Task 2. Find all prime numbers $p$ and natural numbers $n$ for which $\frac{1}{p}=\frac{n}{2010}$?
Solution. First method. From the condition of the problem, we have that $n \cdot p=2010$, that is, $n \cdot p=2 \cdot 3 \cdot 5 \cdot 67$. From here, the value of the number $p$ can be 2, $3,5,67$. So, finally, for $p=2, n=1005$, for $p=3, n=670$, for $p=5$, $n=402$, for $p=67, n=30$.

Second method. The number 2010 can be written in the form

$$
2010=2 \cdot 3 \cdot 5 \cdot 67
$$

so according to this, $p$ can be $2,3,5,67$.

a) for $p=2$ we have $\frac{1}{2}=\frac{n}{2010}$, that is, $n=1005$,

b) for $p=3$ we have $\frac{1}{3}=\frac{n}{2010}$, that is, $n=670$,
c) for $p=5$ we have $\frac{1}{5}=\frac{n}{2010}$, that is, $n=402$,

d) for $p=67$ we have $\frac{1}{67}=\frac{n}{2010}$, that is, $n=30$.