Is there such natural number N, such that a^c has 10 distinct digits, for natural a and c >= N?

Question

Is there such natural number N, such that $a^c$ has 10 distinct digits, for all natural a and all natural $c \ge N$?

Note:

• The question is about the decimal system
• Number a is not a power of 10

For example, $19^{24}$ contains all decimal digits and the same is true for $19^{25}$, and so far.

Answer 1

I strongly believe so, but also believe it will be very difficult to prove. One way to justify the belief is to make a probabilistic argument claiming (most of) the digits of $a^c$ (especially the middle ones, which is most of them) are "random". Then if $a^c$ has $n$ digits the chance any one is missing is about $10 \cdot 0.9^n$. Now sum over $c$ from $N$ to infinity and the chance for a given $a$ is as small as you want by choosing $N$ large enough. Now show you can sum over $a$ and still keep the sum small. The point is that as $a$ gets large $a^c$ has even more digits and even less chance of having one missing. I have done the first part several times, but not the sum over $a$.

Answer 2

$$\left\{ 32043^2,32286^2,33144^2,35172^2,35337^2,35757^2,35853^2,37176^2,37905^2,38772^2 \right\}$$ $$ = \left\{ 1026753849,1042385796,1098524736,1237069584, 1248703569,1278563049,1285437609,1382054976,1436789025,1503267984 \right\}$$

Are these examples of what you meant?