Three bets were made:
(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+BC
B: 2(BA)
C: 2(A+CB)
Since these amounts are equal, after a little manipulation we get equations:
A = 3(Bc), 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 20170906 10:36:37 