april's answer

The number of times t that the key was turned in the q^th cell is equal to the number of divisors of q. Thus if q=p₁^a₁.p₂^a₂...p_k^a_k where the p_i's are distinct primes, then t=(a_i+1)(a₂+1)...(a_k+1). Now if any a_i is odd, t is even and the corresponding cell eventually remained locked. If all the a_i are even, q is a square number, t is odd and the lucky occupants of the "square" cell found that their cells eventually remained open.