A Pre-RMO question! -9

Beginning with 9^1, the last digit of the powers of 9 alternate in a pattern of 9 and 1. Thus the sum of the first even number of powers of nine always ends in 0, thus the sum of the first 100 positive powers of 9 ends in 0. The beginning of the problem has us add 1, giving us n=1. Since the problem asks for n+9, the answer is 10.

$T_{s}=1+9+9^2+9^3+\ldots+9^{100}$

This is the sum of a geometric progression with a common ratio of $9$ and the first term as $1$ . Thus, the sum simplifies to be

$T_{s}=\dfrac{9^{101}-1}{8}$

Now, let $\mu(n)$ denote the unit digit function. Note that $\mu(9^n)=9$ , if $n$ is odd.

Therefore,

$\mu(T_{s})=\mu \left(\dfrac{9^{101}-1}{8} \right)$

$n = \dfrac{\mu(9^{101})-\mu(1)}{\mu(8)}$

$n = \dfrac{9-1}{8} = \dfrac{8}{8} = 1$

Hence, $n+9=\boxed{10}$