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What is the remainder when 52^60 is divided by 31

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What is the remainder when 52^60 is divided by 31  [#permalink]

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New post 24 Feb 2020, 04:30
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Re: What is the remainder when 52^60 is divided by 31  [#permalink]

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New post 24 Feb 2020, 06:35
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Bunuel wrote:
What is the remainder when \(52^{60}\) is divided by 31 ?

A. 0
B. 1
C. 21
D. 23
E. 30

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Working out a pattern will be cumbersome, so let us do by BINOMIAL theorem..

\(52^{60}=(62-10)^{60}\), so remainder will be \((-10)^{60}=100^{30}=(93+7)^{30}\)..
Remainder when \((93+7)^{30}\) is divided by 31 will be \(7^{30}=(7^3)^{10}=343^{10}=(341+2)^{10}=(31*11+2)^{10}\)
Remainder when \((31*11+2)^{10}\) is divided by 31 will be\( 2^{10}=(2^5)^2=32^2=(31+1)^2\)
Thus the remainder = \(1^2\) or 1

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What is the remainder when 52^60 is divided by 31  [#permalink]

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New post 29 Mar 2020, 02:11
2
Bunuel wrote:
What is the remainder when \(52^{60}\) is divided by 31 ?

A. 0
B. 1
C. 21
D. 23
E. 30

Are You Up For the Challenge: 700 Level Questions


52 = 62 - 10

52^60mod31 = (-10)^60mod31 = 10^60mod31 = 7^30mod31 = 2^10mod31 = 1^2mod31 = 1

IMO B
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Re: What is the remainder when 52^60 is divided by 31  [#permalink]

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New post 29 Mar 2020, 05:07
Quote:
52 = 62 - 10

52^60mod31 = (-10)^60mod31 = 10^60mod31 = 7^30mod31 = 2^10mod31 = 1^2mod31 = 1

IMO B


Kinshook Could you please explain what you've exactly done in this part of your solution "10^60mod31 = 7^30mod31 = 2^10mod31 = 1^2mod31"? :please

I understand this part "52^60mod31 = (-10)^60mod31 = 10^60mod31" but then I would have proceded the following way: 2^60*5^60mod31 = 32^12*125^20mod31 = 1^12*1^20mod31 = 1. However, it took me too much time to come up with my solution, it seems to me that your solution is much faster and that you know some tricks in modular arithmetic that I don't know yet... :lol:
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Re: What is the remainder when 52^60 is divided by 31  [#permalink]

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New post 29 Mar 2020, 05:15
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Vendap02 wrote:
Quote:
52 = 62 - 10

52^60mod31 = (-10)^60mod31 = 10^60mod31 = 7^30mod31 = 2^10mod31 = 1^2mod31 = 1

IMO B


Kinshook Could you please explain what you've exactly done in this part of your solution "10^60mod31 = 7^30mod31 = 2^10mod31 = 1^2mod31"? :please

I understand this part "52^60mod31 = (-10)^60mod31 = 10^60mod31" but then I would have proceded the following way: 2^60*5^60mod31 = 32^12*125^20mod31 = 1^12*1^20mod31 = 1. However, it took me too much time to come up with my solution, it seems to me that your solution is much faster and that you know some tricks in modular arithmetic that I don't know yet... :lol:


Hi Vendap02
10^60mod31 = 100^30mod31 = (93+7)^30mod31 = 7^30mod31 = = 343^10mod31 = (341+2)^10mod31 = 2^10mod31 = (31+1)^2mod31 = 1^2mod31 = 1
Hope it is clear now
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Re: What is the remainder when 52^60 is divided by 31  [#permalink]

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New post 29 Mar 2020, 05:21
Great, thank you Kinshook! :)
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Re: What is the remainder when 52^60 is divided by 31   [#permalink] 29 Mar 2020, 05:21
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