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16 Sep 2014, 01:24
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16 Sep 2014, 01:24
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Official Solution:
What is the units digit of \(3^{3^3}\) ?
A. 1
B. 3
C. 6
D. 7
E. 9
If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
\(a^{m^n}=a^{(m^n)}\) and not \({(a^m)}^n\), which on the other hand equals to \(a^{mn}\).
So:
\((a^m)^n=a^{mn}\)
\(a^{m^n}=a^{(m^n)}\) and not \({(a^m)}^n\).
According to the above:
\(3^{3^3}=3^{(3^3)}=3^{27}\)
Now, the units digit of 3 in positive integer power repeats in groups of four: {3-9-7-1} - {3-9-7-1} - ... Hence, the units digit of \(3^{27}\) is the same as the units digit of \(3^3\), which is 7.
Answer: D
What is the units digit of \(3^{3^3}\) ?
A. 1
B. 3
C. 6
D. 7
E. 9
If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
\(a^{m^n}=a^{(m^n)}\) and not \({(a^m)}^n\), which on the other hand equals to \(a^{mn}\).
So:
\((a^m)^n=a^{mn}\)
\(a^{m^n}=a^{(m^n)}\) and not \({(a^m)}^n\).
According to the above:
\(3^{3^3}=3^{(3^3)}=3^{27}\)
Now, the units digit of 3 in positive integer power repeats in groups of four: {3-9-7-1} - {3-9-7-1} - ... Hence, the units digit of \(3^{27}\) is the same as the units digit of \(3^3\), which is 7.
Answer: D
General Discussion
04 Oct 2014, 22:00
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why cant we start from 3^0 instead of 3^1 ?
I did in this way: 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27 then the cycle repeat with 3^4 = 81, 3^5 =243
the pattern here is (1,3,9,7) the answer should be e ?
I did in this way: 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27 then the cycle repeat with 3^4 = 81, 3^5 =243
the pattern here is (1,3,9,7) the answer should be e ?
