Find a general formula for the largest number less than $10^n$ for $n > 1$ that has exactly $8$ divisors.
I was wondering if there was a way to find a general formula for this or if it is possible to create one.
For example, for $n = 2$, $88 = 2^3 \times 11$ is the largest number less than $10^2$ that has $8$ divisors.
Let $x$ have prime factorization $p_1^{a_1}p_2^{a_2} \cdots p_k^{a_k}$.
Let $\tau(x) = \prod_{i=1}^{k} (a_i+1)$, the count-of-divisors function.
There are three cases if $\tau(x) = 8$:
$\tau(x) = (a_1+1) = (8) = 8$
$\tau(x) = (a_1+1)(a_2+1) = (2)(4) = 8$
$\tau(x) = (a_1+1)(a_2+1)(a_3+1) = (2)(2)(2) = 8$
This implies that $x$ is of the form $p^7$, $pq^3$, or $pqr$, where $p,q,r$ are distinct. So we can find the largest way to represent each of these under the limit $10^n$ and then take the maximum of the three.
In practice, it's probably a lot easier to start at $10^n-1$, iterate downward, and compute $\tau(n)$ as you go until you find something with $\tau(x) = 8$.