Let A's initial marbles be a.

He starts by giving a/2 to B.

Let B's initial marbles be b.

He now has b+a/2 before he passes, and (b+a/2)*2/3 after he passes. This is the final quantity everybody winds up with, so call it y. B has passed y/2 to C.

We can solve for C 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 winds up with a/2 + y/5, which equals y.

therefore, a = (8/5)y

B winds up with

(b+a/2)*2/3 = (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.