# 1976 IMO Problems/Problem 3

## Problem

A box whose shape is a parallelepiped can be completely filled with cubes of side $1.$ If we put in it the maximum possible number of cubes, each of volume $2$, with the sides parallel to those of the box, then exactly $40$ percent from the volume of the box is occupied. Determine the possible dimensions of the box.

## Solution

We name a,b,c the sides of the parallelepiped, which are positive integers. We also put \begin{align*} x &= \left\lfloor\frac{a}{\sqrt[3]{2}}\right\rfloor \\ y &= \left\lfloor\frac{b}{\sqrt[3]{2}}\right\rfloor \\ z &= \left\lfloor\frac{c}{\sqrt[3]{2}}\right\rfloor \\ \end{align*} It is clear that $xyz$ is the maximal number of cubes with sides of length $\sqrt[3]{2}$ that can be put into the parallelepiped with sides parallels to the sides of the box. Hence the corresponding volume is $V_2=2\cdot xyz$. We need $V_2=0.4\cdot V_1=0.4\cdot abc$, hence $$\frac ax\cdot \frac by\cdot \frac cz=5\ \ \ \ \ \ \ \ (1)$$ We give the values of $x$ and $a/x$ for $a=1,\dots ,8$. The same table is valid for $b,y$ and $c,z$. $$\begin{tabular}{|c|c|c|} \hline a & x & a/x \\ \hline 1 & 0 & - \\ \hline 2 & 1 & 2 \\ \hline 3 & 2 & 3/2 \\ \hline 4 & 3 & 4/3 \\ \hline 5 & 3 & 5/3 \\ \hline 6 & 4 & 3/2 \\ \hline 7 & 5 & 7/5 \\ \hline 8 & 6 & 4/3 \\ \hline \end{tabular}$$ By simple inspection we obtain two solutions of $(1)$: $\{a,b,c\}=\{2,5,3\}$ and $\{a,b,c\}=\{2,5,6\}$. We now show that they are the only solutions.

We can assume $\frac ax\ge \frac by \ge \frac cz$. So necessarily $\frac ax\ge \sqrt[3]{5}$. Note that the definition of $x$ implies $$x< a/\sqrt[3]2 < x+1,$$ hence $$\sqrt[3]2< a/x < \sqrt[3]2(1+\frac 1x)$$ If $a\ge 4$ then $x\ge 3$ and $\frac ax<\sqrt[3]2(1+\frac 1x)\le \sqrt[3]2(\frac 43)<\sqrt[3]5$ since $2\cdot \frac {4^3}{3^3}<5$. So we have only left the cases $a=2$ and $a=3$. But for $a=3$ we have $a/x=3/2<\sqrt[3]5$ and so necessarily $a=2$ and $a/x=2$. It follows $$\frac by \cdot \frac cz =\frac 52 \ \ \ \ \ \ (2)$$

Note that the definitions of $y,z$ imply $$y< b/\sqrt[3]2 < y+1,\ \ \textrm{and} \ \ z< c/\sqrt[3]2 < z+1.\ \ \ \ (3)$$ Moreover we have from (2) and from $b/y\ge c/z$ that $$\frac by \ge \sqrt{5/2}\ \ \ \ \ (4)$$

If $b=2$ then $b/y=2$ and we would have $c/z=5/4<\sqrt[3]2$, which contradicts $(3)$.

On the other hand, if $b>5$ then $y>4$ and $\frac by<\sqrt[3]2(1+\frac 1y)\le \sqrt[3]2(\frac 54)<\sqrt{5/2}$ since $2^2\cdot \frac {5^6}{4^6}<\frac{5^3}{2^3}$ as $5^3<2^7$. So we have only left the cases $b=3,4,5$. But for $b=3$ we have $b/y=3/2<\sqrt{5/2}$ and for $b=4$ we have $b/y=4/3<\sqrt{5/2}$ and so necessarily $b=5$ and $b/y=5/3$ ($>\sqrt{5/2}$)

So we arrive finally at $a=2,b=5$ and $c/z=3/2$. If $c\ge 8$ then $z\ge 6$ and $\frac cz<\sqrt[3]2(1+\frac 1z)\le \sqrt[3]2(\frac 76)<\frac 32$ since $2\cdot \frac {7^3}{6^3}<\frac{3^3}{2^3}$. On the other hand, for $c\le 7$ there are the only two possible values $c=3$ and $c=6$ which yield the known solutions.