What is the probability that the digit sum of a randomly chosen integer between 0000 and 9999 is divisible by 5? If I have a randomly selected integer between 0000 and 9999, what is the probability that the digit sum of that number is divisible by 5? 
[E.g. 1234 = 1 + 2 + 3 + 4 = 10]
I've started off with knowing that I have 2 options for the last integer, but I'm not sure where to go from there.

In this case we had it easy since the base 10 is a multiple of the divisor 5. What happens if we have $n$ digits in base $B$ and are interested in divisor $d$?
Let $\omega \neq 1$ be a $d$th root of unity, and consider
$$
P(\omega) = \left(\frac{1 + \omega + \omega^2 + \cdots + \omega^{B-1}}{B}\right)^n = \left(\frac{1-\omega^B}{B(1-\omega)}\right)^n.
$$
The coefficient of $\omega^r$ is the probability that the remainder is $r$. We can extract the probability of a zero remainder by going over all roots of unity (using $P(1) = 1$):
$$
\Pr[r=0] = \frac{1}{d} + \frac{1}{d} \sum_{t=1}^{d-1} \left(\frac{1-\omega^{tB}}{B(1-\omega^t)}\right)^n.
$$
It is not immediately clear why this expression is real. The reason is that the terms for $t$ and $d-t$ are complex conjugates (since $\omega^{d-t} = \overline{\omega^t}$).
Perron-Frobenius theory tells us that the norm of the eigenvalues $\frac{1-\omega^{tB}}{B(1-\omega^t)}$ is strictly less than 1, and so the convergence to $1/d$ is exponentially fast.
