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Maths Homework Help
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Please God, can someone help with my daughters homework.
The HCF of 2 numbers is 6. The LCM of the same two numbers is 72.
What are the numbers?
I'm twisted and my head hurts.
Please help :)
The HCF of 2 numbers is 6. The LCM of the same two numbers is 72.
What are the numbers?
I'm twisted and my head hurts.
Please help :)
Answers
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For more on marking an answer as the "Best Answer", please visit our FAQ.You could also try 18 and 24 (18*4 = 24*3 = 72)...
which makes me suspect that either whoever wrote the question forgot to check that there wasn't a second answer, or that they are using a (wrong) definition of Least Common Multiple that stops the LCM of two numbers from also being one of those two numbers (ie, that stops 6 and 72 from also being an answer, which it is). Or perhaps you were meant to find both solutions.
which makes me suspect that either whoever wrote the question forgot to check that there wasn't a second answer, or that they are using a (wrong) definition of Least Common Multiple that stops the LCM of two numbers from also being one of those two numbers (ie, that stops 6 and 72 from also being an answer, which it is). Or perhaps you were meant to find both solutions.
Also I figured I'd try and explain the method for solving problems like this:
1. If you are told the LCM (least common multiple) of two numbers, and the highest common factor or the same two numbers, then step 1 is to write the LCM and HCF as products of prime numbers (ie 2,3,5,7,11,13...).
So, here, 6 = 2x3 and 72 = 2x36 = 2x2x18 = 2x2x2x9=2x2x2x3x3 (finding these factorisations is a matter of patience, but basically all you do is check prime numbers in turn to see if they divide the number, so it is doable).
2. Take the factorisation of the LCM, and split off from it the factors from the highest common factor. That's a bit wordy, but what I mean is that you can write that 72 = 2x2x3x(2x3), and 2x3 = 6 as before. Put these factors aside in a second line:
LCF: 6 = 2x3 (Set 1)
LCF out of HCM: 6 = 2x3 (set 2)
left-over factors: 2, 2, 3
3. The exercise now is to arrange what is left such that all prime factors are used up but no more prime numbers are common to the two numbers you end up building. Here this can be achieved in two ways: giving all the left-over factors to Set 2 (the 6 and 72 solution), or giving all remaining 2's to Set 1 and the last 3 to Set 2 (giving 24 and 18).
That explanation really needs a picture to clear up, I fear... I will try and produce one tomorrow, if you like? But hopefully the above can be followed. It's hard to tell as I'm a bit tired right now.
1. If you are told the LCM (least common multiple) of two numbers, and the highest common factor or the same two numbers, then step 1 is to write the LCM and HCF as products of prime numbers (ie 2,3,5,7,11,13...).
So, here, 6 = 2x3 and 72 = 2x36 = 2x2x18 = 2x2x2x9=2x2x2x3x3 (finding these factorisations is a matter of patience, but basically all you do is check prime numbers in turn to see if they divide the number, so it is doable).
2. Take the factorisation of the LCM, and split off from it the factors from the highest common factor. That's a bit wordy, but what I mean is that you can write that 72 = 2x2x3x(2x3), and 2x3 = 6 as before. Put these factors aside in a second line:
LCF: 6 = 2x3 (Set 1)
LCF out of HCM: 6 = 2x3 (set 2)
left-over factors: 2, 2, 3
3. The exercise now is to arrange what is left such that all prime factors are used up but no more prime numbers are common to the two numbers you end up building. Here this can be achieved in two ways: giving all the left-over factors to Set 2 (the 6 and 72 solution), or giving all remaining 2's to Set 1 and the last 3 to Set 2 (giving 24 and 18).
That explanation really needs a picture to clear up, I fear... I will try and produce one tomorrow, if you like? But hopefully the above can be followed. It's hard to tell as I'm a bit tired right now.