Find the remainder when 4001000 is divided by 17?

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Question

Find the remainder when $400^{1000}$ is divided by 17?

Options:

A . 0

B . 1

C . -1

D . 2

Answer: Option B
:
B

This needs the application of basic remainder, fermat theorem and frequency
Step 1 - Remainder when 400 is divided by 17 is 9
The problem changes to $\frac{9^{1000}}{17}$
From fermat theorem, the Euler's number of 17 is 16.
1000=16k +8
Problem now changes to $\frac{9^{8}}{17}$
Going by frequency method, $\frac{9^{4}}{17}$ gives remainder -1 so $\frac{9^{8}}{17}$ will give remainder 1.
Altenative solution :
We know tha $9^{2} = 81 => (81)^{500} => (- 4)^{500} => 16^{250} => (- 1)^{250} = 1$

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