John, Howard and Ken had a total of 165 pencils. The ratio of Howard's pencils to Ken's pencils was 8 : 9 at first. John and Howard each gave away ¹₂ of their pencils. Given that the three boys had 123 pencils left, how many pencils did John have in the end?

	John	Howard	Ken	Total
Comparing Howard and Ken at first		8 u	9 u
Before	2 p	2x4 = 8 u	9 u	165
Change	- 1 p	-1x4 = - 4 u		- 42
After	1 p	1x4 = 4 u	9 u	123

Total number of pencils that John and Howard gave away
= 165 - 123
= 42

The number of pencils that Howard had at first is repeated.  Make the number of pencils that Howard had at first the same.  LCM of 8 and 2 is 8.

1 p + 4 u = 165 - 123
1 p + 4 u = 42
1 p = 42 - 4 u --- (1)

1 p + 4 u + 9 u = 123
1 p + 13 u = 123
1 p = 123 - 13 u --- (2)

(1) = (2)
42 - 4 u = 123 - 13 u
13 u - 4 u = 123 - 42
13 u - 4 u = 81
9 u = 81

1 u = 81 ÷ 9 = 9

Substitute 1 u = 9 into (1).
1 p = 42 - 4 u
1 p = 42 - 4 x 9
1 p = 42 - 36

1 p = 6

Number of pencils that John had in the end
= 1 p
= 6

Answer(s): 6