Let A ={1,2,3,4,5.......100} Let M be the number of 2 elements subsets {a,b} of A

We can directly see

• $\#\{\text{multiples of 6 in } A\} = \lfloor\frac{100}{6}\rfloor = 16$
• $\#\{\text{multiples of 3 in } A\} = \lfloor\frac{100}{3}\rfloor = 33$
• $\#\{\text{multiples of 2 in } A\} = \lfloor\frac{100}{2}\rfloor = 50$

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 $\binom{16}{2} = 120$ sets $\{a,b\}$ in this case.
• Exactly one of $a$ or $b$ is a multiple of $6$ . Then there are $16 \cdot (100-16) = 1344$ 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 $(33-16)(50-16) = 578$ sets $\{a,b\}$ in this case.

Therefore there are in total $120 + 1344 + 578 = \boxed{2042}$ sets $\{a,b\}$ such that $ab$ is a multiple of $6$ .