Some Unique Numbers

This is not a solution, but rather an explanation of how mathematician D. Wilson arrived at this result. Denote the number as $n$ . Define the number of digits it has as $a$ and its digits $\overline{n_1n_2...n_a}$ . Then we know that

$n \geq 10^{a-1}$

We also know that

$\prod_{k=1}^{a} n_k \leq 9^a$

and

$\sum_{k=1}^{a} n_k \leq 9a$

Then by definition,

$10^{a-1} \leq n=\prod_{k=1}^{a} n_k \cdot \sum_{k=1}^{a} n_k \leq 9^a \cdot 9a$ $10^{a-1} \leq a9^{a+1}$ $a \leq 84$

Therefore,

$\sum_{k=1}^{a} n_k \leq 9a \leq 756$ $\prod_{k=1}^{a} n_k \leq n <10^{85}$

We know that the product of the digits will have the prime factorization $2^w 3^x 5^y 7^z$ . However if $n \mod 10=0$ then $10$ also divides $\prod_{k=1}^{a} n_k$ . Hence, $n$ ends in $0$ and $\prod_{k=1}^{a} n_k=0$ . So $\prod_{k=1}^{a} n_k =2^w 3^x 7^z$ or . $\prod_{k=1}^{a} n_k = 3^x 5^y 7^z$ . This reduces the amount of numbers to check by a considerable amount. It turns out there are only three such numbers, $1,135,144$ . There might be an insight I am lacking that makes further reduction in the numbers to check, so I posted the problem in case someone can provide such insight.