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I am trying to prove or disprove the following bound:

$2+\left(n-\left\lfloor\dfrac n k\right\rfloor k\right)\ge \left\lfloor\dfrac n k\right\rfloor$, where $2<k\le \left\lfloor\dfrac {n-1} 2\right\rfloor$, $n>4$, $k, n$ are coprime, and $n,k\in\mathbb N$.

Any suggestions, solutions, or hints would be appreciated!

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  • $\begingroup$ Coprime doesn't really matter. The point is to see that as $n$ increases, the right side increases without bound, while the left is bounded. $\endgroup$ – Ross Millikan Jul 1 '14 at 21:04
  • $\begingroup$ Perhaps a better question then would be: under what conditions would this inequality hold? $\endgroup$ – user2154420 Jul 1 '14 at 21:08
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    $\begingroup$ If $n=2k+1$, which is the minimum, the right side will be $2$ and the left will be $3$. The left is $2+(n\pmod k)$, so if $n=3k-1$ the left will grow with $k$ and the right will stay $2$ $\endgroup$ – Ross Millikan Jul 1 '14 at 21:13
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It is not true. Take $n$ huge. The left side is no greater than $k+1$, the right is unbounded in $n$. For a specific, $n=101, k=3$, we are comparing $2+(101-33\cdot 3)=4$ and $\lfloor \frac {101}3 \rfloor =33$

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  • $\begingroup$ $k$ has to be greater than 2. $\endgroup$ – user2154420 Jul 1 '14 at 20:55
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    $\begingroup$ Ok, but the logic is the same. $\endgroup$ – Ross Millikan Jul 1 '14 at 20:56
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Hint: $n=16$, $k=3$ and $n=7$, $k=3$. The first case yields $<$, the second one yields $>$.

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  • $\begingroup$ Neither of these work with the conditions on k. $\endgroup$ – user2154420 Jul 1 '14 at 20:59
  • $\begingroup$ @user2154420 edited them to reflect your new conditions. $\endgroup$ – TZakrevskiy Jul 1 '14 at 21:00

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