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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.