Question

The remainder when $4^{45}+4^{44}+4^{43}+4^{42}+4^{41}+...........+4^0$ is divided by 17 is:-

• A. 5
• B. 4
• C. 1
• D. None of these

Answer 1

The correct option is A

5

$S_n=4^0+4^1+4^2+4^3+...................+4^{43}+4^{44}+4^{45}$

$S_n=(1+4+16+64)+4^4(1+4+16+64)+..............+4^{40}(1+4+16+64)+4^{44}+4^{45}$

All the terms except $4^{44}+4^{45}$ are divisible by 17 (as 1 + 4 + 16 + 64 = 85 which is a multiple of 17).

Now, $\frac{4^{45}+4^{44}}{17}=\frac{4^{44}(4+1)}{17}=\frac{{\left(4^4\right)}^{11}\times 5}{17}=\frac{{(255+1)}^{11}\times 5}{17}=\frac{1^{11}\times 5}{17}=\frac{5}{17}.$

Hence the required remainder is 5.