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:
22015≡a(mod20).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...ϕ(20)=8. thus, we have that 22015(mod20)=27(mod20). 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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@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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ϕ(20)=8. thus, we have that 22015(mod20)=27(mod20). 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.
@Calvin Lin ?
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
yes awesome!!