What is the remainder when 3^1+3^2+3^3 ... 3^2009 is divided by 8.

4

Step-by-step explanation:

To find:

The remainder, when 3^0 + 3^1 + 3^2 + ... + 3^2009 is divided by 8

Approach and Working:

Rem (3^0/8) = 1

Rem (3^1/8) = 3

Rem (3^2/8) = 1

Rem (3^3/8) = 3

Rem (3^4/8) = 1

Rem (3^5/8) = 3 and so on ...

We can observe that,

For even powers of 3, the remainder is 1

And for odd powers of 3, the remainder is 3.

Therefore, we can rewrite the given series, in terms of its remainders, as

= 1 + 3 + 1 + 3 + 1 + 3 + ... + 3 = 1005 pairs of (1 + 3) = 1005 x 4 = 4020

Rem (4020/8) = 4