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Let $0 < \delta_1 \leq \delta_2 \leq \delta_3 \leq 1$, and consider the box $B := [0,X^{\delta_1}] \times [0,X^{\delta_2}] \times [0,X^{\delta_3}] \subseteq \mathbb{R}^3$. Let $X > 3$ say. Is it possible to pack $B$ with cubes (box which has equal side lengths) $J_1, \ldots, J_N$ contained in $B$ satisfying the following property:

  1. $(B \cap \mathbb{Z}^3) \subseteq \cup_{j=1}^N (J_i \cap \mathbb{Z}^3)$.

  2. Given any $i \not = j$, $(J_i \cap \mathbb{Z}^3) \cap (J_j \cap \mathbb{Z}^3) = \emptyset.$

  3. $N \leq (\log X)^{C} \frac{X^{\delta_1 + \delta_2 + \delta_3}}{X^{3 \delta_1}} $, where $C$ is a constant independent of $X$.

I would greatly appreciate any comments/references! Thank you very much!

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  • $\begingroup$ What's to stop you taking one huge cube containing everything? $\endgroup$ – Anthony Quas Apr 13 '17 at 3:04
  • $\begingroup$ @AnthonyQuas I fixed the typo. Thank you very much! $\endgroup$ – SJY Apr 13 '17 at 3:06
  • $\begingroup$ I would expect to see some constraint on the side length of the cubes. Otherwise, I still don't see what stops you taking one cube of size $X$ for example. $\endgroup$ – Anthony Quas Apr 13 '17 at 3:10
  • $\begingroup$ @AnthonyQuas I would like the cubes to be contained in $B$ (This was the typo I fixed). So if $\delta_1 = \delta_2 = \delta_3 =1$ then we can take a cube of length $X$ and we are done, but otherwise it will not be contained in $B$. $\endgroup$ – SJY Apr 13 '17 at 3:14
  • $\begingroup$ I see now. Sorry. $\endgroup$ – Anthony Quas Apr 13 '17 at 4:08

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