- --- In primenumbers@yahoogroups.com,

"mikeoakes2" <mikeoakes2@...> wrote:

> (6*a+a1)*(6*b+b1) = 6*c+c1,

Not at my high school, where

> where c = 6*a*b+a1+b1,

> and c1 = a1*b1

c = 6*a*b + a*b1 + b*a1

David - Yes.

I understand your concept.

I have worked with those equations also.

(6 a -1)(6b - 1) = (36 a b - 6 a - 6 b + 1) = 6 ( 6 ab - a - b) + 1

(6 a - 1) (6 b + 1) = 36 a b + 6 a - 6 b - 1 = 6 ( 6 a b + a - b) - 1

(6 a + 1) (6 b - 1) = 36 a b - 6 a + 6 b - 1 = 6 ( 6 a b - a + b) - 1

(6 a + 1) (6 b + 1) = 36 a b + 6 a + 6 b + 1 = 6 ( 6 a b + a + b) + 1

Kermit