a set problem...

We can directly see

• $\#\{\text{multiples of 6 in } A\} = \lfloor\frac{1000}{6}\rfloor = 166$
• $\#\{\text{multiples of 3 in } A\} = \lfloor\frac{1000}{3}\rfloor = 333$
• $\#\{\text{multiples of 2 in } A\} = \lfloor\frac{1000}{2}\rfloor = 500$

Also we note that $ab$ is divisible by $6$ if and only if one of the following disjoint cases hold:

• $a$ and $b$ are both multiples of $6$ . Then there are $\binom{166}{2} = 13695$ sets $\{a,b\}$ in this case.
• Exactly one of $a$ or $b$ is a multiple of $6$ . Then there are $166 \cdot (1000-166) = 138444$ sets $\{a,b\}$ in this case.
• Neither is a multiple of $6$ , but one is a multiple of $2$ and the other is a multiple of $3$ . Then there are $(333-166)(500-166) = 55778$ sets $\{a,b\}$ in this case.

Therefore there are in total $M = 13695 + 138444 + 55778 = 207917$ sets $\{a,b\}$ such that $ab$ is a multiple of $6$ , which gives an answer of $\left\lfloor\frac{M}{10000}\right\rfloor = \boxed{20}$