No

I wouldn't call it trick though. Sum of divisors of number n = p₁^α₁p₂^α₂...p_k^α_k is S = (1 + p₁ + p₁²...p₁^α₁)(1 + p₂ + p₂²...p₂^α₂)...(1 + p_k + p_k²...p_k^α_k). If you want proof, consider small k's for getting logic or even you can induct on k. If k > 1, there are more than one  > 1 multiplications in S, so it wouldn't be prime number. Only case you need to check is k = 1 so only the numbers p^α can be "K-number" where p is prime and α is non-negative integer. I think it is enough to solve the problem.