(1) First, A won from B as much as A had originally.
(2) Second, B won from C as much as B then had left.
(3) Finally, C won from A as much as C then had left.

The three players ended up with equal amounts of money.

Who began with 50 cents?

Final amounts are:

A: A+B-C

B: 2(B-A)

C: 2(A+C-B)

Since these amounts are equal, after a little manipulation we get equations:

A = 3(B-c), and 5C = 4B. A must be a multiple of 3, can't be the answer. If C=50 then B is not an integer. So B is the only one possible.

Edited on September 6, 2017, 10:38 am

Edited on September 6, 2017, 10:39 am

Edited on September 6, 2017, 10:45 am

Edited on September 6, 2017, 10:46 am

Posted by chun on 2017-09-06 10:36:37