Compute $P(\{6|xy\})$ and $P(\{8|xy\})$ if $x$ and $y$ are independent random positive integers (Please edit this post if it is ill-formed) If $x$ and $y$ are independent random positive integers,
 1. How to compute the probability of $(2*3)|xy$ ?
 2. How to compute the probability of $(2^3)|xy$ ?
For both problems it is necessary to set up a probabilistic environment. 

The divisibility of $xy$ by $6$ depends only on the remainders of $x$ and $y$ modulo $6$. It is natural to assume that both $x$ and $y$ are independently and uniformly distributed modulo $6$. In this case they are as well independently and uniformly distributed modulo $2$ and $3$. The probability that $x$ is not divisible by $2$ is ${1\over2}$. Therefore with probability $1-{1\over2}\cdot{1\over2}={3\over4}$ at least one of $x$ or $y$ is divisible by $2$. Similarly, the probability that $x$ is not divisible by $3$ is ${2\over3}$.
Therefore with probability $1-{2\over3}\cdot{2\over3}={5\over9}$ at least one of $x$ or $y$ is divisible by $3$. It follows that with probability $p={3\over4}\cdot{5\over9}={15\over36}$ the product $xy$ is divisible by $6$.

The divisibility of $xy$ by $8$ depends only on the remainders of $x$ and $y$ modulo $8$. It is natural to assume that both $x$ and $y$ are independently and uniformly distributed modulo $8$. The following cases lead to divisibility of $xy$ by $8$:
$$\eqalign{x\in\{1,3,5,7\}\quad&amp;\wedge\quad y=0,\cr
x\in\{2,6\}\quad&amp;\wedge\quad y\in\{0,4\},\cr
x=4\quad&amp;\wedge\quad y\in\{0,2,4,6\},\cr
x=0\quad&amp;\wedge\quad y\in\{0,1,2,3,4,5,6,7\}\ .\cr}$$
Therefore the probability that $xy$ is divisible by $8$ is given by
$$q={1\over2}\cdot{1\over8}+{1\over4}\cdot{1\over4}+{1\over8}\cdot{1\over2}+{1\over8}\cdot1={5\over16}\ .$$
Things might get more easy to compute by realizing that:$$\{6\mid xy\}=\{6\mid x\}\cup\{6\mid y\}\cup(\{2\mid x\}\cap\{3\mid y\})\cup(\{3\mid x\}\cap\{2\mid y\})$$
combined with the application of the principle of inclusion/exclusion.

This approach might be fruitful especially if $x$ and $y$ are independent.

Success not guaranteed, though.