中一數學.........

HCF = 6
a) 12, 18
b) 12, 30
Actually the no. should be 6*a, & 6*b
and a & b should not be multiple of the other.

2007-08-18 13:43:29 補充：
If their HCF = 16a) 32, 48b) 32, 80same theory, can be 16*a & 16*bie. 16*2, 16*316*2, 16*516*3, 16*5..etc.

1 st pair : 12 / 18
2nd pair : 18/24
if the first no. is n (n is not equal to 0 or 6 and 6 is a factor of n )
then the other no . is (n+6)

so n / n+6
can 符合
neither of them is a multiple of the other
find two pairs of possible values of the numbers if their h.c.f is 6


2007-08-18 13:39:29 補充：
if n is = 36n 6 = 42it still符合neither of them is a multiple of the otherfind two pairs of possible values of the numbers if their h.c.f is 6

2007-08-18 13:58:56 補充：
if their hcf is 16then the two no. are n & (n 16)if n is 48 ,then n 16 = 64it still符合neither of them is a multiple of the otherfind two pairs of possible values of the numbers if their h.c.f is 16

2007-08-18 14:00:57 補充：
我的方法是較 applefung927 快 和簡單