"Probability" of Divisibility by Primes Proof .. where "probability" is used loosely.
Let $k$ by a non-negative integer and $p_a$ and $p_b$ be distinct primes. Select a positive integer $x$. What is the probability that $p_a | x$ and $p_b|(x + k)$, or $p_a|(x + k)$ and $p_b|x$ ?
Through experimentation it looks like the probability is $\frac{2}{p_ap_b}$ if $(k, p_ap_b) = 1$, and $\frac{1}{p_a p_b}$ if $(k, p_ap_b) \gt 1$, but I am not sure how to prove it, so I am looking for tips and hints and how the proof might go.
 A: Hint: $p_a|x \land p_b|x+k \leftrightarrow x \equiv m p_a \mod p_ap_b$ where $m$ is defined to be a number with $m p_a \equiv -k \mod p_b$. Analogously, $p_b|x \land p_a|x+k \leftrightarrow x \equiv n p_b \mod p_ap_b$ where $n$ is defined to be a number with $n p_b \equiv -k \mod p_a$. Thus, if $m p_a \equiv n p_b \mod p_ap_b$, the probability is $\frac{1}{p_ap_b}$, and if $m p_a \not \equiv n p_b \mod p_ap_b$, the probaility is $\frac{2}{p_ap_b}$. 
Now $m p_a \equiv n p_b \mod p_ap_b \leftrightarrow (p_b|m \land p_a|n) \leftrightarrow p_b|k \land p_a|k$ and so $\text{Probability}=\frac{1}{p_ap_b}$ iff $p_ap_b|k$.
A: Let's just use $\rm p$ and $\rm q$ for the primes. The key is Sun-Ze aka CRT:
$$\rm p\mid x\wedge q\mid(x+k)\iff \begin{cases}x\equiv 0 & \rm mod~p \\ \rm x\equiv -k & \rm mod~q\end{cases}\iff x\equiv k_{p,q}~~\bmod pq $$
where $\rm k_{p,q}$ is the unique solution to the congruence system in the middle (it can be given by the explicit formula $\rm -k[p^{-1}\bmod q]p$). Similarly $\rm q\mid x\wedge p\mid(x+k)\iff x\equiv k_{q,p}~\bmod pq$, so we see that the possible $\rm x$ are two residue classes if $\rm k_{p,q}\not\equiv k_{q,p}$ but only one residue class if $\rm k_{p,q}\equiv k_{q,p}$.
Finally $\rm k_{p,q}\equiv k_{q,p}\iff 0\equiv-k\bmod p \wedge -k\equiv0\bmod q\iff pq\mid k$. As residue classes mod $\rm pq$ have density $\rm1/(pq)$, we conclude the density of $\rm x$'s  is $\rm 2/pq$ if $\rm pq\nmid k$ and $\rm 1/(pq)$ if $\rm pq\mid k$.
