Among the two-digit natural numbers, which one has the most divisors?
If a number $n$ has a prime factorization of $p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdots \cdots p_{n}^{\alpha_{n}}$, then the number of its divisors is $\left(\alpha_{1}+1\right) \cdot\left(\alpha_{2}+1\right) \cdots \cdots\left(\alpha_{n}+1\right)$; a divisor is indeed of the form $p_{1}^{\beta_{1}} \cdot p_{2}^{\beta_{2}} \cdots \cdots p_{n}^{\beta_{n}}$, where $0 \leq \beta_{i} \leq \alpha_{i}$, meaning the exponent of $p_{1}$ can be $\left(\alpha_{1}+1\right)$ different values, the exponent of $p_{2}$ can be $\left(\alpha_{2}+1\right)$ different values, and so on. Using this, we first determine the maximum number of divisors a two-digit number can have.

Since $2 \cdot 3 \cdot 5 \cdot 7>100$, a two-digit number can have at most 1, 2, or 3 different prime divisors. According to the above, the number of divisors of $p^{a} \cdot q^{b} \cdot r^{c}$ and $2^{a} \cdot 3^{b} \cdot 5^{c}$ is the same, $s$, assuming that $p99$ is not possible. In summary, the sought numbers are:

$$
2 \cdot 3^{2} \cdot 5=90, \quad 2^{2} \cdot 3 \cdot 5=60, \quad 2^{2} \cdot 3 \cdot 7=84, \quad 2^{5} \cdot 3=96, \quad 2^{3} \cdot 3^{2}=72
$$

Note. Many submissions were satisfied with the numbers we also derived (or some of them) without any further explanation or justification (why this is the maximum; is there another number with the same number of divisors). These solutions, in accordance with the principle "the mere announcement of results is not sufficient for a solution" in the scoring competition, received only a portion of the maximum possible points.