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Marble Muse (Posted on 2014-06-23) Difficulty: 3 of 5
At the outset, each of the five friends Ade, Ben, Cal, Dan and Ethan had a positive integer number of marbles.

If Ade gave one-half of his marbles to Ben, and Ben gave a one third of what he then held to Cal, and Cal gave a quarter of what he then held to Dan, and Dan gave one fifth of what he then held to Ethan who then passes on a sixth of his holding to Ade - they would all have an equal number of marbles.

What is the minimum number of marbles that each of the five friends had at the beginning?

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Analytical Solution (spoiler) | Comment 1 of 2
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.

  Posted by Steve Herman on 2014-06-23 17:58:02
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