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If $n$ is a positive integer such that $n\ge 4$ and $\left\lfloor\dfrac{2^n}{n}\right\rfloor$ is a power of $2$; then $n$ is also a power of $2$.

Example $n=4,n=8,16,\cdots,2^k$, then $\dfrac{2^n}{n}=\dfrac{2^{2^k}}{2^k}=2^{2^k-k}$

But why $n\neq 2^k$?

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  • $\begingroup$ If $n$ is no power of two than $\lfloor\frac{2^n}{n}\rfloor$ is strictly smaller than $\frac{2^n}{n},$ because $n$ has a prime factor $p>2$. $\endgroup$ – gammatester Aug 7 '15 at 11:06
  • $\begingroup$ your idea maybe usefull,but I can't use your idea to solve this problem $\endgroup$ – user246688 Aug 7 '15 at 11:08
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If $2^k<n<2^{k+1}$, then $2^n/2^{k+1}<2^n/n<2^n/2^k$, so the number is strictly between two powers of 2. So you just have to check that $2^n/n>1+2^n/2^{k+1}$.
$$\frac{2^n}n-\frac{2^n}{2^{k+1}}\geq\frac{2^n}n-\frac{2^n}{n+1}=\frac{2^n}{n(n+1)}>1$$ and the final inequality follows easily by induction:
Firstly, $2^5=32>30=5(5+1)$.
If $2^n>n(n+1)$ then $2^{n+1}=2(2^n)>2n(n+1)>(n+2)(n+1)$

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$\frac{2^t}{n} = qn+r,$ $0\le r\le\frac{n-1}{n}.$

So $$\left \lfloor \frac{2^t}{n} \right \rfloor = \left \lfloor qn+r \right \rfloor = qn=2^k.$$

It implies that if $p|n$ for some prime $p$ then $p|2^k$ and $p=2$.

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If

$$\left\lfloor \frac{2^n}{n} \right\rfloor = 2^k$$

Then

$$2^k \leq \frac{2^n}{n} < 2^k +1$$

$$n\leq 2^{n-k}< n + \frac{n}{2^k}$$

Now it's clear that :

  • $n>k$, so $2^{n-k}$ is an integer

  • $n \leq 2^k$, or else $\frac{2^n}{n} > 2^k+1$ (this would need a little analysis to prove rigourously, just consider the function $f(x) = 2^x - x(2^k+1)$ )

So you get

$$n\leq 2^{n-k}< n + 1$$

And this imply that $n = 2^{n-k}$

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