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26 Dec 2009, 18:21
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
62% (00:41) correct
38%
(00:53)
wrong
based on 206
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What is the last digit of \(\frac{6^8}{2}\) ?
Try to solve under 10 sec
Try to solve under 10 sec
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:
\(6^7 * \frac{6}{2} = 6*3 = 18 --> 8\)
But we cannot simply divide the last digit by 2. So, we can rewrite as:
\(6^7 * \frac{6}{2} = 6*3 = 18 --> 8\)
26 Dec 2009, 19:43
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walker
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? 
