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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 29 Oct 2018, 17:18
LM wrote:
When 51^25 is divided by 13, the remainder obtained is:

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


Can be written as (52-1)^25

52 is a multiple of 13.

So the remainder is -1^25

= -1

Adding divisor, 13

-1 + 13 = 12

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

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New post 04 Nov 2018, 08:46
Bunuel and VeritasKarishma, what do you think of this method? The way I did was fairly simple as well.

I did:

51^1 / 13 = remainder of 12
51^2 / 13 = remainder of 1

I noticed a pattern, whenever 13 was divided into an odd power of 51 the remainder is 12 and whenever divided by an even power the remainder was 1.

25 is odd, so remainder must be 12.


This has led me to conclude that this works for ALL numbers. A number to an any odd power and that number itself (power of 1) will always have the same remainder when divided by any integer.




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.

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Re: When 51^25 is divided by 13, the remainder obtained is: &nbs [#permalink] 04 Nov 2018, 08:46

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

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