9.1. What is the minimum sum of digits in the decimal representation of the number $f(n)=17 n^{2}-11 n+1$, where $n$ runs through all natural numbers?

# Answer. 2.
Solution. When $n=8$, the number $f(n)$ is 1001, so the sum of its digits is 2. If $f(n)$ for some $n$ had a sum of digits equal to 1, it would have the form $100, \ldots 00$ and would either be equal to 1 or divisible by 10. The function of a real variable $f(x)$ reaches its minimum at $x=\frac{11}{34}1$, and $f(n)$ cannot take the value 1. Furthermore, it is easy to notice that $f(n)$ is always an odd number, so it cannot be divisible by 10. Therefore, the minimum sum of the digits of the number $f(n)=17 n^{2}-11 n+1$ is 2, and it is achieved when $n=8$.

Grading Criteria. Correct answer with verification for $n=8$: 2 points. Proven that $f(n)>1$ and that $f(n)$ cannot take the value 1: 2 points. Proven that $f(n)$ is always an odd number, so it cannot be divisible by 10: 3 points.