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'.

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@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!!

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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## Comments

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TopNewest@Satvik Golechha @Krishna Ar ?

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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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$\phi(20)=8.$ thus, we have that $2^{2015} \pmod {20} = 2^7 \pmod {20}$. Which is nothing but 8!!

Are you joking? (Really, coz u've solved my problem...)..ANyway...Log in to reply

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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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@Calvin Lin ?

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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

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yes awesome!!

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