A positive integer $n$ is called indivisible if from $1<k<n$ and $(k, n)=1$, it follows that $k$ is a prime number.

How many indivisible numbers greater than 2 are there?
Solution. If there exists a prime number $p$ such that $p \nmid n$ and $p < \sqrt{n}$.

$n=3$ and $n=4$ are clearly indivisible.

If $n \geq 5$, then - denoting the increasing sequence of prime numbers by $p_{1}, p_{2}, \ldots$ - the primes less than $\sqrt{n}$ are $p_{1}, p_{2}, \ldots, p_{i}$. If $n$ is an indivisible number, then according to the above, $n=a \cdot p_{1} \cdot p_{2} \cdot \ldots \cdot p_{i}\left(a \in \mathbb{N}^{+}\right)$ and clearly $n \leq p_{i+1}^{2}$, so

$$
p_{1} \cdot p_{2} \cdot \ldots \cdot p_{i} \leq p_{i+1}^{2}
$$

According to Chebyshev's theorem, $p_{i+1} < 2p_{i}$, and for $n > 8$ we have $i \leq 4$. Since $i \geq 1$, there are four cases:
a) $i=1$, then $5 \leq n \leq 9$ (because $\sqrt{n} \leq 3$; and we only examine the new possible $n$'s) and $2 \mid n$ is necessary. The suitable numbers are 6 and 8.
b) $i=2$, then $10 \leq n \leq 25$ and $2 \cdot 3=6 \mid n$ is necessary. The possible numbers are: 12, 18, and 24.
c) $i=3$, then $26 \leq n \leq 49$ and $2 \cdot 3 \cdot 5=30 \mid n$ is necessary, so only $n=30$ fits.
d) $i=4$, then $50 \leq n \leq 121$ and $2 \cdot 3 \cdot 5 \cdot 7=210 \mid n$ is necessary, which is clearly impossible.

According to the above, all the listed numbers are indivisible, so there are 8 indivisible numbers greater than 2: $3,4,6,8,12,18,24,30$.