Exercise 14. Determine the integers $m \geqslant 2, n \geqslant 2$ and $k \geqslant 3$ having the following property: $m$ and $n$ each have $k$ positive divisors and, if we denote $d_{1}<\ldots<d_{k}$ the positive divisors of $m$ (with $d_{1}=1$ and $d_{k}=m$) and $d_{1}^{\prime}<\ldots<d_{k}^{\prime}$ the positive divisors of $n$ (with $d_{1}^{\prime}=1$ and $d_{k}^{\prime}=n$), then $d_{i}^{\prime}=d_{i}+1$ for every integer $i$ such that $2 \leqslant i \leqslant k-1$.
Solution alternative $n^{\circ} 1$ In the same way as in the previous solution, we obtain that $d_{2}=2$. If $m$ is divisible by an odd number $l$, this number is a strict divisor since $m$ is divisible by 2 and thus is even. In particular, $l+1$ is even and divides $n$, so 2 divides $n$, which contradicts the fact that $d_{2}+1$ is the smallest divisor of $n$. In particular, $m$ must necessarily be a power of 2. If $m \geqslant 16$, then 8 and 4 are strict divisors of $m$. Thus, 9 and 5 divide $n$, so 45 divides $n$. In particular, 15 is a strict factor of $n$, so $15-1=14$ is a strict factor of $m$. Thus $m$ is divisible by 7, which contradicts the fact that $m$ is a power of 2.

Therefore, $m$ is a power of 2 satisfying $m<16$ and 2 is a strict factor of $m$. We thus have $m=4$ or $m=8$. If $m=4$, the only strict divisor of 4 is 2, so the only strict divisor of $n$ is 3, so $n=9$. Since the only strict divisor of 4 is 2 and the only strict divisor of 9 is 3, the triplet $(4,9,3)$ works.

If $m=8$, $m$ has strict divisors 2 and 4, $n$ has strict divisors 3 and 5, so $n=15$. Conversely, the only strict divisors of 8 are 2 and 4, and those of 15 are 3 and 5, so $(8,15,4)$ satisfies the statement.

The solution triplets are therefore $(4,9,3)$ and $(8,15,4)$.

Graders' Comment This number theory problem was fairly well solved, with nearly half of the students submitting a paper that scored 6 points or more. There were many solutions: even if most students solving the problem use one of the solutions from the

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