Question

Find the remainder when 1 + 2 + 2^2 + 2^3 + ... + 2^100 is divided by 7.

Thanks in advance!

Answer 1

3.

Explanation:

This is a geometric series so the sum is:

a1 * r^n - 1 / (r - 1)

= 1 * (2^101 -1) / (2-1)

= 2^101 - 1.

Find the remainder when 2^101 is divided by 7:

Note that 101 = 14*7 + 3 so

2^101 = 2^(7*14 + 3) = 2^3 * (2^14)^7 = 8 * (2^14)^7.

By Fermat's Little Theorem (2^14) ^ 7 = 2^14 mod 7 = 4^7 mod 7.

So 2^101 mod 7 = (8 * 4^7) mod 7

= (8 * 4) mod 7

= 32 mod 7

= 4 = the remainder when 2^101 is divided by 7.

So the remainder when 2^101- 1 is divided by 7 is 4 - 1 = 3..