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

Since 1986 = 62*32 + 2, we have

by Euler's theorem, since . Then

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

Then

so the next digit must be 0;

so the next digit is 1;

so the next digit is 1;

so the next digit is 0;

so the next digit is 0; and

so the last digit is 1.

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

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse