The remainder when 445+444+443+442+441+...........+40 is divided by 17 is:-

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The remainder when

4^{45} + 4^{44} + 4^{43} + 4^{42} + 4^{41} + . . . . . . . . . . . + 4^{0}

is divided by 17 is:-

Options:

A . 5

B . 4

C . 1

D . None of these

Answer: Option A
:
A

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{{(4^{4})}^{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.

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