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:
\(2^{2015}\equiv a(mod{20}).\)You have to find 'a'.

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

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

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TopNewest2^2015 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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yes awesome!!

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Good. Nice solution using CRT. Adarsh Kumar

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@Calvin Lin ?

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@Satvik Golechha @Krishna Ar ?

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And it was this problem right?

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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.

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Answer's \(8\)

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

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