2. Find all triples of prime numbers $p, q$ and $r$, for which

$$
\frac{14}{p}+\frac{51}{q}=\frac{65}{r} .
$$
2. First, let us note that for the readers of the fractions in the given equation, the relationship $14+51=65$ holds. Therefore, any triplet of the same prime numbers $p=q=r$ is a solution, and for any solution, if any two of the numbers $p, q, r$ are the same, the third number is also the same. We will therefore further assume that the prime numbers $p, q, r$ satisfying the given equation are mutually distinct (and thus mutually coprime).

After multiplying the equation by the product $p q r$, we get

$$
14 q r + 51 p r = 65 p q,
$$

from which, due to the mentioned coprimality, it follows that

$$
p \mid 14 = 2 \cdot 7, \quad q \mid 51 = 3 \cdot 17 \quad \text{and} \quad r \mid 65 = 5 \cdot 13.
$$

This means that $p \in \{2, 7\}, q \in \{3, 17\}$, and $r \in \{5, 13\}$. Now we can compile and substitute all eight possible triplets $(p, q, r)$ into the equation; we will find that only the triplet $(7, 17, 13)$ satisfies it.

We can shorten the verification by substitution by excluding any of the values $p=2$, $q=3$, or $r=5$. For example, by substituting $r=5$, we get the equation $14 q + 51 p = 13 p q$ after dividing by five, which has no integer solution $p$ for $q=3 (14 + 17 p = 13 p)$ or for $q=17 (14 + 3 p = 13 p)$. Another possibility: from the equation $14 q r + 51 p r = 65 p q$, it follows that $2 p (q - r) = 7 (2 q r + 7 p r - 9 p q)$, so the product $p (q - r)$ is divisible by seven. However, since $q \in \{3, 17\}$ and $r \in \{5, 13\}$ (see above), the difference $q - r$ is not divisible by seven, so the number $p$ is divisible by seven. Similarly, we can justify why $17 \mid q$ and $13 \mid r$.

Another solution. From the given equation, express $r$ in terms of $p$ and $q$:

$$
r = \frac{65 p q}{51 p + 14 q} = \frac{5 \cdot 13 \cdot p \cdot q}{51 p + 14 q}
$$

In the last fraction, we have highlighted the factorization of the numerator into (four) prime factors. Such a fraction will be equal to some prime number $r$ precisely when its denominator is the product of three prime factors from the numerator (other simplifications of the fraction are not possible). We are therefore looking for situations where one of the following cases holds:

$$
\begin{array}{ll}
51 p + 14 q = 5 \cdot 13 \cdot p & \text{and} \quad r = q \\
51 p + 14 q = 5 \cdot 13 \cdot q & \text{and} \quad r = p \\
51 p + 14 q = 5 \cdot p \cdot q & \text{and} \quad r = 13 \\
51 p + 14 q = 13 \cdot p \cdot q & \text{and} \quad r = 5
\end{array}
$$

By simple manipulation of the equations, we find that the first two cases occur only in the situation where $p = q$ (in which case $p = r$ as well). The last two cases lead to

$$
q = \frac{3 \cdot 17 \cdot p}{5 p - 14}, \quad \text{resp.} \quad q = \frac{3 \cdot 17 \cdot p}{13 p - 14}
$$

from which, by analogous reasoning about the simplification of the fractions (the case $p = q$ can already be excluded) and considering the obvious inequalities $5 p - 14 < 17 p$ and $13 p - 14 < 17 p$, we get the equations

$$
5 p - 14 = 3 p, \quad \text{resp.} \quad 13 p - 14 = 3 p.
$$

The first equation has the solution $p = 7$ (to which $q = 17$ and $r = 13$ correspond), and the second equation has no integer solution.

Answer: All solutions $(p, q, r)$ are triplets $(p, p, p)$, where $p$ is any prime number, and the triplet $(7, 17, 13)$.

For a complete solution, award 6 points; 2 points for finding the solution $(p, p, p)$, 1 point for finding the solution $(7, 17, 13)$, and 1 to 3 points (depending on the completeness of the reasoning) for justifying that no other solutions exist (this will mainly involve divisibility arguments).