×

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

2 years, 3 months ago

Sort by:

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 · 2 years, 3 months ago

yes awesome!! · 2 years, 3 months ago

Good. Nice solution using CRT. Adarsh Kumar · 2 years, 3 months ago

@Calvin Lin ? · 2 years, 3 months ago

@Satvik Golechha @Krishna Ar ? · 2 years, 3 months ago

And it was this problem right? · 2 years, 3 months ago

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. · 2 years, 3 months ago

Answer's $$8$$ · 2 years, 3 months ago

Yes,I actually know the answer but could you please provide a step-by-step solution? · 2 years, 3 months ago

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 years, 3 months ago

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. · 2 years, 3 months ago

Yes, but it still provides the answer! ^_^. I can't think of any other method. · 2 years, 3 months ago

You should always have CHINESE REMAINDER THEOREM in mind while solving this type of problems · 2 years, 3 months ago

well,thanx for your help but if you can think of any other method please post!! · 2 years, 3 months ago

Sure · 2 years, 3 months ago