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What is the remainder when 6^x is divided by 10

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What is the remainder when 6^x is divided by 10  [#permalink]

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New post 28 Nov 2017, 10:14
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Question Stats:

67% (00:45) correct 33% (00:54) wrong based on 71 sessions

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What is the remainder when \(6^x\) is divided by 10?

1. x is a positive integer
2. x is a positive even integer

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Re: What is the remainder when 6^x is divided by 10  [#permalink]

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New post 28 Nov 2017, 21:14
1 alone is sufficient. If X is 0, then the remainder is 1. In all other cases the remainder is 6

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What is the remainder when 6^x is divided by 10  [#permalink]

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New post 28 Nov 2017, 21:29
pushpitkc wrote:
What is the remainder when \(6^x\) is divided by 10?

1. x is a positive integer
2. x is a positive even integer

Source: Experts Global


Answer should be D.

Statement 1: x is a positive interger. Meaning X > 0 and is an Integer.

Now, \((6)^P\) (where P is a positive Integer) will always give the last digit as 6 and hence remainder will always remain 6. --> SUFFICIENT

Statement 2: This is just a filtered part of Statement 1. Meaning in Statement 1, I already said we will get a single solution for X > 0 and X as an integer. Now, this includes both even and Odd. This, Statement 2 is also sufficient.

Hence, D.
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Re: What is the remainder when 6^x is divided by 10  [#permalink]

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New post 28 Nov 2017, 22:05
6 raised to any positive integer power will Always give an integer that ends in '6' (meaning the unit place digit of that number would be 6). So such an integer, when divided by 10, would always give a remainder 6.

Here, each statement alone suffices to tell us that last digit of 6^x would be 6, hence remainder would be 6.

Hence D answer
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Re: What is the remainder when 6^x is divided by 10  [#permalink]

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New post 29 Nov 2017, 09:00
Yeah true. My bad.
I jumped to the conclusion after reading statement 1.

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Re: What is the remainder when 6^x is divided by 10 &nbs [#permalink] 29 Nov 2017, 09:00
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