The sum of all two-digit numbers divisible by $5$ is

The two-digit numbers divisible by $5$ form an Arithmetic Progression (AP):
$$10,15,20,25,\dots ,95$$

Here, the first term $a=10$, the common difference $d=5$, and the last term $l=95$.
Find the number of terms ($n$):

$$l=a+(n-1)d⟹95=10+(n-1)5$$

$$85=5(n-1)⟹n-1=17⟹n=18$$

Now, compute the sum ($S_n$):

$$S_n=\frac{n}{2}(a+l)=\frac{18}{2}(10+95)=9\times 105=945$$