# An easy-to-state elusive combinatorial problem (revisited)

The following question is tackled with based completely on David Speyer's argument on a related problem:

Let $x,y,z\in \mathbb{Q}:x\ge1,y\ge1,z\ge1$. What should be the minimum value of $s\in \mathbb{R}: n∈[1,s]$ to ensure that $5p−4≤nx≤5p−1$, $5q−4≤ny≤5q−1$ and $5r-4\le nz\le 5r-1$ for all $x,y,z$ and some $p,q,r \in \mathbb{N}$?

Without loss of generality, let $x\ge y\ge z$. Now we break the problem into cases.

Case 1 (the main case): $z\ge9/4$. In this case, there exists an integer $l$ such that $z\le 5l−4<5l−1≤4z$. As $r$ ranges from $(5l−4)/z$ to $(5l−1)/z$, the value of $ry$ increases by $3(y/z)\ge3$ and $rx$ increases by $3(x/z)\ge3$. Therefore, for some $r$ in this range, $ry$ and $rx$ must lie in an interval of the form $(5m−4,5m−1)$. For this $r$, we have $r∈[1,4]$ and $(rx,ry,rz)$ of the desired form.

So, from now on, assume $z\le 9/4$. Since this means $1≤z≤4$, if $5l−4≤y≤5l-1$ and $5m−4≤x≤5m-1$, we are done.

Does this argument hold for the case shown?

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I'm not convinced. For the values allowed for r, less than half might work for x, and a separate less than half might work for y. You need to argue in the main case that that does not happen (there is at least one r that works for both x and y), and I do not see that argument yet. –  The Masked Avenger Jun 30 '13 at 2:51
If $ry$ increases by $3(y/z)\ge 3$ wouldn't that imply an increment in $rx$ by $3(x/y)\ge 3$ as well? –  Maaz-ul-Haq Jun 30 '13 at 2:59
Yes, but if the first 2 don't work for x and the last 2 don't work for y, then 3 is not big enough. –  The Masked Avenger Jun 30 '13 at 3:50