Find a possible pair of numbers given HCF and a factor of the LCM I was just going through a GCSE paper with a student and I came across a question that I'm struggling to find a good method for. The question was this:

Martin thinks of two numbers.
The Highest Common Factor of the two numbers is 6   The Lowest Common
  Multiple is a multiple of 15
Write a pair of numbers that Martin could be thinking of.

It took me an embarrassingly long time to come up with a solution but I still don't have a method that will work quickly for different numbers. What would be the best way to come up with a solution?
 A: Both numbes must be multiples of $6$ and at least one must be a multiple of $5$ (because $5$ is a prime dividing $15$). So the first attempt is $a=30$ and $b=6$ and works
A: Two possible numbers $a$ and $b$ have $HCF= 6$ and $LCM = 15$
Use prime factors and Venn Diagrams.  Intersection = HCF and Union = LCM
$$6 = 2 \times  3$$
$$15 = 3 \times 5$$
Intersection $= 2,3$
Union $= 2,3,5$
Therefore a possible pair when separated out correctly could be:
$$a = 2\times 3$$
$$b = 2\times 3\times 5$$
This is probably the smallest possible pair of such numbers.  You could potentially find more by adding more prime numbers outside the Intersection but making sure they aren't repeated in both sets.
Hope this helps.
A: To guarantee the GCD to be $6$, we choose two multiples of $6$ which differ by $6$-($1$).
Since the LCM is divisible by $15$ (and thus $5$), choose a multiple of $5$ and $6$, say $30$. Applying($1$), we get either ($24,30$) or ($30,36$), where both of them satisfy the given condition. 
A: Considering prime factors will help. 
There are infinitely many such pairs.
$$a=2^p3^q5^rw,\ \ b=2^s3^t5^uz$$
where $p,q,r,s,t,u$ are non-negative integers and $w,z$ are coprime positive integers neither divisible by $2,3$ nor $5$ satisfy your condition if and only if
$$\min(p,s)=1$$
$$\min(q,t)=1$$
$$\min(r,u)=0$$
$$\max(r,u)\ge 1$$
A: I too was struggling with this problem and arrived at a general solution for finding a pair of numbers given the HCF and the LCM. It is essentially the reverse of finding the LCM of two numbers by the use of a prime factor tree. Consider the following example:
The LCM of two numbers ( A & B) is 120 and the HCF is 4.
Step 1: find the prime factors of the LCM
i.e  PF of LCM are 2 2 2 3 5
However note that the HCF of the numbers A & B is 4 and so we could re-write the PF of the LCM as    2 4 3 5 
We now proceed as if we were finding the LCM of the two numbers but in reverse to produce several possible values of A & B (note the factor 4 appears in both columns as it is (highest) common factor.
     A.............. B

(24)   2 3 4............4 5    (20)
(60)   3 5 4............4 2    (8)
(40)   2 5 4........... 4 3    (12)
So three possible solutions are A=24, B=20; A=60, B=8; A = 40, B=12.
A: this question is incorrect!
To have an LCM of 15 the numbers can only be 1 and 15 or 3 and 5
The given answer of 6 and 30 would give a HCF of 6 and and LCM of 30  
A: It's wrong. It's not possible to find out 2 numbers whose HCF is 6 and LCM is 15,  because HCF must be a factor of LCM. And 6 is not a factor of 15.
