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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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26 Dec 2009, 17:21
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What is the last digit of $$\frac{6^8}{2}$$ ?

Try to solve under 10 sec
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26 Dec 2009, 18:43
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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 , so maybe there is a shortcut?
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26 Dec 2009, 19:04
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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:

$$6^7 * \frac{6}{2} = 6*3 = 18 --> 8$$
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26 Dec 2009, 19:11
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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:

$$6^7 * \frac{6}{2} = 6*3 = 18 --> 8$$

There was a shortcut. +1.
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26 Dec 2009, 21:35
2 lions in the same jungle.

Thats interesting ....
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26 Dec 2009, 21:41
GMAT TIGER wrote:
2 lions in the same jungle.

Thats interesting ....

but you are the only tiger

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26 Dec 2009, 22:03
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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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17 Oct 2013, 16:15
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:

$$6^7 * \frac{6}{2} = 6*3 = 18 --> 8$$

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

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

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18 Feb 2014, 00:47
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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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05 Mar 2014, 11:55
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:

$$6^7 * \frac{6}{2} = 6*3 = 18 --> 8$$

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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05 Mar 2014, 22:30
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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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05 Mar 2014, 23:25
dbiersdo wrote:
Is there a reason we don't have any possible options to choose from in the original post?

This is walker's own question, not an official one, so no options.
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Re: What is the last digit of 6^8/2 ? [#permalink]

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26 Jan 2015, 06:10
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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28 Feb 2016, 19:15
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Re: What is the last digit of 6^8/2 ? [#permalink]

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21 Sep 2017, 21:09
Hello from the GMAT Club BumpBot!

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Re: What is the last digit of 6^8/2 ?   [#permalink] 21 Sep 2017, 21:09
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