Assume we have two uniform distributions $X=U(\mathbb{Z}_m)$ and $Y=U(\mathbb{Z}_n) \bmod m$, for $m,n \in \mathbb{N}$.
The statistical distance is defined as: $$ \Delta(X, Y) = \frac{1}{2} \sum_{a \in \mathbb{Z}_m} | \Pr[X = a] - \Pr[Y = a] | \enspace. $$
Question: What is the statistical distance between $X$ and $Y$ expressed as a function of $m$ and $n$?
$$ \newcommand{\Z}{\mathbb{Z}} \newcommand{\abs}[1]{\left|#1\right|} $$
The statistical distance between $X=U(\Z_m)$ and $Y=U(\Z_n) \bmod m$ is $$ \Delta(X, Y) = \abs{ \frac{(n \bmod m)^2}{mn} - \frac{n \bmod m}{n} } \enspace . $$
Writing out the definition, we have
$$ \Delta(X, Y) = \frac{1}{2} \sum_{a \in \Z_m} \abs{\Pr[X=a] - \Pr[Y=a]} \enspace. $$
For $a \in \Z_m$, we have $\Pr[X=a] = 1/m$ and $$ \Pr[Y=a] = \frac{n- (n \bmod m)}{nm} \text{ , if $a\geq n \bmod m$ ,} \\ \Pr[Y=a] = \frac{n- (n \bmod m) + m}{nm} \text{ , if $a < n \bmod m$ .} $$
Denote $r = (n \bmod m)$. It follows that $$ \Delta(X, Y) = \frac{1}{2} \left( \left( \sum_{a \in \Z_m: a < r} \abs{\frac{1}{m} - \frac{n - r + m}{nm}} \right) + \left( \sum_{a \in \Z_m: a \ge r} \abs{\frac{1}{m} - \frac{n- r}{nm}} \right) \right) = \frac{1}{2} \left( r \abs{\frac{1}{m} - \frac{n- r + m}{nm}} + (m - r) \abs{\frac{1}{m} - \frac{n- r}{nm}} \right) = \abs{ \frac{r^2-rm}{nm} } \enspace. $$