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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).

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    $\begingroup$ I suspect you can work this out for yourself if you just write down what it means (using base ten for "concatenation"). As a warm-up exercise, would it be possible in base two? The argument is less obvious in base ten, but actually not more difficult once you figure out how to express a power of two as the base ten concatenation of two such powers. $\endgroup$ – hardmath Feb 28 '15 at 0:43
  • $\begingroup$ It's perhaps worth nothing that there are triple-concatenations, such as $128$ and (if you allow intervening zeroes) $1024$ and $2048$. I think there was a question here about such things some time ago. $\endgroup$ – Blue Feb 28 '15 at 2:14
  • $\begingroup$ I may have misunderstood your example, but based on the example of 256, then it seems like 24, 28, 216, 232,264, 2128, 2256, 2512, 21024, etc... all qualify $\endgroup$ – frogfanitw Feb 4 '16 at 7:05
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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.

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  • $\begingroup$ I really like your approach, but I'm not sure I see why $r - s + k = 0$. Unless both $r - s$ and $k$ need to be 0, in which case I agree. $\endgroup$ – pjs36 Feb 28 '15 at 0:59
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    $\begingroup$ @pjs36 $2^{r-s}10^k$ must be odd, but still an integer. Since this is $2^{r-s+k}5^k$, then $r-s+k=0$. $\endgroup$ – ajotatxe Feb 28 '15 at 1:02
  • $\begingroup$ While clearly $r,s \lt t$, I think the cases $r \ge s$ and $r \lt s$ have to be considered separately. $\endgroup$ – hardmath Feb 28 '15 at 1:10
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    $\begingroup$ @hardmath I suspect it's actually not too hard to show that $r\gt s$ - this is a consequence of the fact that the length of the period of $2^m\bmod 10^n$ is 'sufficiently long'; in this case, it only has to be larger than $\approx 3n$, which should be easily provable (EDIT and is: see exploringbinary.com/… ) $\endgroup$ – Steven Stadnicki Feb 28 '15 at 2:13
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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 $n \leq k$.

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