Power of 2 with equal number of decimal digits?

Question

Does there exist an integer $n$ such that the decimal representation of $2^n$ have an equal number of decimal digits $\{0,\dots,9\}$, each appearing 10% of the time?

The closest I could find was $n=1,287,579$ of which $2^n$ has 387,600 digits broken down as

0  38,808   10.012%
1  38,735    9.993%
2  38,786   10.007%
3  38,751    9.997%
4  38,814   10.014%
5  38,713    9.987%
6  38,731    9.992%
7  38,730    9.992%
8  38,709    9.986%
9  38,823   10.016%

score 3 · Accepted Answer · answered Apr 25 '19 at 00:18

3

No. If each digit appears $x$ times, then the sum of all the digits will be $45x$; this implies $3|2^n$ which cannot be the case.

answered Apr 25 '19 at 00:18

Abraham Zhang

894

1

Very elegant solution. – kccu Apr 25 '19 at 00:23
2

Agreed. Very nice and simple proof. Thank you. For fun tried other bases which were now multiplies of three. Found 32043^2 contains all digits once. – sheppa28 Apr 25 '19 at 17:20

Power of 2 with equal number of decimal digits?

1 Answers1