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M25-31

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Bunuel
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M25-31 [#permalink] New post   16 Sep 2014, 01:24
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Question Stats:

59% (00:35) correct 41% (00:41) wrong based on 476 sessions
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What is the units digit of \(3^{3^3}\) ?

A. 1
B. 3
C. 6
D. 7
E. 9
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D
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Bunuel
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Re M25-31 [#permalink] New post   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
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langtuprovn2007
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Re: M25-31 [#permalink] New post   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 ?
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Re: M25-31 [#permalink] New post   05 Oct 2014, 02:16
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langtuprovn2007 wrote:
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 ?


That's just wrong. If this were the case, then all numbers in power which is a multiple of cyclicity would have the last digit of 1. You should always start with power of 1.

For more check UNITS DIGITS, EXPONENTS, REMAINDERS PROBLEMS in our Special Questions Directory.

Hope it helps.
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Miracles86
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Re: M25-31 [#permalink] New post   21 Apr 2018, 05:18
Hi guys,


when we divide 27 by 3, that yields a remainder of 3 (maximum possible remainder since R < Divisor)

Now when:

3^1 = 3 R=0
3^2 = 9 R=1
3^3 = 27 R=2
3^4 = 81 R=3

Now since the remainder was 3, I've associated to the last digit of 1.

Can anyone please tell me where I've went wrong in my thought proccess?

Thanks
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Re: M25-31 [#permalink] New post   21 Apr 2018, 05:31
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Miracles86 wrote:
Hi guys,


when we divide 27 by 3, that yields a remainder of 3 (maximum possible remainder since R < Divisor)

Now when:

3^1 = 3 R=0
3^2 = 9 R=1
3^3 = 27 R=2
3^4 = 81 R=3

Now since the remainder was 3, I've associated to the last digit of 1.

Can anyone please tell me where I've went wrong in my thought proccess?

Thanks


You should divide the power, which is 27, by cyclicity number, which is 4 to find the remainder: 27 divided by 4 gives the remainder of 3. Hence, the units digit of 3^27 is the same as the units digit of 3^3, which is 7.
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Re: M25-31 [#permalink] New post   05 Jan 2022, 09:19
I think this is a high-quality question and I agree with explanation.
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Bunuel
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Re: M25-31 [#permalink] New post   05 Oct 2023, 11:05
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Duplicate of M09-32. Unpublished.
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Re: M25-31 [#permalink]
05 Oct 2023, 11:05
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