Determine the smallest integer $k \geqslant 2$ satisfying the following property. For any partition of the set $\{2,3, \ldots, k\}$ into two parts, at least one of the two parts contains three numbers $a, b, c$, not necessarily distinct, such that $a b=c$.

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(solved by Auguste Ramondou)

Let $\{A, B\}$ be the partition of $\llbracket 2, k \rrbracket$. We call $A$ the part that contains 2.

We try to construct these parts in such a way that none of them contains two numbers and their product:

- Since $2 \in A$ and $2 \times 2 = 4$, 4 must be in $B$ (if $k \geq 4$).
- Since $4 \in B$, similarly, $16 \in A$ (if $k \geq 16$).
- Then $A$ contains 2 and 16, so $8 = \frac{16}{2} \in B$.

But $2 \times 16 = 4 \times 8 = 32$, so if $k \geq 32$, no matter which part 32 is placed in, it will be the product of two numbers in the part.

Thus, the smallest integer satisfying the desired property is necessarily less than or equal to 32. Furthermore, if we take $k \leq 32$, we observe that the partition which places in $A$ all prime numbers between 2 and $k$, plus 16 and 24, and in $B$ all the rest, ensures that neither part contains the product of two of its elements.

The minimum sought is therefore exactly 32.