can anyone tell me what is trick behind this problem?
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No
I wouldn't call it trick though. Sum of divisors of number n = p1α1p2α2...pkαk is S = (1 + p1 + p12...p1α1)(1 + p2 + p22...p2α2)...(1 + pk + pk2...pkαk). 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 pα can be "K-number" where p is prime and α is non-negative integer. I think it is enough to solve the problem.