# Moderate Number System Solved QuestionAptitude Discussion

 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

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.