# Problem 1. (2 points)

Let $p_{1}, p_{2}, \ldots, p_{97}$ be prime numbers (not necessarily distinct). What is the largest integer value that the expression

$$
\sum_{i=1}^{97} \frac{p_{i}}{p_{i}^{2}+1}=\frac{p_{1}}{p_{1}^{2}+1}+\frac{p_{2}}{p_{2}^{2}+1}+\ldots+\frac{p_{97}}{p_{97}^{2}+1}
$$

can take?
# Answer: 38

Solution: Note that the function $\frac{x}{x^{2}+1}$ decreases when $x \geqslant 2$, which means $\sum_{i=1}^{97} \frac{p_{i}}{p_{i}^{2}+1} \leqslant 97 \cdot \frac{2}{2^{2}+1}=$ $97 \cdot 0.4=38.8$. Therefore, the answer cannot be greater than 38.

At the same time, $\frac{3}{3^{2}+1}=0.3$, which is 0.1 less than the value for $p=2$. Therefore, if in the above maximum sum, eight twos are replaced by threes, the result will be exactly 38.