Power Factoring Contest Question The question was as follows:
Compute the smallest positive integer $n$ such that $n^n$ has at least $1,000,000$ positive divisors.
I did some work, finding that if $n=2^a*3^b*5^c*7^d$ then the $n^n= 2^{na}*3^{nb}*5^{nc}*7^{nd}$ and that the number of positive divisors is $(an+1)(bn+1)(cn+1)(dn+1)$ I wasn't sure how to proceed after that point.
More Information: This was a team round, with 20 minutes to do it (and other problems) with no calculators allowed.
 A: Already $n=2\cdot 3\cdot 17= {102}$ gives us  $(n+1)^3>10^6$ divisors.
Any number with at least four distinct prime divisors is $\ge 2\cdot 3\cdot 5\cdot 7>102$ and can thus be ignored: We need to check only up to three prime divisors.
If $n=p^kq^m$, we get $(kn+1)(nm+1)$ divisors (the case $m=0$ is not excluded). To make this $\ge 10^6$, one of the factors, say $kn+1$, needs to be $\ge 1000$. With $n<102$, we conclude $k>9$, but then $n\ge2^9>102$: We need only check numbers with at least, hence with exactly three prime divisors.
So $n=p^aq^br^c$ with $p,q,r$ distinct primes, $a,b,c\in\mathbb N$. 
Then $a\ge3$ would imply $n\ge 2^3\cdot 3\cdot 5=120>102$. Hence alle exponents are $\le2$. If $a=b=2$, then $n\ge 2^23^25=180>102$, hence at most one exponent is $>1$, i.e. $n=pqr$ or $n=p^2qr$.
As in the first paragraph, $n=pqr$ gives us $(n+1)^3$ divisors, so we need $n\ge 99$, but $n=99, 100, 101$ are not of this form.
Remains the case $n=p^2qr$ with $(2n+1)(n+1)^2$ divisors. As $n<102$, we find $pqr\le 50$, so $pqr=2\cdot3\cdot 5=30$ or $pqr=2\cdot 3\cdot 7=42$ and that's it. Thus the $n$ we need to test are $60, 90, 84$. As $(2\cdot 60+1)(60+1)^2<10^6$, the minimal $n$ is indeed $$n=84.$$
