Kimberly, Cathy and Elyse each had some balls and decided to play a game with their balls. In round 1, Kimberly lost ¹₃ of her balls to Cathy. In round 2, Cathy lost ¹₃ of her total number of balls to Elyse. In round 3, Elyse lost ¹₃ of her total number of balls to Kimberly. In the end, Kimberly, Cathy and Elyse had 116, 72 and 68 balls respectively. How many balls did Cathy have at first?

	Kimberly	Cathy	Elyse
		67
Before 1	3 b
Change 1	- 1 b	+ 1 b
After 1	2 b	108
Before 2	82	3 u
Change 2		- 1 u	+ 1 u
After 2		2 u	102
Before 3			3 p
Change 3	+ 1 p		- 1 p
After 3			2 p
Comparing Kimberly, Cathy and Elyse in the end	116	72	68

Working backwards.

2 p = 68

1 p = 68 ÷ 2 = 34

At the start of Round 3:
Number of balls that Elyse had
= 3 p
= 3 x 34
= 102

2 u = 72

1 u = 72 ÷ 2 = 36

At the start of Round 2:
Number of balls that Cathy had
= 3 u
= 3 x 36
= 108

At the start of Round 2:
Number of balls that Kimberly had
= 116 - 34
= 82

2 b = 82

1 b = 82 ÷ 2 = 41

At the start of Round 1:
Number of balls that Cathy had
= 108 - 1 b
= 108 - 41
= 67

Answer(s): 67