MATHEWS: NoZeros  NoZeros

© Walter Schneider 2000 (last updated 30/1/2003)

1033 (1 Decillion) is the largest known power of ten being the product of two zerofree numbers (i.e. numbers without the digit zero):

1033 = 233 · 533 = 8 589 934 592 · 116 415 321 826 934 814 453 125.

All known exponents of 10 with this property are (Madachy, 1979)

1, 2, 3, 4, 5, 6, 7, 9, 18 and 33.

We are searching for exponents k such that 2k and 5k does not contain the digit zero. A computer search yields the following results for powers nk (n=2,3,5,7):

Powers nk without Digit Zero
n Searched
for k up to
Maximum k Number of k's k's
2 108 86 35 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 15, 16, 18, 19, 24, 25, 27, 28, 31, 32, 33, 34, 35, 36, 37, 39, 49, 51, 67, 72, 76, 77, 81, 86
3 108 68 22 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 19, 23, 24, 26, 27, 28, 31, 34, 68
5 108 58 15 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 17, 18, 30, 33, 58
6 108 44 13 1, 2, 3, 4, 5, 6, 7, 8, 12, 17, 24, 29, 44
7 108 35 9 1, 2, 3, 6, 7, 10, 11, 19, 35

Open Questions

(1) Is 1033 the largest solution?
(2) Are there other solutions for 2 ≤ n ≤ 9?
(3) Does for each n exist a largest k?
(4) Is this also true for each base?

It's for example not known if 2k for k > 15 and base 3 always contains the digit 0. If true, this would mean that the persistence of a number in base 3 is always less or equal 3.

It seems that for high powers not only the digit 0, but also each digit or even combinations of digits are present. The following table shows that in base 10 high powers seem to be pandigital (i.e. each digit is present).

Powers nk not pandigital
n Searched for k up to Largest k
2 105 168
3 105 106
5 105 65
6 105 64
7 105 61

Zerofree Powers

What happens when we fix the exponent which means that we search for zerofree squares, cubes, ..., n-th powers? The following pattern shows that there are infinitely many zerofree squares:

342 = 1156,
3342 = 111556,
33342 = 11115556,
etc.

For zerofree cubes D. Hickerson found the following formula which yields a zerofree cube for n ≥ 5 and n = 2 (mod 3):

f(n) = (2 · 105n - 104n + 17 · 103n-1 + 102n + 10n - 2) / 3.

The formula was slightly simplified by Lew Baxter:

f(n) = (2 · 105n - 104n + 2 · 103n + 102n + 10n + 1) / 3.

For higher powers no formula is known and the question is still open if there are infinitely many zerofree powers!

References

Guy, Richard K.: Unsolved Problems in Number Theory,
Second Edition, Springer-Verlag 1994, Problem F25.