HMMT November

If $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$, we must have $\left(n \alpha_{1}+1\right)\left(n \alpha_{2}+1\right) \ldots\left(n \alpha_{k}+1\right)=861=3 \cdot 7 \cdot 41$. If $k=1$, we have $n \mid 860$, and the only prime powers dividing 860 are $2,2^{2}, 5$, and 43 , which are not solutions. Note that if $n \alpha_{i}+1=3$ or $n \alpha_{i}+1=7$ for some $i$, then $n$ is either $1,2,3$, or 6 , which are not solutions. Therefore, we must have $n \alpha_{i}+1=3 \cdot 7$ for some $i$. The only divisor of 20 that is divisible by $p_{i}^{n / 20}$ for some prime $p_{i}$ is 20 , and it is indeed the solution.