wizardofmaths
Dear GMAT Tutor
I have been learning remainder theroem for quite some time.I am confused in a question of cyclicity.
The question is: find the remainder of 2 raised to the power 8 divided by 96
1)now confusing part is that if I am to find the cyclicity of 2 when divided by 12(I reduced 2 raised to the power 8 divided by 96
to 2 raised to the power 5 divided by 12 i.e, I reduced it by the factor of 8 which i will multiply by the final answer later)
then here it is
2, 4, 8, 4, 8, 4, 8.......
so how am i supposed to count the cyclicity from 2 or from 8 if from 2 then what will be the answer or start from 8 then what will be the answer?
PLEASE DON'T USE ANY OTHER METHOD TO SOLVE IT. I request you to solve the question with cyclicity method only and keep 12 in the denominator.
2) what will we do when after dividing cyclicity to the power we get 0 remainder?
1) The cyclicity of \(2^n\) when divided by \(12\) is described as followings.
\(2^1 = 2\) : \(2\)
\(2^2 = 4\) : \(4\)
\(2^3 = 8\) : \(8\)
\(2^4 = 16\) : \(4\)
\(2^5 = 32\) => \(2 * 4 = 8\) : \(8\)
\(2^6\) : \(8 * 2 = 16\) : \(4\)
...
The remainders are \(2, 4, 8, 4, 8, ...\) as you mentioned.
The remainder of \(2^5\) when it is divided by \(12\) is \(8\).
Therefore, you should count from \(2\).
2) We will have \(0\) remainders afterwards,
The following is an example.
Consider \(2^n\)'s remainders when they are divided by \(8\).
\(2^1 = 2\) : \(2\)
\(2^2 = 4\) : \(4\)
\(2^3 = 8\) : \(0\)
\(2^4 = 16\) : \(0 * 2 = 0\)
\(2^5 = 32\) : \(0 * 2 = 0\)
...