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

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

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27 Apr 2016, 11:23
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78% (00:42) correct 22% (00:43) wrong based on 142 sessions

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

A. 0
B. 1
C. 2
D. 4
E. 5
[Reveal] Spoiler: OA

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Kudos [?]: 133204 [0], given: 12439

Manager
Joined: 16 Feb 2016
Posts: 53

Kudos [?]: 20 [0], given: 26

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

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27 Apr 2016, 20:18
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

Kudos [?]: 20 [0], given: 26

Math Expert
Joined: 02 Aug 2009
Posts: 5234

Kudos [?]: 5894 [0], given: 118

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

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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

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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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

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

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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.

Simple and straight to the point!

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

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