Partitioning into Divisors

Set $a = 2^{a_1} \cdot 5^{a_2}, b = 2^{b_1}\cdot 5^{b_2}, c= 2^{c_1}\cdot 5^{c_2}$ . We first count the number of triples without restriction. We seek the number of non-negative integer solutions to $a_1 + b_1 + c_1 = 6, a_2 + b_2 + c_2 = 6$ . This is equivalent to arranging 6 1's with 2 0's (dividing blocks), thus there are ${6 + 2\choose 2}$ ways to solve either of the two equations. By the rule of product, there are ${ 8 \choose 2} \cdot {8 \choose 2} = 784$ solutions in total.

If $a =b$ , we seek the number of non-negative integer solutions to $2a_1 + c_1 = 6, 2a_2 + c_2 = 6$ . For each equation, there are 4 solutions, corresponding to $a_i=0, 1, 2, 3$ . By the product rule, there are $4 \cdot 4 = 16$ solutions in all. The cases $b=c$ and $c=a$ are exactly the same. If $a=b=c$ , there is exactly 1 solution.

Let $A, B,$ and $C$ be the set of triples where $b=c, c=a,$ and $a=b$ , respectively. By the Principle of Inclusion and Exclusion, $| A \cup B \cup C | = |A| + |B| + |C| - |A\cap B| - |B\cap C| - |C\cap A| + |A\cap B \cap C| = 16 + 16 + 16 - 1 - 1 - 1 + 1 = 46$ . Hence, the number of triples with pairwise distinct values is $784 - 46 = 738$ .