急!!!HELP!!! 數學功課-英文題 ( 只有一題 )

Since their H.C.F is 16, both numbers must be multiples of 16. Moreover, other factors of the both numbers must not have any common factor except 1. If you let the first number be 16a and the second be 16b (where a,b are positive integers), then the H.C.F of a and b is 1. Furthermore, since neither of them is a multiple of the other, neither a nor b could equal to 1.

Therefore, you can take a=2 and b=3, then a possible set of numbers is {32, 48}.
If you take a=3 and b=5, then another possible set is {48,80}.


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2013-07-24 23:30:10 補充：
In other words, a and b must be co-prime integers so as to obtain the required pairs

Suppose the two number are 16a and 16b,
since they are not multiple of each other,
i.e. a and b are also not .
Possible pairs: a=2 b=3 , 32,48 / a=3 b=4 48,64