Question

Ind the last 7 digits in the binary expansion of $27^{1986}$. (give your answer in binary.)

Answer 1

Since 1986 = 62*32 + 2, we have

$27^{1986}\equiv\left(27^{62}\right)^{32}\cdot27^2\equiv27^2\pmod{2^6}$

by Euler's theorem, since $\varphi(64)=32$. Then

$27^{1986}\equiv27^2\equiv729\equiv25\pmod{64}$

which tells us the 7th digit from the right is 1.

Then

$25\equiv25\pmod{2^5}$

so the next digit must be 0;

$25\equiv9\pmod{2^4}$

so the next digit is 1;

$9\equiv1\pmod{2^3}$

so the next digit is 1;

$1\equiv1\pmod{2^2}$

so the next digit is 0;

$1\equiv1\pmod2$

so the next digit is 0; and

$1\equiv0\pmod1$

so the last digit is 1.

So the last 7 digits of this number in binary are 1011001.