How many powers of 2 are easy to double? 
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Is 2048 the highest power of 2 with all even digits (base ten)? 

Numbers written in base $10$ are easiest to double when their digits lie in the range $0, \ldots, 4$, so that no carries are performed.  For instance, ...


*

*$1$ is easy to double, yielding $2$;

*$2$ is also easy to double, yielding $4$;

*$4$ is also also easy to double, yielding $8$;

*but $8$ is not easy to double, since $8+8 = 16$ involves a carry.


The only numbers $2^n$ which are easy to double that I'm aware of are the three listed above, along with $2^5 = 32$ and $2^{10} = 1024$.  My suspicion is that $2^{10}$ is the last such power of $2$, and that's my question:

My question is: Is $2^{10} = 1024$ the last example of a power of $2$ which is easy to double?  If it isn't the last example, how many more are there?  Is there a(n infinite) family of examples which is easy to produce?

There's an theorem of Kummer which can be used to solve a "base $p$ analogue" of this question for $p$ a prime:

Theorem (Kummer): In $p$-adic arithmetic, the number of times a carry is performed when adding the numbers $n$ and $m$ (in base $p$) is equal to the number of times $p$ divides the binomial coefficient $\binom{n+m}{m}$.

I haven't found any analogue that holds in base $10$.  An essential obstacle is that $\binom{5}{3} = 10$ shows that $2 + 3$ has carries when written both in base $2$ and base $5$, but not in base $10$.  Because of this sort of phenomenon, I suspect that this isn't the right way to approach my problem, but it helps form a complete question and I thought someone might find it cute.
 A: Here's another potential approach to the problem: the powers of $2$ eventually fall into a cycle $\bmod\,10^k$ for each $k$, and this cycle isn't 'full' in the sense that it only has $4\cdot 5^{k-1}$ elements (see http://www.exploringbinary.com/cycle-length-of-powers-of-two-mod-powers-of-ten/ ); we only need all of these elements to include a digit $\geq 5$ for some $k$ to guarantee that there will always be a carry for any powers of $2$ larger than that cycle, and then the rest could easily be checked by hand.  If we pretend that these elements are all independent, then the odds that any given one is 'high-digit-free' are $\frac{1}{2^k}$, which seems like good odds; unfortunately, there are so many numbers that have to avoid the 'high-digit-free' zone that it's impossible to reasonably bound the non-collision probability away from $1$ (if we only had $2^k$ elements then we would have probability roughly $\left(1-2^{-k}\right)^{2^k}\approx e^{-1}$ of avoiding a collision; with as many elements as we have, the probability that all of our cycle elements are good is roughly $e^{-2.5^k}$), so in fact it's probable that every cycle of powers of $2\bmod\,10^k$ contains some 'bad' (carry-free) set of trailing digits, even if no actual powers of $2$ do.  Still, this would be easy enough to check for all $k$ up to some relatively modest bound, say $k=16$ or so.
A: Too long for a comment.
The question you are asking is basically the following: how many powers of two only conatin the digits 0,1,2,3 and 4. Or similarly, which power of two has all the digits even $2^{n+1}$ has this property if $2^n$ is easy to double. 
It is very easy to prove that there are infinitely many powers of two which don't have this property. Actually, using the irrationality of $\log_2(10)$, it is easy to prove that given any $n$-digit number, there are infinitely many powers of two which start with that number. And you can also get a lower bound on the number of powers up to $N$ of two which can start with a certain combination, but I highly doubt that this can lead you anywhere. The best result you can hope for with this approach is is that between $1$ and $N$ at least $p\%$ are not easy to double, where you can probably get high percentages but never 100 $\%$.
A: If you do it in binary, all of them are very easy to double: Just append another $0$. For example, $\{1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, \ldots\}$.
