Highest common factor: "The highest common factor of two numbers is 6. The lowest common multiple of the same two numbers is 72. What are the number?"

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Ehands1221

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Hello,

I am an adult learner who has recently started working through an old maths book in my spare time.

I am stuck on a particular question that follows:

"The highest common factor of two numbers is 6. The lowest common multiple of the same two numbers is 72. What are the number?"

The answer should be: 18 and 24, if someone could give me some advice that would be great.

Thanks.
 
Two numbers whose HCF is 6 can be written as 6x and 6y, where x and y are relatively prime.

Their product is (6x)(6y) = 36xy; what is their LCM? If that is equal to 72, what must xy be? What can x and y be? What are 6x and 6y?
 
Thanks for your prompt response.

I think 36xy is the LCM, so I could set that equal to 72 and get xy = 2. I guess x and y could be either 2 or 1? Then 6(2) = 12; 6(1) = 6?
 
Not quite. I left one step of the work unstated.

The LCM of 6x and 6y is not their product; this is true in general of any two numbers that are not relatively prime. If you think about some of the ways you can find an LCM, you should see that. In fact, one way to find the LCM of a and b is LCM(a, b) = ab/HCF(a, b).

Do you see that 6xy is a multiple of both 6x and 6y, and is less than 36xy? What can you conclude?
 
I am clutching at straws a little bit, but using ab/HCF, the LCM of the two numbers is 36/6 = 6? But then we are given the LCM in the question, so is it of any use?
 
Ehands1221 said:
I am clutching at straws a little bit, but using ab/HCF, the LCM of the two numbers is 36/6 = 6? But then we are given the LCM in the question, so is it of any use?
Click to expand...
That's the LCM of 6 and 6. We want the LCM of 6x and 6y, right?
 
So maybe 36xy/6 = 6xy. This gives us the LCM of 6x and 6y. 6xy = 72; xy = 12? Thank you so much for your help so far.
 
Now x and y have no common factors.

12 = 1*12 = 2*6 = 3*4. What next?
 
Ehands1221 said:
I am an adult learner who has recently started working through an old maths book in my spare time.
I am stuck on a particular question that follows:
"The highest common factor of two numbers is 6. The lowest common multiple of the same two numbers is 72. What are the number?"
The answer should be: 18 and 24, if someone could give me some advice that would be great.
Click to expand...
I am an adult instructor, so I have never understood not giving clear rules.
Suppose \(A=2^3\cdot3^2\cdot 5~\&~B=2\cdot3^4\cdot 7^2) \) then
GCD(A,B) stands for Greasiest Common Divisor, the largest divisor of both.
LCM(A,B) stands for Least Common Multiple , the least multiple of both.
Thus to find \(GCD(A,B)=2\cdot 3^2\) find the common factors and the highest common power of each.
Also \(LCM(AB)=2^3\cdot 3^4\cdot 5\cdot 7^2\) is the product of each factor and the greatest power of each factor.
The common factor must have factors common to both.
The common multiple must have factors in both.
 
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Thank you everyone for your help.

I have written out the factors of 72 and I can see that the answers: 18 and 24 are there which is promising?

The GCD 6 and 72 is 6? Whilst the LCM is 72?
 
As Jomo suggested, you needed to find a pair of relatively prime numbers (my x and y) whose product is 12. There are two such pairs, 1*12 and 3*4. These lead to solutions (18, 24) and (6, 72). In both cases, the GCD (called HCF in your context, which is fine) is 6, because that is the greatest number that evenly divides both numbers, and the LCM is 72, because that is the smallest number that is a multiple of both. The answer of 6, 72 feels "trivial" in some sense, but is a valid answer.
 
Ehands1221 said:
Why do you think the textbook has not included the answer (6,72)?
Click to expand...
Either they didn't think carefully, or they consider that a "trivial" solution that should not be mentioned. Possibly they said that somewhere else. (I'm assuming you quoted the entire problem, and there was nothing in the context that would eliminate such an answer.)

But I think they just forgot to consider it (as I did at first, because I hadn't written anything out).
 
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