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Choose any natural number $x$ and let it has $y$ digits in the demical system. Then, is there a natural number $z$ where the first $y$ numbers of $2^z$ are $x$?

For example, if we let $x=20$, then the answer is 'yes' because $2^{11}=2048$ and $20$ was appeared.

What will be the result if we generalize this problem to any number systems, not only demical system? (ex. binary system, etc...)

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Sure. Fix your block $N$. Then for each $k$ The greatest power of $2$ which which does not exceed $N\times 10^k$ is $$\lfloor \log_2 N\times 10^k\rfloor = \lfloor \log_2 N + k\log_2 10\rfloor$$ We note that $N\times 10^k$ ends in $k$ zeroes, so we can add $10^{k}-1$ to it without changing the first block of $N$. Thus we want to find a power of $2$ between $N\times 10^k$ and $(N+1)\times 10^k-1$ (Note: we can drop the $-1$ because $(N+1)\times 10^k$ is clearly not a power of $2$, unless $k=0$). It follows that we want to find $k$ such that $$\lfloor \log_2 N + k\log_2 10\rfloor<\lfloor \log_2 (N+1) + k\log_2 10\rfloor$$ As the $\log_2 10$ is irrational, the integer multiples of it are dense $\pmod 1$ so this is always possible.

Example: $N=7$. We can't take $k=0$, because that $-1$ interferes. If we require $k>0$ we search numerically and find that the first time these two floors are unequal comes when $k=13$. In that case we see that the left hand is $45$ and the right is $46$. Thus we are lead to expect that $2^{46}$ should begin with $7$ and indeed $2^{46}=70368744177664$.

To generalize to other bases: Note that we depend critically on the irrationality of $\log_2 10$. Indeed, $\log_{10}10 $ is rational and no power of $10$ begins with $3$. But this irrationality is enough.

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  • $\begingroup$ Oh, this problem can be generalized to power of w, not only power of 2. Thank you $\endgroup$ – suliman Jan 26 '17 at 16:35

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