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6 in any integer power >0 has the last digit 6. Integer ending with 6 divided by 2 can have 3 or 8 as last digit. But as \(6^8\) will be divisible by 4 for sure, divided by 2 it must give the even number, hence 8 is the correct answer.

Another solution: \(\frac{6^8}{2}=\frac{2^8*3^8}{2}=2^7*3^8\)

Cyclisity of both \(2\) and \(3\) is \(4\), hence \(2^7\) has the same last digit as \(2^3=8\) and \(3^8\) has the same last digit as \(3^4=81\), hence \(8*1=8\).

I used first approach, took me 11 secs , so maybe there is a shortcut? _________________

Re: What is the last digit of 6^8/2 ? [#permalink]

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18 Feb 2014, 01:47

1

This post received KUDOS

6 multiplied by 6 any times would give the last digit 6 itself However 6 multiplied by 3 would give 8 in the last place. So has been factorized accordingly _________________

Re: What is the last digit of 6^8/2 ? [#permalink]

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05 Mar 2014, 23:30

Is there a reason we don't have any possible options to choose from in the original post?

How my brain processed this:

Step 1 --> 6^(Any Power) will end in a 6 (i.e. 6*6 = 36, 6*6*6 = 216, etc.). To me this makes sense to do first based off order of operations. Step 2 --> Take any result an divide by 2. Quick and easy would be 36/2 = 18. Any answer will be 8!

Where you can trip yourself up is assuming you take 6/2=3, and to me if you stick to order of operations you can avoid this mistake.

Re: What is the last digit of 6^8/2 ? [#permalink]

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28 Feb 2016, 20:15

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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