   SEARCH HOME Math Central Quandaries & Queries  Question from matthew, a student: Hi, please help me with this, Find two numbers with HCF of 3 and LCM of 180 Hi Mathew. Let's say the numbers are A and B.

If the HCF is 3, then the only prime factor they share is 3 (because 3 itself is prime).

The LCM = HCF x (non-common-prime-factors from each of A and B).

By factoring 180, I find that 180 = 2 x 2 x 3 x 3 x 5.

Now I remove the HCF and have 2 x 2 x 3 x 5 left. I have to distribute these to A and B.

I know that A and B cannot both have factors of 2, because then the LCM would be 6 instead of 3 (3 x 2 = 6). So the 2 x 2 must be due to just one of the original numbers, so I'm left to distribute factors of 3, 4 and 5. I could give them all to one of the original numbers, or split them up. There are several ways of doing this:

A = 3 x 3, B = 3 x 4 x 5
or
A = 3 x 4, B = 3 x 3 x 5
or
A = 3, B = 3 x 3 x 4 x 5
or
A = 3 x 5, B = 3 x 3 x 4

Each of these will work. You could flip them around (switching B with A), but you'd still have the same two original numbers.

Cheers,
Stephen La Rocque and Penny Nom     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.