Japan Junior Mathematical Olympiad

If one of $a,b,c$ is less than or equal to $2$, then we see all of the cubes used will be colored and therefore, the condition of the problem will not be satisfied. So, we assume that each of $a,b,c$ is greater than or equal to $3$. We see that the uncolored cubes form an $(a-2)\times (b-2)\times (c-2)$ rectangular parallelepiped enclosed within the given rectangular parallelepiped. Therefore, it is enough to determine the number of distinct (modulo permutations) triples $(a,b,c)$ of positive integers for which $$\frac{abc}{(a-2)(b-2)(c-2)}=2,a,b,c\ge 3$$ are satisfied.

We may assume, without loss of generality, that $a\ge b\ge c$ hold. Since the function $\frac{x}{x-2}=1+\frac{2}{x-2}$ is monotone decreasing for $x\ge 3$, we have $\frac{a}{a-2}\le \frac{b}{b-2}\le \frac{c}{c-2}$, and therefore, we have $2=\frac{abc}{(a-2)(b-2)(c-2)}\le {\left(\frac{c}{c-2}\right)}^3$. As ${\left(\frac{10}{8}\right)}^3=\frac{125}{64}<2$, we must have $c\le 9$. On the other hand, if $c\le 4$, then we get $\frac{c}{c-2}\ge 2$ from which it follows that $\frac{ab}{(a-2)(b-2)}\le 1$, but this is impossible since $1<\frac{a}{a-2}\le \frac{b}{b-2}$. Thus, we conclude that $c=5,6,7,8,9$ must be satisfied.

i. When $c=5$: In this case, we have $5ab=6(a-2)(b-2)$, which is transformed into $(a-12)(b-12)=120$. From $a\ge b\ge 5$, we get $a-12\ge b-12\ge -7$ and we see that it is not possible for $a-12$ and $b-12$ to be negative simultaneously. Therefore, the number of distinct pairs $(a,b)$ to satisfy the condition of the problem under the case in consideration equals the one-half of the numbers of the positive factors of $120=2^3\cdot 3\cdot 5$, which equals $\frac{1}{2}\{(3+1)\times (1+1)\times (1+1)\}=8$.

ii. When $c=6$: In this case, we have $3ab=4(a-2)(b-2)$, which is transformed into $(a-8)(b-8)=48$. Since $a\ge b\ge 6$ implies that $a-8\ge b-8\ge -2$, we see that it is not possible to have $a-8$ and $b-8$ be negative simultaneously. Thus the number of pairs $(a,b)$ to satisfy the condition of the problem in this case is the one-half of the positive factors of $48=2^4\cdot 3$, which equals $\frac{1}{2}\{(4+1)\times (1+1)\}=5$.

iii. When $c=7$: In this case, we have $7ab=10(a-2)(b-2)$, which is transformed to $(3a-20)(3b-20)=280$. If we note that $a\ge b\ge 7$, $3a-20\equiv 1(\mathrm{mod}3)$ and $3b-20\equiv 1(\mathrm{mod}3)$, then we see that only possibilities for $(3a-20,3b-20)$ are $(12,2)$, $(70,4)$, $(40,7)$, $(28,10)$, and the number of pairs $(a,b)$ to satisfy the condition of the problem in this case equals $4$.

iv. When $c=8$: In this case, we have $2ab=3(a-2)(b-2)$, which is transformed to $(a-6)(b-6)=24$. From $a\ge b\ge 8$, we see that the pairs $(12,2)$, $(8,3)$, $(6,4)$ are the only possibilities for $(a,b)$ to satisfy the condition of the problem in this case. Thus the number of such $(a,b)$ in this case is $3$.

v. When $c=9$: In this case, we get $9ab=14(a-2)(b-2)$, which is transformed to $(5a-28)(5b-28)=18\cdot 28$. From $a\ge b\ge 9$, $5a-28\equiv 2(\mathrm{mod}5)$ and $5b-28\equiv 2(\mathrm{mod}5)$, we see that there are no pairs $(a,b)$ satisfying the condition of the problem in this case.

Summarizing the results for the cases considered above, we get $8+5+4+3=20$ for the number of triples $(a,b,c)$ to satisfy the condition of the problem.