$\mathrm{A}=\{1,2,3,4,5,6\}$

$\mathrm{R} =\{( \mathrm{x} , \mathrm{y} ): \mathrm{y} $ is divisible by $\mathrm{x} \}$

We know that any number $(\mathrm{x})$ is divisible by itself.

So, $(\mathrm{x}, \mathrm{x}) \in \mathrm{R}$

$\therefore \mathrm{R}$ is reflexive.

Now, $(2,4)\in \mathrm{R}$ $[$ as $4$ is divisible by $2]$

But, $(4,2) \notin \mathrm{R}$ . $[$ as $2$ is not divisible by $4]$

$\therefore \mathrm{R}$ is not symmetric.

Let $( \mathrm{x} , \mathrm{y} ),\,( \mathrm{y} , \mathrm{z} ) \in \mathrm{R} .$ Then, $\mathrm{y}$ is divisible by $\mathrm{x}$ and $\mathrm{z}$ is divisible by $\mathrm{y}$

$\therefore$  $ \mathrm{z}$ is divisible by $\mathrm{x}$ $\Rightarrow(\mathrm{x}, \mathrm{z}) \in \mathrm{R}$

$\therefore $  $\mathrm{R}$ is transitive.

Hence, $\mathrm{R}$ is reflexive and transitive but not symmetric.