Find the remainder when \(1 + 2 + 2^2 + 2^3 + \dots + 2^{100}\) is divided by 7.
1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + ...... + 2^100 =
( 1 + 2 + 2^2) + 2^3 ( 1 + 2 + 2^2) + 2^6 ( 1 + 2 + 2^2) + ........ + 2^96 + 2^97 + 2^98+ 2^99 + 2^100 =
( 7) + 8(7) + 64 ( 7) + ...... + 2^96 ( 1 + 2 + 2^2) + 2^99 + 2^100
7 + 8(7) + 64 (7) +.........+ 2^96 ( 7) + 2^99 + 2^100
Every factorization is divisible by 7
Looking at the last two terms
Note that 2^99 + 2^100 = 2^99 ( 1 +2) = 3 * 2^99 = 3 * (2^3)^33
And
2^3 mod 7 =1
So
(2^3)^33 mod 7 = 1
So
3 * (2^3)^33 mod 7 =
3 (1) = 3 = the remainder