# [seqfan] Re: Multiply my digits

Chris Thompson cet1 at cam.ac.uk
Thu Aug 1 17:01:27 CEST 2019

```On Jul 27 2019, Lars Blomberg wrote:

>Another attack: since the 7-smooth numbers are (relatively) rare, I generated
>all with n digits and then successively multiply digits until a 1-digit number
>appears. No digit 0 is allowed.
>
>I find the longest sequences as follows:
>n is the number of digits in the next-to-last number, the concatenation of
>all prime factors in this number is the last number (whose digits may be
>permuted and still give a solution).
>
>n=9
>2,12,126,1792,1741824,368947264,22222277777777
>n=16
>6,16,128,288,23328,2939328,1792336896,3416267673274176,2222223333333333333333333333333337
>n=15
>8,18,192,124416,1194891264,727326941773824,22222222222222222222222222222222222222233377
>Interestingly, no other solutions were found for n <= 50.

As the number of n-digit 7-smooth numbers is proportional to n^4, and the
fraction of n-digit numbers with no zero is (9/10)^n, it is a fairly natural
heuristic conjecture that there are only a finite number of 7-smooth numbers
with no zero digit in their decimal expansion.

In fact, as one cannot have the exponents of 2 and 5 both non-zero, the
count of candidates actually behaves like n^3 rather than n^4, and a
back-of-the-envelope calculation suggest that the largest zero-free
7-smooth number might have around 150 digits.

--
Chris Thompson
Email: cet1 at cam.ac.uk

```