Stuart has drawn a pair of concentric circles, as shown.  He draws chords $\overline{AB}$, $\overline{BC}, \ldots$ of the large circle, each tangent to the small one.  If $m\angle ABC=75^\circ$, then how many segments will he draw before returning to his starting point at $A$? [asy]
size(100); defaultpen(linewidth(0.8));
real rad1 = 1/Sin(37.5);
draw(Circle(origin,1)); draw(Circle(origin,rad1));
pair A = rad1*dir(190), B = rad1*dir(190 + 105), C = rad1*dir(190 + 2*105), D = rad1*dir(190 + 315);
draw(A--B--C--D,EndArrow(size=5));
label("$A$",A,W); label("$B$",B,S); label("$C$",C,E);
[/asy]
We look at $\angle ABC$. $\angle ABC$ cuts off minor arc $\widehat{AC}$, which has measure $2\cdot m\angle ABC = 150^\circ$, so minor arcs $\widehat{AB}$ and $\widehat{BC}$ each have measure $\frac{360^\circ-150^\circ}{2}=105^\circ$.

Stuart cuts off one $105^\circ$ minor arc with each segment he draws.  By the time Stuart comes all the way around to his starting point and has drawn, say, $n$ segments, he will have created $n$ $105^\circ$ minor arcs which can be pieced together to form a whole number of full circles, say, $m$ circles.  Let there be $m$ full circles with total arc measure $360^\circ \cdot m$.  Then we have \[105^\circ \cdot n = 360^\circ \cdot m.\]  We want to find the smallest integer $n$ for which there is an integer solution $m$.  Dividing both sides of the equation by $15^\circ$ gives $7n=24m$; thus, we see $n=24$ works (in which case $m=7$).  The answer is $\boxed{24}$ segments.