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.

Similarly,

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.