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Last digit [#permalink]

26 Dec 2009, 18:21

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based on 2 sessions

What is the last digit of \frac{6^8}{2} ?

Try to solve under 10 sec

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Re: Last digit [#permalink]

26 Dec 2009, 19:43

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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CEO

Joined: 17 Nov 2007

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Re: Last digit [#permalink]

26 Dec 2009, 20:04

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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 :wink:

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Re: Last digit [#permalink]

26 Dec 2009, 20:11

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:

There was a shortcut. +1.
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Re: Last digit [#permalink]

26 Dec 2009, 22:35

2 lions in the same jungle.

Thats interesting .... :wink:

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Re: Last digit [#permalink]

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]

26 Dec 2009, 23:03

(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

Re: Last digit [#permalink] 26 Dec 2009, 23:03

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