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Number Theory : Remainders of division by 6

Finding remainders when sum of powers of 9 are divided by 6

Question

What is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is divided by 6?

(1) 3
(2) 2
(3) 0
(4) 5

Correct Choice is (3) and Correct Answer is 0

Explanatory Answer

6 is an even multiple of 3. When any even multiple of 3 is divided by 6, it will leave a remainder of 0. Or in other words it is perfectly divisible by 6.

On the contrary, when any odd multiple of 3 is divided by 6, it will leave a remainder of 3. For e.g when 9 an odd multiple of 3 is divided by 6, you will get a remainder of 3.

9 is an odd multiple of 3. And all powers of 9 are odd multiples of 3.
Therefore, when each of the 8 powers of 9 listed above are divided by 6, each of them will leave a remainder of 3.

The total value of the remainder = 3 + 3 + .... + 3 (8 remainders) = 24.
24 is divisible by 6. Hence, it will leave no remainder.

Hence, the final remainder when the expression 9^1 + 9^2 + 9^3 + ..... + 9^8 is divided by 6 will be equal to '0'.

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