}*-8\)
now\((15+1)^{30}\) will leave 1 as remainder, so total remainder = 1*-8 = -8..
as correctly done, we cannot have negative remainder so remainder = 10-8 = 2
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Re: What is the remainder when 3^123 is divided by 5?  [#permalink]

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New post 27 Apr 2016, 23:34
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I believe there is an easier way to do as follow:
3^1 = 3; 3^2 = 9, 3^3 = 7, 3^4 = 1; So basiccally 3 has a cycle of 4.
Therefore 123: 4 = 30 with remainder of 3.
3^3 will have the unit digit of 7, thereby dividing by 5 will leave a remainder of 2.

C.
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Re: What is the remainder when 3^123 is divided by 5?  [#permalink]

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New post 28 Apr 2016, 22:56
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gnahanut wrote:
I believe there is an easier way to do as follow:
3^1 = 3; 3^2 = 9, 3^3 = 7, 3^4 = 1; So basiccally 3 has a cycle of 4.
Therefore 123: 4 = 30 with remainder of 3.
3^3 will have the unit digit of 7, thereby dividing by 5 will leave a remainder of 2.

C.


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Re: What is the remainder when 3^123 is divided by 5?  [#permalink]

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New post 26 Sep 2017, 08:07
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Re: What is the remainder when 3^123 is divided by 5? &nbs [#permalink] 26 Sep 2017, 08:07
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