6. (NET 4) ${ }^{\mathrm{IMO} 3} \mathrm{~A}$ rectangular box can be filled completely with unit cubes. 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 \(b_{i} = \left[a_{i} / \sqrt[3]{2}\right]\), the condition of the problem is equivalent to \(\frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}} = 5\). We list some values of \(a, b = [a / \sqrt[3]{2}]\) and \(a / b\):

| \(a\) | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| \(b\) | 1 | 2 | 3 | 3 | 4 | 5 | 6 | 7 | 7 |
| \(a / b\) | 2 | 1.5 | 1.33 | 1.67 | 1.5 | 1.4 | 1.33 | 1.29 | 1.43 |

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 \(\frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}} < (5 / 3)^{3} < 5\), 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 \((2,5,6)\).