Find the minimum of $ab$ if $a^a\cdot b^b$ has exactly 98 trailing zeroes.
We see that $a^a\cdot b^b=2^{98}\cdot 5^{98}\cdot k$ for odd integral k. Then $a^a=2^{pa}5^{qa}\cdots$ and $b^b=2^{rb}5^{sb}\cdots$. Thus $pa+rb=98=sb+qa$. Usually $a\cdot b$ is minimized when $a=b$, but there is no constraints that I could create. Thanks!
For positive integers $p,n$, let $\nu_p(n)$ represent the largest integer $k$ such that $p^k\mid n$. So we're looking for $a,b$ such that $\nu_{10}(a^ab^b)=98$. As John Omielan notes in the comments, that means that we need one of $\nu_2(a^ab^b)$ and $\nu_5(a^ab^b)$ to be equal to $98$ and the other one equal to or greater than it.
A sufficiently motivated reader can quickly verify that $\nu_p(ab)=\nu_p(a)+\nu_p(b)$ if $a$ and $b$ are non-zero and that $\nu_p(a^n)=n\nu_p(a)$.
Let us consider $\nu_5(a^a)$. This is $0$ if $5\not\mid a$. On the other hand, if $5\mid a$, then $\nu_5(a^a)=a\nu_5(a)$ must also be a multiple of $5$. Thus, $\nu_5(a^a)$ is always a multiple of 5, and there are therefore no solutions to $\nu_5(a^ab^b)=\nu_5(a^a)+\nu_5(b^b)=98$.
So we will need an exact solution to $\nu_2(a^ab^b)=98$. This is at least possible. In fact, $\nu_2(98^{98})=98\nu_2(98)=98$, since $98=2\cdot49$. So one batch of solutions to our problem would be $a=98$ and $b$ a sufficiently large odd multiple of 5 that $\nu_5(b^b)\ge98$. (These $b$ would be $75, 105, 115, 125, ....$)
Is there a solution where both $a$ and $b$ are even? At least one of those numbers would need to be divisible by $10$, but they could not both be (as then $\nu_2(a^ab^b)$ would also a multiple of 10 instead of $98$). The only possibility that would contribute enough powers of $5$ without too many powers of $2$ is if $b=50$. $\nu_5(50^{50})=50\cdot2=100$ and $\nu_2(50^{50})=50\cdot1=50$, so we'd be set if there were an $a$ such that $\nu_2(a^a)=48$. But there is no such $a$: for instance $\nu_2(12^{12})=12\cdot2=24$ and $\nu_2(24^{24})=24\cdot3=72$.
Therefore, our optimal solution is $75^{75}98^{98}$, so $ab=7350$.