Divisibility time

Let n n is an odd number and is a multiple of 11. Find the remainder when 9 + 9 2 + + 9 n 9+9^2+\cdots + 9^n is divided by 6.


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The answer is 3.

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2 solutions

Chew-Seong Cheong
Jun 11, 2016

We note that:

9 3 ( m o d 6 ) 9 2 3 × 3 9 3 ( m o d 6 ) 9 3 3 × 3 9 3 ( m o d 6 ) . . . . . . 9 k 3 ( m o d 6 ) where k = 1 , 2 , 3... \begin{aligned} 9 & \equiv 3 \pmod{6} \\ 9^2 & \equiv 3 \times 3 \equiv 9 \equiv 3 \pmod{6} \\ 9^3 & \equiv 3 \times 3 \equiv 9 \equiv 3 \pmod{6} \\ ... & \quad ... \\ \implies 9^k & \equiv 3 \pmod{6} \quad \text{where } k=1,2,3... \end{aligned}

k = 1 n 9 k { 3 ( m o d 6 ) if n is odd. 0 ( m o d 6 ) if n is even. \begin{aligned} \implies \sum_{k=1}^n 9^k & \equiv \begin{cases} 3 \pmod{6} & \text{if } n \text{ is odd.} \\ 0 \pmod{6} & \text{if } n \text{ is even.} \end{cases} \end{aligned}

Since n n is odd, the remainder is 3 \boxed{3} .

The sum is divisible by 3 but not divisible by 2. Hence, it will leave a remainder of 3 when divided by 6.

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