If one places cubes with volume 2 in the box such that their edges are parallel to the edges of the box, one can fill exactly $40 \%$ of the box. Determine all possible (interior) sizes of the box.
6. Suppose $a_{1} \leq a_{2} \leq a_{3}$ are the dimensions of the box. If we set $b_{1}=$ $\left[a_{i} / \sqrt[3]{2}\right]$, the condition of the problem is equivalent to $\frac{a_{1}}{b_{1}} \cdot \frac{\alpha_{2}}{\alpha_{2}} \cdot \frac{a_{2}}{\alpha_{2}}=5$. We list some values of $a, b=[a / \sqrt[3]{2}]$ and $a / b$ :
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$a$ & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$b$ & 1 & 2 & 3 & 3 & 4 & 5 & 6 & 7 & 7 \\
\hline$a / b$ & 1.5 & 1.33 & 1.67 & 1.5 & 1.4 & 1.33 & 1.29 & 1.43 \\
\hline
\end{tabular}

We note that if $a>2$, then $a / b \leq 5 / 3$, and if $a>5$, then $a / b \leq 3 / 2$. If $a_{1}>2$, then $\frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{\sigma_{2}} \cdot \frac{a_{1}}{b_{1}}<(5 / 3)^{3}<5$, a contradiction. Hence $a_{1}=2$. If also $a_{2}=2$, then $a_{3} / b_{3}=5 / 4 \leq \sqrt[3]{2}$, which is impossible. Also, if $a_{2} \geq 6$, then $\frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}} \leq(1.5)^{2}<2.5$, again a contradiction. We thus have the following cases:
(i) $a_{1}=2, a_{2}=3$, then $a_{3} / b_{3}=5 / 3$, which holds only if $a_{3}=5$;
(ii) $a_{1}=2, a_{2}=4$, then $a_{3} / b_{3}=15 / 8$, which is impossible;
(iii) $a_{1}=2, a_{2}=5$, then $a_{3} / b_{3}=3 / 2$, which holds only if $a_{3}=6$.

The only possible sizes of the box are therefore $(2,3,5)$ and $(2,5,6)$.