Let $p,q,r$ be distinct prime numbers and let
\[A=\{p^aq^br^c\mid 0\le a,b,c\le 5\} \]
Find the least $n\in\mathbb{N}$ such that for any $B\subset A$ where $|B|=n$, has elements $x$ and $y$ such that $x$ divides $y$.

[i]Ioan Tomescu[/i]
1. **Understanding the Problem:**
   We need to find the smallest \( n \in \mathbb{N} \) such that any subset \( B \subset A \) with \( |B| = n \) contains elements \( x \) and \( y \) where \( x \) divides \( y \). Here, \( A \) is defined as:
   \[
   A = \{ p^a q^b r^c \mid 0 \le a, b, c \le 5 \}
   \]
   where \( p, q, r \) are distinct prime numbers.

2. **Counting the Elements in \( A \):**
   The set \( A \) consists of all possible products of the form \( p^a q^b r^c \) where \( a, b, c \) range from 0 to 5. Therefore, the total number of elements in \( A \) is:
   \[
   |A| = (5+1)(5+1)(5+1) = 6 \times 6 \times 6 = 216
   \]

3. **Applying the de Bruijn-Tengbergen-Kruyswijk Theorem:**
   According to the de Bruijn-Tengbergen-Kruyswijk theorem, a maximal antichain in a product poset \( [n_1] \times [n_2] \times \cdots \times [n_k] \) is at the level of middle rank \( \lfloor (n_1 + n_2 + \cdots + n_k) / 2 \rfloor \) or \( \lceil (n_1 + n_2 + \cdots + n_k) / 2 \rceil \).

4. **Applying the Theorem to Our Case:**
   In our case, the product poset is \( [5] \times [5] \times [5] \). The middle rank is:
   \[
   \lfloor (5 + 5 + 5) / 2 \rfloor = \lfloor 15 / 2 \rfloor = 7
   \]
   Therefore, we need to consider the elements where the sum of the exponents \( a + b + c = 7 \).

5. **Counting the Elements with Sum of Exponents Equal to 7:**
   We need to count the number of solutions to the equation \( a + b + c = 7 \) where \( 0 \le a, b, c \le 5 \). This can be done using the stars and bars method, but we must ensure that the solutions are within the given range.

   The valid solutions are:
   \[
   (5,0,2), (5,1,1), (5,2,0), (4,0,3), (4,1,2), (4,2,1), (4,3,0), (3,0,4), (3,1,3), (3,2,2), (3,3,1), (3,4,0), (2,0,5), (2,1,4), (2,2,3), (2,3,2), (2,4,1), (2,5,0), (1,1,5), (1,2,4), (1,3,3), (1,4,2), (1,5,1), (0,2,5), (0,3,4), (0,4,3), (0,5,2)
   \]
   There are 27 such elements.

6. **Conclusion:**
   Since the maximal antichain has 27 elements, any subset \( B \) of \( A \) with \( |B| = 28 \) must contain at least one pair \( (x, y) \) such that \( x \) divides \( y \).

The final answer is \( \boxed{ 28 } \).