Are there arbitary large non-trivial powers NOT containing all digits?

Asked Feb 23 '17 at 20:41

Active Feb 23 '17 at 20:41

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Are there arbitary larger numbers of the form $a^b$ with positive integers $a,b>1$ and $a\ne 0\mod 10$ NOT containing all digits ?

Here :

Biggest powers NOT containing all digits.

is either a very similar or the same question, but the question is so old that I decided to restate it.

My current best example :

The number $1955^{39}$ has $129$ digits, but does not contain the digit $1$.

If we assume that the digits are uniformly distributed in $[0,9]$ and considering the growth of the non-trivial powers, can we expect that we have enough chances to get , lets say , a $1000$-digit example ?

In other words, are there enough non-trivial powers with a given decimal-expansion-length, that we can expect that the answer to the question is yes ?

edited Apr 13 '17 at 12:21

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asked Feb 23 '17 at 20:41

Peter

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How have you obtained those examples? – mrnovice Feb 23 '17 at 20:48
brute force using PARI/GP – Peter Feb 23 '17 at 20:49
1

Doesn't $a=10^k-1,b=2$ give you infinitely many? You can likely use the same with $b=3$, and $b=4$. Do you want $b$ to be arbitrarily large? – Thomas Andrews Feb 23 '17 at 21:02
@ThomasAndrews Correct, I overlooked this family. Do you know further restrictions to avoid such families ? – Peter Feb 23 '17 at 21:26
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Similar stuff with $b=2$ and $a=c10^k\pm 1$ for reasonably simple $c$ (like $c=1,2,3,4,5.) No idea what you'd need to do in general to rule out these sorts of cases. – Thomas Andrews Feb 23 '17 at 23:27

Are there arbitary large non-trivial powers NOT containing all digits?

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