Question

The remainder when $3^0+3^1+3^2+\cdots +3^{200}$ is divided by 13 is:

• A. 12
• B. 3
• C. none of these

Answer 1

$S_n=3^0+3^1+3^2+\cdots +3^{200}{S}_n=\frac{3^0(3^{201}-1)}{(3-1)}=\frac{3^{201}-1}{2}Now,\frac{3^{201}-1}{2}=\frac{(3^3{)}^{67}-1^3}{2}=\frac{\left[\right(27{)}^{67}-1]}{2}$
Now since $a^n-b^n$ is divisible by (a - b) therefore $\left[\right(27{)}^{67}-1]$ is divisible by (27 - 1) = 26.
Hence it is also divisible by 13.
Thus the remainder is zero since the given expression is divisible by 13.