## Problem 1

Determine the smallest sum of the digits of the number $3 n^{2}+n+1$, where $n \in \mathbb{N}, n \geq 2$.
## Solution (Prof. Cristian Mangra, Bucharest):

For $n=8$, we obtain the number 201 which has the sum of digits 3.

We will show that numbers of the form $3 n^{2}+n+1$ cannot have the sum of digits equal to 1 or 2. Since numbers of the form $3 n^{2}+n+1$ are odd, they cannot be written as $10^{k}$ or $2 \cdot 10^{k}$, where $k \in \mathbb{N}^{*}$.

If $3 n^{2}+n+1=10^{k}+1$, then $n(3 n+1)=10^{k}$. Since $(n, 3 n+1)=1$, we obtain the cases:

- $n=5^{k}, 3 n+1=2^{k}$, clearly impossible.
- $n=2^{k}, 3 n+1=5^{k}$, impossible, since, for $k=1$ we get $7=5$, and, for $k \geq 2$ we have $5^{k}>4^{k}>3 \cdot 2^{k}+1$.