Bulgarian Spring Tournament

For each prime $p$, the numbers of the interval $n\in [p^2,p^3)$ are not solutions, because the degree of $p$ in the decomposition of the LCM is 2 and so contributes by a factor of 3 to the number of divisors. Therefore this number is not a power of 2. From Bertrand's postulate for a prime $p$ there is a prime $q$ with $p<q<2p$, respectively $p^2<q^2<4p^2<p^3<q^3$ for $p\ge 5$; also $3^2<5^2<3^3<5^3$. Therefore, each interval $[p^2,p^3)$ and $[q^2,q^3)$ for consecutive primes $3\le p<q$ intersects. We obtained that $n<9$ and a direct check shows that only $n=1,2,3$ and $8$ are solutions. $\square$