What is the remainder when X^Y is divided by Z , Where X = 32, Y = (32

rvgmat12

\(32^{32^{32}}\) can be written as \((26+6)^{32*32^{31}}\)

As we are looking at only the remainder, so in the expansion all terms except last will be divisible by 13

Last term = \((6)^{2*16*32^{31}}\)

\((36^{16*32^{31}}=(39-3)^{16*32^{31}}\)

Again the last term = \((-3)^{16*32^{31}}=(3)^{16*32^{31}}=3^{2^4*2^{5*31}}=3^{2^{159}}\)

Now \(3^{2^1}=9\) and will leave a remainder of 9 when divided by 13.
\(3^{2^2}=3^4=81=13*6+3\) and will leave a remainder of 3 when divided by 13.
Next, \(3^{2^3}=3^8=(3^4)^2\) and as \(3^4\) leaves a remainder 3, \((3^4)^2\) will leave a remainder of \((3)^2\) when divided by 13.

Thus the pattern is 9,3,9,3….
\(3^{2^{odd}}\) will leave 9 and \(3^{2^{even}}\) will leave 3 as the remainder.

Thus 9 is the answer.

Options are wrong.