Does there exist a power of 2 which is the concatenation of two powers of 2?

Question

I am curious whether there exists a power of $2$,

$z = d_1d_2\ldots d_n$ (where $d_i$ is the $i$-th digit of $z$), such that $z_1 = d_1d_2\ldots d_j$ and $z_2 = d_{j+1}d_{j+2}\ldots d_n$, $1\le j\le n$, are both powers of 2.

For example, imagine 56 was a power of 2, then $2^8 = 256$ would be an example of such a $z$, because 256 can be split up as 2|56, where 2 and 56 are powers of 2. Of course, 56 is not an actual power of 2, so 256 is not really an example of such a z.

I have checked all powers of 2 up to those with 1000 digits, but I can't think how to prove or disprove the existence of $z$.

(This is not a homework assignment or anything, I'm just curious).

Answer 1

From $$2^r\cdot10^k+2^s=2^t$$ we get $$2^{r-s}\cdot 10^k+1=2^{t-s}$$

Since $s$ is clearly lesser than $t$, LHS must be even, so $r-s+k=0$ or $s=r+k$. Now, $$2^r\cdot10^k+2^{r+k}=2^t$$ or $$2^k(5^k+1)=2^{t-r}$$ This implies that $5^k+1$ is a power of $2$, but $5^k+1\equiv 1+1\equiv 2\pmod 4$, so $k=0$. But if $k=0$, there's no "concatenation", in the sense you mean.

I have used critically the fact that we usually write numbers in base $10$, so the question remains open for other bases.

Answer 2

If there were such an example, then you would have two numbers $x,y$ such that $x$ and $y$ are both powers of $2$ and yet $z = 10^n x + y$ is also a power of two for some $n$. Let $2^k = y$. If $n > k$ Then $z/y = 5^n 2^{n-k}x + 1$, which is odd since $n > k$. Thus $z/y$ is not a power of two so $z$ cannot be a power of two. I'm not sure what happens when $n \leq k$.