Let the set \(P= \{2,3,4,..., 25\}\). For each \(k ∈ P\), define \(Q(k)= \{x ∈ P\) such that \(x > k\) and \(k\) divides \(x\}\). Then the number of elements in the set \(P - \bigcup_{k=2}^{25}…

Nine elements remain. The union removes precisely the composite numbers, since each composite in $P$ has a proper divisor in $P$, while no prime does. The remaining primes are $2,3,5,7,11,13,17,19,23$. The count $15$ gives the number removed.