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What is the last digit of 6^8/2 ?

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

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New post 26 Dec 2009, 18:21
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What is the last digit of \(\frac{6^8}{2}\) ?

Try to solve under 10 sec ;)
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Re: Last digit [#permalink]

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walker wrote:
What is the last digit of \(\frac{6^8}{2}\) ?

Try to solve under 10 sec ;)


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 :-D , so maybe there is a shortcut?
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Re: Last digit [#permalink]

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the last digit of \(6^n\) is always 6 (6*6=36).

But we cannot simply divide the last digit by 2. So, we can rewrite as:

\(6^7 * \frac{6}{2} = 6*3 = 18 --> 8\) :wink:
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Re: Last digit [#permalink]

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Re: Last digit [#permalink]

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Re: Last digit [#permalink]

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New post 26 Dec 2009, 22:41
GMAT TIGER wrote:
2 lions in the same jungle.

Thats interesting .... :wink:


but you are the only tiger :wink:
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Re: Last digit [#permalink]

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(6^8)/2 = (6/2) * 6*6*6*6*6*6*6
= 3 * 6*6*6*6*6*6*6
= 18 * 6 *6*6*6*6*6

if 18 is multiplied by any no. of 6 last digit will be 8


hence ans 8

I think this is shortest method
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Re: Last digit [#permalink]

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New post 17 Oct 2013, 17:15
walker wrote:
the last digit of \(6^n\) is always 6 (6*6=36).

But we cannot simply divide the last digit by 2. So, we can rewrite as:

\(6^7 * \frac{6}{2} = 6*3 = 18 --> 8\) :wink:


Hi @Walker, would you kindly explain why you factorized it that way?
Thanks
Cheers
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Re: What is the last digit of 6^8/2 ? [#permalink]

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New post 18 Feb 2014, 01:47
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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
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Re: Last digit [#permalink]

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New post 05 Mar 2014, 12:55
jlgdr wrote:
walker wrote:
the last digit of \(6^n\) is always 6 (6*6=36).

But we cannot simply divide the last digit by 2. So, we can rewrite as:

\(6^7 * \frac{6}{2} = 6*3 = 18 --> 8\) :wink:


Hi @Walker, would you kindly explain why you factorized it that way?
Thanks
Cheers
J :)


To say Cyclicity should not be divided. . so 6 was taken out and then divided by 2 and then multiplied with the cyclicity 6

Tats y its factorized in that way.

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

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

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

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New post 26 Jan 2015, 07:10
When do we need to separate cyclicity like that and check Bunuel?
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Re: What is the last digit of 6^8/2 ? [#permalink]

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Re: What is the last digit of 6^8/2 ?   [#permalink] 28 Feb 2016, 20:15
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