9,10,11
Induction (other).
]

Let $d(N)$ denote the number of divisors of $N$ (the numbers 1 and $N$ are also considered divisors). Find all such $N$ for which the number $P=\frac{N}{d(N)}$ is prime.
Let $N=p_{1}^{k_{1}} \ldots p_{s}^{k_{s}}$, be the prime factorization of $N$. Without loss of generality, we can assume that $P=p_{1}$. Then

$d(N)=\left(k_{1}+1\right)\left(k_{2}+1\right) \ldots\left(k_{\mathrm{s}}+1\right)$ and $\frac{N}{d(N)}=\frac{p_{1}^{k_{1}}}{k_{1}+1} \cdot \ldots \cdot \frac{p_{s}^{k_{s}}}{k_{s}+1}=p_{1}$. It is not difficult to verify that

if $p>3, k \geq 2$, then $\frac{p^{k}}{k+1}>p$, and if $p=3, k=2$, then $\frac{p^{k}}{k+1}=p$;

if $p=3, k>2$, then $\frac{p^{k}}{k+1}>p$;

if $p \geq 2$, then $\frac{p^{k}}{k+1} \geq 1$, and if $p>2$ or $k>1$, the inequality is strict;

if $p=2, k \geq 4$, then $\frac{p^{k}}{k+1}>p$.

Consider two cases:

1) $N$ is divisible by $p_{1}^{2}$, that is, $k_{1} \geq 2$. By the inequalities given, $\frac{p_{1}^{k_{1}}}{k_{1}+1} \cdot \ldots \cdot \frac{p_{s}^{k_{s}}}{k_{s}+1}$ is not greater than $p_{1}$

only when $p_{1}=3, k_{1}=2$ or $p_{1}=2, k_{1}=2$ or 3.

In the first case, the product already has a factor equal to $p_{1}$, so all other factors must be equal to one. Therefore, either $N=9$ or $N=2 \cdot 9=18$.

In the second case, if $k_{1}=3$, then $\frac{p_{1}^{k_{1}}}{k_{1}+1}=2$, and thus $N=8$. If $k_{1}=2$, then $\frac{p_{1}^{k_{1}}}{k_{1}+1}=4 / 3$. A three appears in the denominator. Therefore, $N$ must be divisible by 3. The number $N=3 \cdot 2^{2}$ works, and since $\frac{3^{k}}{k+1}>1.5$ for $k>1$, the number 3 cannot appear in a higher power in the factorization of $N$.

Thus, in this case, the numbers $N=8,9,12$, and 18 work.
2) $k_{1}=1$. Then the first factor is $p_{1 / 2}$. Therefore, the product of the remaining factors is 2. If the number $N$ has a prime divisor not less than 5, then the corresponding factor is greater than 2, which is impossible. Therefore, $N=p \cdot 2^{s^{3} r}$ and $r \leq 1$ (otherwise $\frac{3^{r}}{r+1}>2$) and $s \leq 3$ (otherwise $\frac{2^{*}}{s+1}>2$). By checking all these cases, we find that the numbers $N=24,8 p, 12 p, 18 p$ work.

## Answer

$8,9,12,18,24,8 p, 12 p, 18 p$, where $p$ is any prime number greater than 3.