The sum $S_{n}=9+19+29+39+\cdots+a_{n}$ denotes the sum of the first $n$ natural numbers ending in 9. What is the smallest value of $n$ for which $S_{n}$ is greater than $10^{5}$?
Note that $S_{n}$ is the sum of the first $n$ terms of an arithmetic progression whose first term is $a_{1}=9$ and the common difference is $r=10$. Substituting these data into the formula $a_{n}=a_{1}+(n-1) r$ we get $a_{n}=9+10(n-1)$. On the other hand, note that:

$$
\begin{aligned}
& 9=9+0 \cdot 10 \\
& 19=9+1 \cdot 10 \\
& 29=9+2 \cdot 10 \\
& 39=9+3 \cdot 10 \\
& \ldots \\
& a_{n}=9+(n-1) \cdot 10
\end{aligned}
$$

Then we have:

$$
\begin{aligned}
S_{n} & =9+19+29+\cdots+a_{n} \\
& =(9+0)+(9+10)+(9+2 \cdot 10)+\cdots+[9+(n-1) \cdot 10] \\
& =9 n+[1 \cdot 10+2 \cdot 10+3 \cdot 10+\cdots+(n-1) \cdot 10] \\
& =9 n+[1+2+3+\cdots+(n-1)] \cdot 10 \\
& =9 n+\frac{n(n-1)}{2} \cdot 10 \\
& =9 n+5 \cdot n(n-1) \\
& =5 n^{2}+4 n
\end{aligned}
$$

Since we want $S_{n} \geq 10^{5}$, we need to find the smallest positive integer $n$ such that $5 n^{2}+4 n \geq 10^{5}$, or equivalently, $5 n^{2}+4 n-10^{5} \geq 0$. Solving the equation $5 x^{2}+4 x-10^{5}=0$ we get

$$
x=\frac{-4 \pm \sqrt{16+20 \cdot 10^{5}}}{10}
$$

The positive root is $x_{1}=\frac{-4+\sqrt{16+20 \cdot 10^{5}}}{10} \simeq 141.02$.

But, $5 x^{2}+4 x-10^{5}$ is positive outside the roots. Since we are looking for the smallest positive integer such that $5 x^{2}+4 x-10^{5} \geq 0$, then $n=142$.