Question on Number System

Question

Find the remainder of $\frac{9^{1}+9^{2}+...+9^{n}}{6}$ where $n$ is multiple of 11.

I am getting $0$ or $3$. But given answer is 3. Can anyone check?

Comments

i also think ans could be {0,3}, because at last we are left with,

(3*kmod6)mod6....where k ϵ Z+

this equation gives both 0 and 3 on varying k..

Answer 1

(9¹ +9² +...+9ⁿ )mod 6 = (3¹+3²+...+3ⁿ)mod 6

=(3+9+9*3+.....+3ⁿ)mod 6

every term has remainder 3

so n=11

(3*11)mod 6=33 mod 6=3

Comments

@Dharmendra Lodhi....what if n=22??

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each number (9 9^2, 9^3...) divided by 6 gives remainder 3,

3*9 =27

27/6, remainder = 3

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when n is even remainder is 0

when n is odd remainder is 3