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17^27 has a units digit of:

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Intern
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Joined: 15 Sep 2018
Posts: 4
17^27 has a units digit of:  [#permalink]

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New post Updated on: 15 Sep 2018, 23:58
1
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

88% (00:28) correct 12% (00:35) wrong based on 25 sessions

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17^27 has a units digit of:

(a) 1
(b) 2
(c) 3
(d) 7
(e) 9

Originally posted by Aviv29 on 15 Sep 2018, 08:09.
Last edited by Bunuel on 15 Sep 2018, 23:58, edited 1 time in total.
Renamed the topic and edited the question.
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Re: 17^27 has a units digit of:  [#permalink]

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New post 15 Sep 2018, 08:30
Aviv29 wrote:
17^ 27 has a units digit of:

(a)1
(b)2
(c)3
(d)7
(e)9

Some can explain to me how to solve this question?
Thank you for the support :-)


\(7^1 = 7\)
\(7^2 = 9\)
\(7^3 = 3\)
\(7^4 = 1\)

\(7^5 = 7\)
\(7^6 = 9\)

So, 7 has Cyclicity of 4

Now, \(17^{27} = 17^{4*6}*7^{3}\)

Since , \(7^4 = 1\) thus \(7^{24} = 1\) and \(7^3\) = Units digit 3

Thus, \(17^{27}\) will have units digit as 3, Answer must be (C)
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Re: 17^27 has a units digit of:  [#permalink]

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New post 15 Sep 2018, 08:33
Aviv29 wrote:
17^ 27 has a units digit of:

(a)1
(b)2
(c)3
(d)7
(e)9

Some can explain to me how to solve this question?
Thank you for the support :-)


OA:C
7 has the cyclicity of 4 (7,9,3,1).
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
7^6 = 117649
and so on

\(17^{27}= 17^{6*4+3}\)

So the unit digit of \(17^{27}\) would be \(3\).

please check below link for more info about cyclicity concept
https://gmatclub.com/forum/cyclicity-on ... 13019.html
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Re: 17^27 has a units digit of:  [#permalink]

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New post 15 Sep 2018, 09:02
Great - Thank you both very much!
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Re: 17^27 has a units digit of: &nbs [#permalink] 15 Sep 2018, 09:02
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