Find all positive integers $n=p_{1} p_{2} \cdots p_{k}$ which divide $\left(p_{1}+1\right)\left(p_{2}+1\right) \cdots\left(p_{k}+1\right)$, where $p_{1} p_{2} \cdots p_{k}$ is the factorization of $n$ into prime factors (not necessarily distinct).

Answer: All numbers $2^{r} 3^{s}$ where $r$ and $s$ are non-negative integers and $s \leq r \leq 2 s$.
Let $m=\left(p_{1}+1\right)\left(p_{2}+1\right) \cdots\left(p_{k}+1\right)$. We may assume that $p_{k}$ is the largest prime factor. If $p_{k}>3$ then $p_{k}$ cannot divide $m$, because if $p_{k}$ divides $m$ it is a prime factor of $p_{i}+1$ for some $i$, but if $p_{i}=2$ then $p_{i}+1<p_{k}$, and otherwise $p_{i}+1$ is an even number with factors 2 and $\frac{1}{2}\left(p_{i}+1\right)$ which are both strictly smaller than $p_{k}$. Thus the only primes that can divide $n$ are 2 and 3 , so we can write $n=2^{r} 3^{s}$. Then $m=3^{r} 4^{s}=2^{2 s} 3^{r}$ which is divisible by $n$ if and only if $s \leq r \leq 2 s$.