Answer:

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For any given values of hcf and lcm, here is a technique that will always get you a correct answer:

First, take the lcm and divide it by the hcf.

720÷12=60

Factor the number you got from that division into prime factors.

60=22⋅3⋅5

To find the number of possible pairs, count the number of primes, ignoring multiplicity. For this example, there are 3 primes: 2, 3, and 5.

The number of possible pairs will be 2 raised to the power of (the number of primes minus 1). So in this case, the number of possible pairs is: 23−1=4

.To determine what those pairs are, start with the hcf and multiply it by one of your primes (INCLUDE the multiplicity this time). For this example, I will start with the factor of 2, so 12⋅22=48

. Use that and the hcf to make a set of 2 NUMBERS:

{48,12}

Now, take another one of the primes, and multiply one of the numbers in the set by it. There are 2 WAYS of doing this, so it will give you 2 sets of numbers, I will choose 3 as the NEXT prime:

{{3⋅48,12},{48,3⋅12}}={{144,12},{48,36}}

From here, you repeat the process. For each set of 2 numbers, multiply one of them by one of the remaining primes, which means each set of 2 will be duplicated. The only prime left for me to deal with is 5 in this case:

{{5⋅144,12},{144,5⋅12},{5⋅48,36},{48,5⋅36}}={{720,12},{144,60},{240,36},{48,180}}

Therefore, for this problem, the possible pairs of numbers are 720 and 12, 240 and 36, 180 and 48, and 144 and 60.