Brazilian Math Olympiad

a. For example, $900 = 2^2 \cdot 3^2 \cdot 5^2$, which has $(2+1) \cdot (2+1) \cdot (2+1) = 27$ positive divisors.

b. No, there doesn't. Let $n$ be a number with at least $200$ divisors. If the $i$-th divisor is $d$, then the $i$-th to last divisor is $\frac{n}{d}$. Let $m$ be the $100$th divisor. So $m \ge 100$ and $\frac{n}{m} > m \iff n > m^2 = 10000$. Close enough, but how do we fix this? First notice that if $97$, $99$ and $100$ are all divisors of $n$, then $n \ge \text{lcm}(97, 99, 100) > 11000$. So the key observation is considering the $98$th, $99$th and $100$th divisors. Let $k, \ell$ and $m$ be such divisors. Notice that if $m \ge 105$ we are done because then $n > m^2 = 11025 > 11000$. So $98 \le k < \ell < m \le 104$. But $\text{gcd}(x, y) \le |x - y|$ implies that $n \ge \text{lcm}(k, \ell, m) \ge \frac{k \cdot \ell \cdot m}{(\ell-k)(m-\ell)(m-k)} \ge \frac{98 \cdot 99 \cdot 100}{(104-98) \cdot 3^2} > 11000$. Here we used the fact that if $x+y \le 2t$ then $xy \le t^2$ applied to $x = \ell - k, y = m - \ell$ and $t = 3$.