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Updated on: 07 Sep 2012, 06:38
Originally posted by Legendaddy on 07 Sep 2012, 06:17.
Last edited by Bunuel on 07 Sep 2012, 06:38, edited 1 time in total.
Last edited by Bunuel on 07 Sep 2012, 06:38, edited 1 time in total.
Renamed the topic.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
55% (00:47) correct
45%
(00:39)
wrong
based on 283
sessions
History
Date
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07 Sep 2012, 06:47
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Legendaddy
If exponentiation is indicated by stacked symbols, the rule is to work from the top down: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\). For more check Number Theory chapter of Math Book: math-number-theory-88376.html
Hence, \(2^{3^4}=2^{(3^4)}=2^{81}\).
Now, the units digit of 2 in positive integer power repeats the following pattern {2, 4, 8, 6}:
2^1=2;
2^2=4;
2^3=8;
2^4=16;
2^5=32 (2 again),
Therefore the units digit of \(2^{81}\) will be the same as the units digit of 2^1 (81=4*20+1), so 2.
Answer: B.
P.S. PLEASE READ CAREFULLY AND FOLLOW: rules-for-posting-please-read-this-before-posting-133935.html (RULE #3). THANK YOU.
07 Sep 2012, 10:00
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Legendaddy
This kind of "what is last digit" problems can be easily solved using rule of cyclicity.
We need to remember that cyclicity of 2, 3, 7, 8 is 4. This means original number repeats after 4th power, hence \(2^5\) has first digit 2.
Cyclicity of 4, 9 is 2.
1, 5, 6 will remain same always.
With this basic knowledge we can proceed to answer this question.
\(2^{3^4}\)
\(=2^{81}\)
\(->2^1\)
So, last digit is 2.
With this method, I am sure one can answer such question within 1 minute.
