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
$$\frac{abc}{(a-2)(b-2)(c-2)} = 2, \quad a, b, c \ge 3$$
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 $a \ge b \ge 7$, $3a-20 \equiv 1 \pmod{3}$ and $3b-20 \equiv 1 \pmod{3}$, 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 $a \ge b \ge 9$, $5a-28 \equiv 2 \pmod{5}$ and $5b-28 \equiv 2 \pmod{5}$, 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.