A+B=C AxB=C 咁A,B,C 代表咩呢?

You have to solve A+B=AB, i.e. A+B-AB=0

by simple algebra:
A+B-AB = 0
(A-1)(B-1) - 1 = 0
==> B = 1/(A-1) + 1

Thus, you can have many combinations of A, B, C
E.g.: A = 5, B = 1.25, C = 6.25

However, if you restrict the solutions to be integers, The only combinations is (A, B, C) = (0,0,0) or (2,2,4)

It can be explained as follows:
To make B to be integer, 1/(A-1) must be an integer, i.e.
|A-1| <= 1 ==> A = 0, 1, 2
If A = 0, B = 0 and C = 0
If A = 1, B and C is undefined
If A = 2, B = 2 and C = 4

A = 2
B = 2
C = 4

1) 2(a) + 2(b) = 4 (c)

2) 2(a) x 2(b) = 4 (c)

2009-07-23 13:57:53 補充：
also , as the answer from emily_9512, A , B , C would be 0 , too

冇可能(A代表=1, B代表=0
,C代表=1 )∴A+B=C-->1+1=0
AxB=C--->1x0=0=-=不是=C(1)

..
A=0
B=0
C=0
∴A+B=C -->0+0=0
AxB=C-->0x0=0 =-=

2009-07-23 13:51:37 補充：
打錯=-="

冇可能(A代表=1, B代表=0
,C代表=1 )∴A+B=C-->1+0=1
AxB=C--->1x0=0=-=不是=C(1)

..
A=0
B=0
C=0
∴A+B=C -->0+0=0
AxB=C-->0x0=0 =-=

A代表=1
B代表=0
C代表=1