[email protected] wrote:What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is divided by 6?

A. 0
B. 3
C. 2
D. 5
E. None of the above

Ans-B

First recognize that a number that's divisible by 6 must be even AND must be divisible by 3.
Also recognize that the product of any number of odd integers will always be odd. So, since 9 is odd, we know that:
9^1 is odd, and it is divisible by 3
9^2 is odd, and it is divisible by 3
9^3 is odd, and it is divisible by 3
9^4 is odd, and it is divisible by 3
.
.
.
9^9 is odd, and it is divisible by 3

So, 9^1 + 9^2 + 9^3 +...+ 9^9 = odd + odd + odd .... + odd = (9)(odd) = ODD
Also, 9^1 + 9^2 + 9^3 +...+ 9^9 is divisible by 3.

So, 9^1 + 9^2 + 9^3 +...+ 9^9 is divisible by 3 and it's odd.
So, if we subtracted 3 (an odd number) from 9^1 + 9^2 + 9^3 +...+ 9^9, we'd get an even number that's divisible by 3.
In other words, 9^1 + 9^2 + 9^3 +...+ 9^9 - 3 is divisible by 6.
So, when we divide 9^1 + 9^2 + 9^3 +...+ 9^9 by 6, we must get a remainder of 3

Answer: B

Cheers,
Brent