John, Valen and Warren had a total of 106 coins. The ratio of Valen's coins to Warren's coins was 6 : 5 at first. John and Valen each gave away ¹₂ of their coins. Given that the three boys had 73 coins left, how many coins did John have in the end?

	John	Valen	Warren	Total
Comparing Valen and Warren at first		6 u	5 u
Before	2 p	2x3 = 6 u	5 u	106
Change	- 1 p	-1x3 = - 3 u		- 33
After	1 p	1x3 = 3 u	5 u	73

Total number of coins that John and Valen gave away
= 106 - 73
= 33

The number of coins that Valen had at first is repeated.  Make the number of coins that Valen had at first the same.  LCM of 6 and 2 is 6.

1 p + 3 u = 106 - 73
1 p + 3 u = 33
1 p = 33 - 3 u --- (1)

1 p + 3 u + 5 u = 73
1 p + 8 u = 73
1 p = 73 - 8 u --- (2)

(1) = (2)
33 - 3 u = 73 - 8 u
8 u - 3 u = 73 - 33
8 u - 3 u = 40
5 u = 40

1 u = 40 ÷ 5 = 8

Substitute 1 u = 8 into (1).
1 p = 33 - 3 u
1 p = 33 - 3 x 8
1 p = 33 - 24

1 p = 9

Number of coins that John had in the end
= 1 p
= 9

Answer(s): 9