One of the first semi-real unsolved mathematics problem that I ever tried to solve was this one: is there a number greater than 1, in base 10, such that the product of the digits of its square is equal to itself? For instance, 4 squared is 16, and the product of the digits of 16 is 6; if 6 was equal to 4 we'd have a solution, but it's not so we keep looking.

Suppose *n* is a solution to the problem. It's very easy to prove a couple of things
about *n*:

*n*is even;- the only prime factors of
*n*are 2, 3 and 7.

This narrows down the search field a lot; you don't have to check every single positive integer. I wrote programs and ran them for a long time back in the 80's searching for a solution to this. In doing this, it becomes clear right away that arbitrary positive integers of any size tend to have zeros in their digits when you square them. One might think that there is a largest square with no zero digits. Alas, 66666666...666666^2 never has a zero digit no matter how many 6's you put in there.

I've searched for a solution, checking all n=2^{i}3^{j}7^{k}
with 0 ≤ i,j,k ≤ 500. At the same time, this checks all *n* ≤ 10^{150}.
Results of this are here. This shows the indices at which you
get non-zero digit products, and includes the little gp code needed.

The largest non-zero result in that search is from *n*=2^{112}3^{13}7, with
*n*^{2}= 3357899791173323669918785556264153583368288482168381437766224257343594296179163136.

As far as I know, the result is still unsolved. I certainly expect there to be no
such *n*, but it really is of no importance at all.

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