LCM for particular set of factors

[Master Blaster]

Hi

I'm trying to help my son (Year 7) with a question regarding factors and multiples. The question states that 60 (for example) has a lot of factors, including the numbers 1, 2, 3, 4, 5 & 6. There are two parts to the question:

1. Find the lowest number that has factors of 1, 2, 3, 4, 5, 6, 7, 8, 9 & 10.

2. Express, in written form, how you can determine the smallest number that contains a particular set of factors.

I asked my son if they have learned prime factorisation yet.
They haven't, but it appears in a unit of work later in his schooling.

For Question 1:

I told my son that if he multiplies all the factors together (1 through 10), that will find a number that is divisible by all the factors, although not the lowest number.

I also told him that its possible to remove factors that are doubling up. For example, he can remove 2 and 5 (provided he keeps 10), as the number 10 is already divisible by 2 and 5. Similarly, 3 can be removed because of 6, and 4 can be removed because of 8. That leaves 5, 6, 7, 8, 9 & 10. I know that other numbers can be removed, but without using prime factorisation (which he hasn't learned at school, yet), I wasn't able to explain how to do it.

Also, I am at a loss to answer Question 2, which is to explain all the above in written form.

 

[Dr.Peterson]

Hi

I'm trying to help my son (Year 7) with a question regarding factors and multiples. The question states that 60 (for example) has a lot of factors, including the numbers 1, 2, 3, 4, 5 & 6. There are two parts to the question:

1. Find the lowest number that has factors of 1, 2, 3, 4, 5, 6, 7, 8, 9 & 10.

2. Express, in written form, how you can determine the smallest number that contains a particular set of factors.

I asked my son if they have learned prime factorisation yet.
They haven't, but it appears in a unit of work later in his schooling.

For Question 1:

I told my son that if he multiplies all the factors together (1 through 10), that will find a number that is divisible by all the factors, although not the lowest number.

I also told him that its possible to remove factors that are doubling up. For example, he can remove 2 and 5 (provided he keeps 10), as the number 10 is already divisible by 2 and 5. Similarly, 3 can be removed because of 6, and 4 can be removed because of 8. That leaves 5, 6, 7, 8, 9 & 10. I know that other numbers can be removed, but without using prime factorisation (which he hasn't learned at school, yet), I wasn't able to explain how to do it.

Also, I am at a loss to answer Question 2, which is to explain all the above in written form.

I'd like to know what methods he has learned for finding an LCM. But I would suggest doing it one step at a time.

To have factors 1 and 2, clearly it must be a multiple of 2. Bring in 3, and you need the LCM of 2 and 3, which is 6. To also be a multiple of 4, it must be a multiple of the LCM of 6 and 4, which is 12. Just keep going.

 

[pka]

Hi

I'm trying to help my son (Year 7) with a question regarding factors and multiples. The question states that 60 (for example) has a lot of factors, including the numbers 1, 2, 3, 4, 5 & 6. There are two parts to the question:

1. Find the lowest number that has factors of 1, 2, 3, 4, 5, 6, 7, 8, 9 & 10.
2. Express, in written form, how you can determine the smallest number that contains a particular set of factors.

Your son might explore this link . It can used to factor any integer.

 

[Master Blaster]

I'd like to know what methods he has learned for finding an LCM. But I would suggest doing it one step at a time.

To have factors 1 and 2, clearly it must be a multiple of 2. Bring in 3, and you need the LCM of 2 and 3, which is 6. To also be a multiple of 4, it must be a multiple of the LCM of 6 and 4, which is 12. Just keep going.

Hi, and thanks for the reply.

I believe the methods he has been taught for finding an LCM include:


1. By observation after writing out the multiples of each factor.

2. Finding LCM of any two numbers, by multiplying the numbers together and dividing by their HCF.

 

[Dr.Peterson]

I believe the methods he has been taught for finding an LCM include:

1. By observation after writing out the multiples of each factor.

2. Finding LCM of any two numbers, by multiplying the numbers together and dividing by their HCF.

I think either of those should be suitable for the step-by-step approach I suggested.

 

[Master Blaster]

I think either of those should be suitable for the step-by-step approach I suggested.

Thanks again.
Now that I read your previous post again, I see the point you make.

Just to confirm, the process you outline for 3 or more numbers is to start with finding the LCM of the lowest two numbers, then work up from there one number at a time.

It didn't occur to me previously, but (for example):

Given LCM (2,3) = 6
Then, LCM (2,3,4) = LCM (6,4) = 12

And,
Given LCM (2,3,4) = 12
Then, LCM (2,3,4,5) = LCM (12,5) = 60
etc

This didn't appear obvious at first, but makes sense when thinking it through.

There was nothing in the textbook that alluded to how to calculate LCM for a large group of factors.
Perhaps if there was something in the textbook hinting at the process, the kids could learn faster?

 

[Dr.Peterson]

There was nothing in the textbook that alluded to how to calculate LCM for a large group of factors.
Perhaps if there was something in the textbook hinting at the process, the kids could learn faster?

If I were helping in person, I'd be looking through the book to see what has been said. This is really easier to think through in terms of prime factors, but perhaps something else in how the concept was defined would be of use.

My guess is that this problem is meant to get students thinking about how things work, and maybe even lead toward the discovery of the usefulness of primes. For example, after 60, the example that was provided, the next divisor you need is 7, and they might realize that nothing so far is a multiple of 7, so they have to multiply 60 by 7 to get the smallest multiple of 1 through 7.

 

[pka]

1. Find the lowest number that has factors of 1, 2, 3, 4, 5, 6, 7, 8, 9 & 10.

If I were tutoring one of our grandchildren on the above, I would show this.
That has all those factors, but that is clearly not the least such number.
If we had \(2^3\) we get factors \(2,~4,~\&~8\) if we add \(3^2\) we get \(3,~6,~\&~9\).
By adding \(5\) we get factors \(5~\&~10\) all we need a \(7\). LOOK HERE
Try to find a smaller number.

 

[Otis]

… without using prime factorisation (which he hasn't learned at school, yet) …

Good idea!

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