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Q.

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

 A.

3

 B.

2

 C.

0

 D.

5

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Solution:
Option(C) is correct

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 is divided by 6, each of them will leave a remainder of 3.

The total value of the remainder $= 3 + 3 + .... + 3 \text{ (8 remainders)} = 24$.
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'.


(1) Comment(s)


Amit Jaiswal
 ()

each power of 9 will leave 3 as a remainder when it is devided by 6.

here it will leave 3 for 8 times. so that,

$3*8=24$ which is completely divisible by 6.