SELECTION and TRAINING SESSION

We call the number $m+k$ from the problem condition convenient if it has at least $n$ distinct prime divisors, otherwise we call the number $m+k$ inconvenient. It is easy to see that we may consider inconvenient numbers only. Take one of them, $m+k=q_1^{a_1}{q}_2^{a_2}\dots q_l^{a_l}$, where all $q_i$ are primes, $l\le n-1$. Since $$q_1^{a_1}{q}_2^{a_2}\dots q_l^{a_l}=m+k>n^{n-1}\ge n^l,$$ there exists an $i$ with $q_i^{a_i}>n$, so we can choose $p_k=q_i$. In a similar way we can choose prime divisors for other inconvenient numbers. It remains to note that for different inconvenient numbers the chosen prime divisors are different.