HELP!!!

Yesterday I encountered this problem while I was solving a problem.But don't worry the problem has been solved.Could somebody give a solution to this: 22015a(mod20).2^{2015}\equiv a(mod{20}).You have to find 'a'.

Note by Adarsh Kumar
5 years ago

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@Satvik Golechha @Krishna Ar ?

Adarsh Kumar - 5 years ago

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Answer's 88

Krishna Ar - 5 years ago

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Yes,I actually know the answer but could you please provide a step-by-step solution?

Adarsh Kumar - 5 years ago

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@Adarsh Kumar Are you joking? (Really, coz u've solved my problem...)..ANyway...ϕ(20)=8. \phi(20)=8. thus, we have that 22015(mod20)=27(mod20)2^{2015} \pmod {20} = 2^7 \pmod {20}. Which is nothing but 8!!

Krishna Ar - 5 years ago

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@Krishna Ar I think you made a slight mistake as you have applied Euler's Totient theorem which can only be applied when the GCD(a,n)=1.

Adarsh Kumar - 5 years ago

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@Adarsh Kumar Yes, but it still provides the answer! ^_^. I can't think of any other method.

Krishna Ar - 5 years ago

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@Krishna Ar well,thanx for your help but if you can think of any other method please post!!

Adarsh Kumar - 5 years ago

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@Adarsh Kumar Sure

Krishna Ar - 5 years ago

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@Krishna Ar and I think that results are important but the method is more important.

Adarsh Kumar - 5 years ago

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@Krishna Ar You should always have CHINESE REMAINDER THEOREM in mind while solving this type of problems

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And it was this problem right?

Krishna Ar - 5 years ago

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Yes,it was this problem but I figured out that the answer would not be 10 so I solved that problem but, I do not know how to find the remainder.

Adarsh Kumar - 5 years ago

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@Calvin Lin ?

Adarsh Kumar - 5 years ago

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2^2015 is a multiple of 4 And 2^2015 leaves a remainder of 3 modulo 5 ( Fermat's little theorem) so, on dividing by 20, I.e.4*5, remainder is 8 as krishna said

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Good. Nice solution using CRT. Adarsh Kumar

Krishna Ar - 5 years ago

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yes awesome!!

Adarsh Kumar - 5 years ago

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