[Number Theory]
Problem: Compute the smallest positive integer $n$ such that $n^{n}$ has at least 1,000,000 positive divisors.

Solution: Let $k$ denote the number of distinct prime divisors of $n$, so that $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}, a_{i}>0$. Then if $d(x)$ denotes the number of positive divisors of $x$,

$$
d\left(n^{n}\right)=\left(a_{1} n+1\right)\left(a_{2} n+1\right) \cdots\left(a_{k} n+1\right) \geq(n+1)^{k}
$$

Note that if $n \geq 99$ and $k \geq 3$, then $d\left(n^{n}\right) \geq 100^{3}=10^{6}$, so $102=2 \cdot 3 \cdot 17$ is an upper bound for the solution. Look for values less than 99, using two observations: (1) all $a_{i} \leq 6$



(because $p^{7}>99$ for all primes); and (2) $k \leq 3$ (because $2 \cdot 3 \cdot 5 \cdot 7>99$ ). These two facts rule out the cases $k=1$ (because $(*)$ yields $\left.d \leq(6 n+1)^{1}<601\right)$ and $k=2$ (because $\left.d\left(n^{n}\right) \leq(6 n+1)^{2}<601^{2}\right)$.

So $k=3$. Note that if $a_{1}=a_{2}=a_{3}=1$, then from $(*), d\left(n^{n}\right)=(n+1)^{3}<10^{6}$. So consider only $n<99$ with exactly three prime divisors, and for which not all exponents are 1 . The only candidates are 60,84 , and 90 ; of these, $n=84$ is the smallest one that works:

$$
\begin{aligned}
& d\left(60^{60}\right)=d\left(2^{120} \cdot 3^{60} \cdot 5^{60}\right)=121 \cdot 61 \cdot 61<125 \cdot 80 \cdot 80=800,000 \\
& d\left(84^{84}\right)=d\left(2^{168} \cdot 3^{84} \cdot 7^{84}\right)=169 \cdot 85 \cdot 85>160 \cdot 80 \cdot 80=1,024,000
\end{aligned}
$$

Therefore $n=\mathbf{8 4}$ is the least positive integer $n$ such that $d\left(n^{n}\right)>1,000,000$.

Final answer: 84