Answer:
4
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