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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=60720÷12=60
Factor the number you got from that division into prime factors.
60=22⋅3⋅560=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=423−1=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=4812⋅22=48. Use that and the hcf to make a set of 2 numbers:
{48,12}{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}}{{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}}{{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
First, take the lcm and divide it by the hcf.
720÷12=60720÷12=60
Factor the number you got from that division into prime factors.
60=22⋅3⋅560=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=423−1=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=4812⋅22=48. Use that and the hcf to make a set of 2 numbers:
{48,12}{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}}{{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}}{{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
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