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.

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )