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How many $N$ of the form $2^n,\text{ with } n \in \mathbb{N}$ are there such that no digit is a power of $2$?

For this one the answer given is the $2^{16}$, but how could we prove that that this is the only possible solution? and what about the general case of $x^n, \text{ with } x,n \in \mathbb{N}$?

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@Ilmari:Yes, but the question is how to prove that conclusively. – Quixotic Jan 2 '12 at 11:35
The last n digits in a sequence of powers of 2 will form a repeating cycle eventually, so if a brute force of all possible last 5 digits of powers of 2 finds only 65536, then it will be known that all possible solutions end in 65536. – Angela Richardson Jan 2 '12 at 11:49
We could extend Angela Richardson's idea by checking the last 6 digits of the number and finding the cycle. This could easily be done by computer. I see the contest-math tag though. No idea how you'd do this without a computer. – Mike Jan 2 '12 at 12:22
2^12506= 64 mod(10^6) – Raymond Manzoni Jan 2 '12 at 13:34
I think there are some unsolved problems nearby. For example, I think it has not been proved that for all sufficiently large $n$ there's a zero in the decimal representation of $2^n$. According to it's conjectured that 86 is the highest power of 2 with no zero. – Gerry Myerson Jan 2 '12 at 16:14

Define the acceptable digits to be 0, 3, 5, 6, 7, and 9; and define the score of a number $n$ to be the number of trailing acceptable digits in the decimal expansion of $n$ (with no leading zeroes). So for instance 65536 has a score of 5, and $2^{96} = 79228162514264337593543950336$ has a score of 7 (and this is the smallest power of $2$ with a score greater than 5).

I did a computer search for high-scoring powers of $2$ up to $2^{332192}$ (i.e. those with less than 100000 decimal digits). The highest-scoring was $2^{57072}$, with a score of only 25 (it ends with ...40535076966633036050333696).

So your conjecture is plausible, but it looks like one of those things that could be very difficult to prove.

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Downvoted only because it's more a comment than an answer. – Greg Martin Jan 4 '12 at 7:48
@Greg: You think I should have put all this in a comment? Don't be silly. – TonyK Jan 4 '12 at 7:54

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