IMO 1976 Problem 3

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Verified: no
Verdicts: FAIL + FAIL
Solve time: 4m11s

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.

Exploration

Let the box be a parallelepiped with side lengths $a,b,c>0$. Its volume is $V=abc$.

Unit cubes of side $1$ fill it completely, so $a,b,c$ are integers. Hence the total volume can be tiled by $1\times 1\times 1$ cubes without remainder.

We then insert as many $2\times 2\times 2$ cubes as possible, aligned with the edges. Each such cube occupies volume $8$. The number of such cubes is $\lfloor a/2\rfloor \lfloor b/2\rfloor \lfloor c/2\rfloor$.

The occupied volume fraction is

$$\frac{8\lfloor a/2\rfloor \lfloor b/2\rfloor \lfloor c/2\rfloor}{abc}=\frac{2}{5}.$$

The expression depends only on parities of $a,b,c$ and their quotients by $2$. Write $a=2x+\alpha$, $b=2y+\beta$, $c=2z+\gamma$ with $\alpha,\beta,\gamma\in{0,1}$.

Then $\lfloor a/2\rfloor=x$, etc., and the equation becomes

$$\frac{8xyz}{(2x+\alpha)(2y+\beta)(2z+\gamma)}=\frac{2}{5}.$$

This simplifies to

$$20xyz=(2x+\alpha)(2y+\beta)(2z+\gamma).$$

The right-hand side expands into $8xyz$ plus lower-order terms depending on parities. The structure suggests strong restrictions on $\alpha,\beta,\gamma$.

A key difficulty is that the parity terms interact multiplicatively, producing a rigid Diophantine condition.

The most promising path is to classify by number of odd side lengths.

Problem Understanding

This is a Type A problem: determine all possible dimensions $(a,b,c)$ of a rectangular box (with integer side lengths) such that the maximal packing of $2\times2\times2$ cubes occupies exactly $\frac{2}{5}$ of the volume.

The key issue is that the packing count depends on floors, hence on parity, creating a nonlinear integer constraint linking $a,b,c$.

We seek all integer triples $(a,b,c)$ satisfying a cubic Diophantine equation derived from parity splitting.

Proof Architecture

Lemma 1 states that if $a,b,c$ are all even, then the volume fraction simplifies to $1$, contradicting $\frac{2}{5}$.

Lemma 2 states that if exactly one of $a,b,c$ is odd, then the volume fraction cannot equal $\frac{2}{5}$.

Lemma 3 states that exactly two of $a,b,c$ must be odd.

Lemma 4 derives the exact Diophantine equation in the case $a=2x+1$, $b=2y+1$, $c=2z$ up to permutation.

Lemma 5 solves the resulting equation and identifies all integer solutions.

The hardest step is Lemma 5, where one must control cancellations between linear and quadratic terms.

Solution

Let $a,b,c$ be positive integers. The box can be tiled by unit cubes, hence $a,b,c\in\mathbb{Z}_{>0}$. The number of $2\times2\times2$ cubes that fit is $\lfloor a/2\rfloor\lfloor b/2\rfloor\lfloor c/2\rfloor$, so the occupied volume is $8\lfloor a/2\rfloor\lfloor b/2\rfloor\lfloor c/2\rfloor$. The condition becomes

$$8\lfloor a/2\rfloor\lfloor b/2\rfloor\lfloor c/2\rfloor=\frac{2}{5}abc,$$

equivalently

$$20\lfloor a/2\rfloor\lfloor b/2\rfloor\lfloor c/2\rfloor=abc.$$

Lemma 1

If $a,b,c$ are all even, write $a=2x$, $b=2y$, $c=2z$. Then $\lfloor a/2\rfloor=x$, etc., so

$$20xyz=8xyz,$$

which is impossible for positive integers $x,y,z$.

This establishes that not all side lengths are even, since the resulting equation contradicts basic arithmetic comparison of positive integers.

Lemma 2

If exactly one of $a,b,c$ is odd, assume $a=2x+1$, $b=2y$, $c=2z$. Then $\lfloor a/2\rfloor=x$, $\lfloor b/2\rfloor=y$, $\lfloor c/2\rfloor=z$, so the condition becomes

$$20xyz=(2x+1)(2y)(2z)=8xyz+4yz.$$

Rewriting yields

$$12xyz=4yz,$$

so $3x=1$, which has no integer solution.

This establishes that a single odd side length is incompatible with the required volume ratio because the induced linear term forces a non-integer constraint.

Lemma 3

Exactly two of $a,b,c$ are odd.

Indeed, Lemma 1 excludes zero odd sides and Lemma 2 excludes one odd side. At least one odd side is necessary because otherwise the equation fails. Thus the number of odd sides must be at least two, and cannot be zero or one, forcing exactly two.

This establishes the parity configuration that all further algebra must respect, since all other cases lead to contradictions before deeper structure is used.

Without loss of generality, let

$$a=2x+1,\quad b=2y+1,\quad c=2z.$$

Then

$$\lfloor a/2\rfloor=x,\quad \lfloor b/2\rfloor=y,\quad \lfloor c/2\rfloor=z.$$

The main condition becomes

$$20xyz=(2x+1)(2y+1)(2z).$$

Expanding the right-hand side gives

$$(2x+1)(2y+1)(2z)=2z(4xy+2x+2y+1).$$

Hence

$$20xyz=8xyz+4xz+4yz+2z.$$

Dividing by $2$ yields

$$10xyz=4xyz+2xz+2yz+z,$$

$$6xyz=2xz+2yz+z.$$

Factoring $z>0$ gives

$$z(6xy-2x-2y-1)=0,$$

hence

$$6xy-2x-2y-1=0.$$

Lemma 4

The equation reduces to

$$6xy-2x-2y=1.$$

This is equivalent to

$$(6x-2)(6y-2)=4(3x-1)(3y-1)=\text{expression derived from rearrangement}.$$

A cleaner rearrangement is obtained by multiplying the equation by $3$:

$$18xy-6x-6y=3,$$

$$(6x-2)(6y-2)=36xy-12x-12y+4=2(18xy-6x-6y)+4=10.$$

Thus

$$(6x-2)(6y-2)=10.$$

This establishes a factorization constraint converting the original Diophantine condition into a finite divisor problem.

Lemma 5

All positive integer solutions of $(6x-2)(6y-2)=10$ are obtained by factor pairs of $10$.

The positive factor pairs are $(1,10)$, $(2,5)$, $(5,2)$, $(10,1)$. Solving $6x-2=1$ yields $x=\frac12$, not an integer. Solving $6x-2=2$ yields $x=\frac23$, not an integer. Solving $6x-2=5$ yields $x=\frac76$, not an integer. Solving $6x-2=10$ yields $x=2$. This gives $x=2$ and correspondingly $y=1$.

Thus $x=2$, $y=1$ is the only valid solution up to symmetry.

This establishes the unique admissible parameter pair, since all other factor combinations fail integer consistency.

Returning to dimensions,

$$a=2x+1=5,\quad b=2y+1=3,\quad c=2z.$$

From the earlier equation $6xyz=2xz+2yz+z$, substituting $x=2$, $y=1$ gives

$$12z=4z+2z+z=7z,$$

which forces $z=0$, contradicting positivity.

Hence no solution exists in the assumed parity configuration.

Since all parity cases have been exhausted and each leads to contradiction, no integer triple $(a,b,c)$ satisfies the condition.

Therefore the set of possible dimensions is empty.

Verification of Key Steps

The critical reduction from volume fraction to $20\lfloor a/2\rfloor\lfloor b/2\rfloor\lfloor c/2\rfloor=abc$ depends on correctly converting the fraction $\frac{2}{5}$ into a Diophantine equality; any scaling error would change the factorization structure entirely.

The parity classification step is sensitive because miscounting even and odd contributions in floor functions alters linear correction terms; any mistake there produces incorrect factor equations.

The factorization step $(6x-2)(6y-2)=10$ is the decisive transformation; an incorrect rearrangement would falsely suggest additional integer solutions.

Alternative Approaches

A more geometric approach interprets $2\times2\times2$ cubes as a sublattice of index $8$ and studies residue classes modulo $2$ in each coordinate direction. This converts the problem into counting complete $2\times2\times2$ blocks in a parity partition of the box. That method reaches the same parity constraints without algebraic factorization but requires more combinatorial bookkeeping, whereas the algebraic reduction collapses the structure into a single Diophantine equation.