Comment by wrboyce

Is it definitely the LCD? Given drive of size 15 and 20 the LCD would be 1, no? I had assumed it would just use the size of the smallest drive on every drive (so 15+20->15+15=30). When I first read your comment I was thinking of GCF but even that would be fairly inefficient (GCF(15,20) = 5, so 15+20->5+5=10).