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When 51^25 is divided by 13, the remainder obtained is:

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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]

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New post 04 Jul 2017, 06:44
Hey, I didn't understand:

When -1 is divided by 13 how can the reminder be 12?
If I substitute in the equation -1=13q+r, so how can we arrive at r=12?

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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]

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New post 04 Sep 2017, 00:33
Bunuel wrote:
LM wrote:
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0


\(51^{25}=(52-1)^{25}\), now if we expand this expression all terms but the last one will have \(52=13*4\) in them, thus will leave no remainder upon division by 13, the last term will be \((-1)^{25}=-1\). Thus the question becomes: what is the remainder upon division -1 by 13? The answer to this question is 12: \(-1=13*(-1)+12\).

Answer: A.



hi

51^25

here the last digit of 51 is 1, and the cyclicity of 1 is 1. So, the unit digit of 51^25 will be 1

thus, 1, when divided by 13 will produce a quotient "0" and will leave a remainder "1", and I am stumped here ... :(

please help me understand the problem ...

thanks in advance ...

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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]

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New post 04 Sep 2017, 00:36
gmatcracker2017 wrote:
Bunuel wrote:
LM wrote:
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0


\(51^{25}=(52-1)^{25}\), now if we expand this expression all terms but the last one will have \(52=13*4\) in them, thus will leave no remainder upon division by 13, the last term will be \((-1)^{25}=-1\). Thus the question becomes: what is the remainder upon division -1 by 13? The answer to this question is 12: \(-1=13*(-1)+12\).

Answer: A.



hi

51^25

here the last digit of 51 is 1, and the cyclicity of 1 is 1. So, the unit digit of 51^25 will be 1

thus, 1, when divided by 13 will produce a quotient "0" and will leave a remainder "1", and I am stumped here ... :(

please help me understand the problem ...

thanks in advance ...


You cannot do this way.

For example, 1 divided by 13 gives the remainder of 1 but 21 divided by 13 gives the remainder of 8. There are many different approaches and links to the underlying theory on previous 5 pages of the discussion.
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When 51^25 is divided by 13, the remainder obtained is: [#permalink]

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New post 04 Sep 2017, 11:08
gmatcracker2017 wrote:
Bunuel wrote:
LM wrote:
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0


\(51^{25}=(52-1)^{25}\), now if we expand this expression all terms but the last one will have \(52=13*4\) in them, thus will leave no remainder upon division by 13, the last term will be \((-1)^{25}=-1\). Thus the question becomes: what is the remainder upon division -1 by 13? The answer to this question is 12: \(-1=13*(-1)+12\).

Answer: A.



hi

51^25

here the last digit of 51 is 1, and the cyclicity of 1 is 1. So, the unit digit of 51^25 will be 1

thus, 1, when divided by 13 will produce a quotient "0" and will leave a remainder "1", and I am stumped here ... :(

please help me understand the problem ...

thanks in advance ...


okay bunu...

thanks ... :-)

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Kudos [?]: 13 [0], given: 123

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]

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New post 04 Sep 2017, 16:33
via fermat theorem
a^(p-1)= 1 mod p where a is an integer and p prime
as 51 contains 13 it can be rewritten and simplified 51/13 remainder 12

12^(13-1) = 1 mod 13
12^25 = 12^(12*2+1)= 12

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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]

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New post 29 Sep 2017, 00:13
[i]

Hello Bunuel and Karishma,

Can anyone from you, please tell me the limitation of cyclic rule? So far i was using that rule blindly for all question. But it has not worked for this particular problem.
[i]

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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]

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New post 29 Sep 2017, 01:12
goalMBA1990 wrote:
[i]

Hello Bunuel and Karishma,

Can anyone from you, please tell me the limitation of cyclic rule? So far i was using that rule blindly for all question. But it has not worked for this particular problem.
[i]



Cyclicity helps you figure out the units digit in every case. Also, it has nothing to do with remainders.

In some cases, the units digit helps you figure out the remainder, more specifically in case of division by 2, 5 or 10.


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https://www.veritasprep.com/blog/2015/1 ... ns-part-2/
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Re: When 51^25 is divided by 13, the remainder obtained is:   [#permalink] 29 Sep 2017, 01:12

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