My earlier solution is marginally easier if I start with B's passes. It then goes like this:
Let y be the amount that everybody winds up with.
B has y after his pass, so he must have (3/2)y immediately before his pass. He passes y/2 to C.
We can solve for c (C's initial qty) in terms of y.
(c + y/2)*3/4 = y
so c = (4/3)y - y/2 = (5/6)y
c passes y/3 to D.
d = (5/4)y - y/3 = (11/12)y
d passes y/4 to E
e = (6/5)y - y/4 = (19/20)y
e passes y/5 to A.
So A, if a initially starts with qty a, he winds up with a/2 + y/5, which equals y.
therefore, a = (8/5)y, and A passes (4/5)y to B.
B winds up with
(b+(4/5)y)*2/3 = y
b = (3/2)y - (4/5)y = (7/10)y
y therefore is divisible by 6,12,20,5 and 10, whose LCM is 60.
so the minimum possible y is 60.
A starts with 96, passes 48, and ultimately receives 12, ending with 60.
B starts with 42, receives 48, passes 30, ending with 60.
C starts with 50, receives 30, passes 20, ending with 60.
D starts with 55, receives 20, passes 15, ending with 60.
E starts with 57, receives 15, passes 12, ending with 60.