Lesson video

Hi, there.

My name is Ms. Lambell.

You've made a really good decision, deciding to join me today to do some maths.

Come on then let's get started.

Welcome to today's lesson.

The title of today's lesson is Checking and Securing Understanding of HCF and LCM, and that's within the Unit: Arithmetic Procedures and Index Laws.

By the end of this lesson, you'll be able to calculate the lowest common multiple or the highest common factor efficiently.

HCF is an abbreviation for the highest common factor.

LCM is an abbreviation the lowest common multiple.

During this lesson, we will be using prime factorization, which is a method to find the prime factors of an integer.

After we have used prime factorization, we will then write numbers as a product of their prime factors.

Today's lesson is split into three learning cycles.

In the first one, we will review finding the HCF and the LCM.

In the second one, we will use a prime factor form to help us with this, and then in the final learning cycle, we will look at solving problems. Let's get going with that first one.

So we'll look at reviewing the HCF and the LCM.

What is the HCF of 24 and 60? How could we go about answering that question? Well, we could list the factor pairs of both numbers.

The factor pairs of 24 are one and 24, two and 12, three and eight and four and six.

And then we can list the factor pairs for 60, and these are one and 60, two and 30, three and 20, four and 15, five and 12, and six and 10.

We want to know what the highest common factor is.

So we are looking to identify the highest number that appears in both lists.

The highest one that is common, which in this case is 12.

The HCF of 24 and 60 is 12.

What is the lowest common multiple of 15 and 18? We can list the multiples of 15, and we can list the multiples of 18.

Here is a list of multiples of 15.

Obviously, that list continues, and here are my multiples of 18.

90 and 90, that's the lowest number that appears in both lists.

So 90 is the lowest common multiple of 15 and 18.

Group A has 40 students.

Group B has 56 students.

They need to get into the largest teams they can, but each group must have the same number in each team.

Let's list the factors of 40, one and 40, two and 20, four and 10, five and eight.

The factors 56, one and 56, two and 28, four and 14 and seven and eight.

Now we can identify the highest common factor, which is eight.

The largest teams they would be able to get into is teams of eight.

Oak Academy are having a school fete and they will be selling cheeseburgers.

They want to ensure that they have nothing left over.

What is the smallest number of cheeseburgers they can make? Burgers come in boxes of 18, rolls in packs of 12, and cheese slices in packs of 30.

We need to work out the smallest number of complete cheeseburgers that they can make.

I'm going to list my multiples of 18, 12, and 30.

Here are my multiples of 18.

Why did I stop at 90? I know that the number of cheese slices is a multiple of 10 as the one's digit is zero and so I'm stopping at 90 'cause I think this could be my answer.

Now I'm going to list my multiples of 12.

I don't get to 90 and I'm gonna list my multiples of 30.

We can clearly see now that 90 appears for burgers and cheese slices, but not for rolls.

We need to continue until we have a number that is common to all three.

So I continue, and I can now see the smallest number of cheeseburgers that they can make is 180.

We've identified here the lowest common multiple.

Which of the following is the LCM of 12 and 18? Which is the lowest common multiple of 12 and 18? And it's C.

If I were to list the multiples of 12 and 18, 36 would be the first number that appeared in both lists.

Which of the following is the HCF of eight and 20? Here the answer is four.

If I rotate my factor pairs, the highest factor in both would be four.

Now it's your turn, you're going to work out the LCM and HCF of the following.

Pause the video and then come back when you are ready.

And here are your answers.

A, the LCM was 24, HCF four.

B, LCM 45, HCF three.

C, LCM 108, the HCF was six.

D, the LCM was 168 and the HCF was 14, and E the LCM was 1000 and the HCF was 25.

Now let's move on to that second learning cycle.

We are now going to use prime factors to help us find the HCF and the LCM.

This is gonna be particularly useful if our numbers are a lot larger because we don't really want to be writing out all of the factors of very large numbers.

We're going to find the HCF of 60 and 102.

Now we know that prime factors of an integer can be found using prime factorization, and when we have integers written as a product of their prime factors, we can find any factors, and therefore we are able to find the HCF of two or more numbers.

I'm going to find the HCF as 60 and 102, like I said.

60 written as product of its prime factors is two squared multiplied by three multiplied by five, and 102 is two multiplied by three multiplied by 17.

We need to identify the highest factors which are common to both products.

Sophia says this will be the highest exponent of each common factor.

Do you agree with Sophia? Sophia, unfortunately, is incorrect.

You need to identify the lowest exponent, which is common to both.

Why is it the lowest exponent and not the highest? For example, here two squared is a factor of 60, but only two is a factor of 102.

Sofia now says, it will be the product of two and three, which is six.

The highest common factor of 60 and 102 is six.

We're now gonna find the highest common factor of 10800 and 14040.

You can see now why we would not want to be listing out all of the factor pairs of those two numbers.

Here are each of them written as a product of their prime factors.

Identify the highest factors which are common to both products.

The common prime factors are two, three, and five.

We need to ensure that we have identified the lowest exponent of each.

So the LCM is, we've got exponent of four and an exponent of three, and so we choose the lowest exponent.

Moving on to the powers of three, we've got an exponent of three and exponent of three, so the lowest exponent is three, so we're gonna multiply that by three cubed.

Now onto the powers of five, we've got an exponent of two.

Remember, if you can't see an exponent, it's an exponent one, and so the lowest is five.

This gives us the HCF of 10800 and 14040 is 1080.

It's the product of two cubed multiplied by three cubed multiplied by five.

Find the HCF of 60 and 102.

We can identify the highest factors which are common to both products.

The common factors are two and three.

Ensure we have identified the lowest exponent of each.

So the common factors of two and three, the lowest exponent of each was one, but we don't need to write it, so the HCF of 60 and 102 was six.

This now double checks with what we got earlier.

Let's try this one.

Again, here they are written as a product of their prime factors.

Identify the highest factors which are common to both products.

The common prime factors are two and three.

Ensure we identify the one with the lowest exponent, so it's going to be two squared multiplied by three.

The HCF of 180 and 168 is 12.

Sophia is working out the HDF of 35 and 175.

35 is five multiplied by seven, and 175 is five squared multiplied by seven.

What mistake has Sophia made? The highest common factor of 35 and 175 is five squared multiplied by seven, which is 175.

Pause a video and decide what mistake Sophia has made, and even better, you can give me the correct answer.

She's used the highest exponent of each common factor and not the lowest.

How could Sophia have known her answer must be wrong? A factor of a number cannot be greater than the number itself.

So the HCF cannot be greater than 35 in this case.

What is the lowest common multiple of 24 and 40? We know that a factor multiplied by a factor gives a product, and we know that a factor multiplied by a factor also gives us the multiple.

For an integer to be common multiple of both numbers, it must be the product of the highest common factor.

Here we've got two cubed multiplied by three, that's 24, and we've got two cubed multiplied by five, which is 40.

We can see that two cubed is clearly common to both.

We need to multiply that factor by another factor, and the factor needs to be the same in both numbers.

So these two boxes here need to be the same, and so therefore I need multiply by five in the top one and multiplied by three in the bottom one.

This means that the LCM of 24 and 40 is two cubed multiplied by three, multiplied by five, which is 120.

You can also use a Venn diagram to help you.

Two cubed is common to both, so it goes in the intersection.

Three was a prime factor of 24, but not 40, and five was a prime factor of 40 but not 24.

Hopefully, you've noticed now that actually the calculation I did for LCM, two cubes multiplied by three multiplied by five is actually the same as the product of all of the numbers in the Venn diagram.

Find the lowest common multiple of A and B.

What integers are A and B? To work that out, we're just going to, on our calculator, do two cubes multiplied by three square multiplied by five, which is 360, and two squared multiplied by three multiplied by five squared multiplied by seven, which is 2,100.

Here are the two products of prime factors and what we're going to do is we are going to do it numerically, but we're also going to do it with a Venn diagram so that you can see how the two are linked.

We've got two cubes and two squared.

What's common to both? The highest common factor of both is two squared.

It's common to both, so it's going to go in the intersection of the Venn diagram.

The top one, however, also has another factor of two, so we need to make sure that we've got that other factor of two and put that into our Venn diagram.

Now let's look at our powers of three.

Three squared and three.

They both have a factor of three, and so in the Venn diagram that goes in the intersection, but the top number has an additional factor of three, so we're going to write that and then put that into our Venn diagram.

Now moving on to powers of five, common to both is five.

Put that in both of them, and then in the Venn diagram in the intersection.

And the bottom one this time has an extra factor of five, so let's put that and put it into our diagram.

And then we've got seven, it's not common to both of them and so it's going to go just on the right hand number and just on the right hand side of the Venn diagram.

We're looking to make sure that we are creating a multiple of both of the numbers.

We can see the common part is two squared multiplied by three multiplied by five, and I need to multiply that by the same thing in both of my products.

So two multiplied by three and then five multiplied by seven.

But these two things need to be the same.

So here I need to introduce the factor of five and seven, and on the right hand number, I need to introduce the factor of two and three.

The LCM of A and B is two cubed multiplied by three squared multiplied by five squared multiplied by seven, and you can get this either from the calculations or from the Venn diagram, and this is 12,600.

I'm going to find the LCM of A and B.

I'm going to use the Venn diagram.

Two squared and two cubed.

Two squared is common to both, so it goes in the middle, and then there is an additional factor of two for B.

Now let's look at the powers of three.

Three is common to both, but B also has an additional factor of three.

Five squared is only in A and seven is only in B.

To find the lowest common multiple, we find the product of all of the numbers in the Venn diagram.

The LCM of those two numbers is 12,600, and now it's over to you for this one.

Here's your question, pause the video, when you are ready, come back and we'll check your answer.

Good luck.

How did you get on? Let's check.

That's what your Venn diagram should have looked like and then that would give you an answer of 3,240.

Andeep has two lengths of ribbon, one is 315 centimetres long and one is 240 centimetres long.

He needs to cut them into the longest equal strips he can with no waste.

What is the longest length of ribbon that he can make? We can also use a Venn diagram to help us see the common factors.

Here are 315 and 240 written as a product of their prime factors.

We can put these into our Venn diagram, two to the power of four only appears it's in 240.

Then we look at our powers of three.

We can see that three is common to both.

The 315 has an additional factor of three.

Five is common to both, and seven is just a factor of 315.

We can then find the length of the ribbon.

The longest length of ribbon here is 15.

That's the product of the intersection.

Here we were finding the highest common factor.

A machine makes X chocolates every minute.

The chocolates can be packed into boxes of 10, 12, and 28.

What is the smallest value of x? Here are each of those numbers written as a product of prime factors and here is a Venn diagram.

Let's start with the powers of two.

Two is common to all of them, so it's going to go in the centre, and there is an additional two that is common to the 12 and the 28, it's going to go here.

And then we can see that there are no other common factors, and so we just need to put these in the correct place in the diagram.

Five is only a factor of 10, three is only a factor of 12, and seven is only a factor of 28.

The smallest value of x is 420, that's the lowest common multiple.

Which Venn diagram and answer is correct for the LCM of two cubed multiplied by three squared multiplied by five squared, and two squared multiplied by three cubed multiplied by five.

Pause the video, make your decision, and come back when you are ready.

What did you decide? The correct answer was C.

Now for Task B, you can use your preferred method, so whichever method you like that we've talked about, and then you are going to work out the HCF and the LCM of A and B in the following.

Pause the video, good luck with this, and then return when you are done.

And question number two, Oak Academy are having a school fete and they will be selling cheeseburgers.

They want to ensure that they have nothing left over.

What's the smallest number of cheeseburgers they can make? Burgers come in boxes of 15, rolls in packs of 12, and cheese slices in packs of 25.

Pause the video and work out how many burgers we can make.

Remember, we want to know the smallest number.

And question number three.

Three schools send their year 10 students to an event.

They must get into the largest teams they can without mixing schools.

What's the largest team size they can use? Again, pause the video and come back when you're ready.

Great work.

Now let's check those answers.

One A, the answer was HCF was 12, LCM was 3,300, and B, HCF 252 and LCM was 3,528.

Onto question two, we were there identifying the lowest common multiple and that was 300.

And question number three, you needed to find the highest common factor which was 12.

And now we can move on to that final learning cycle, which is problems involving HCF and LCM.

X is an integer.

The LCM of x and 105 is 420, and the HCF of x and 105 is 15.

We need to find the value of x.

This type of problem is much easier to do if we use a Venn diagram.

Here's my Venn diagram.

So on the left I've got my prime factors of x and on the right my prime factors of 105.

We know that the HCF is the product of the intersection.

Remember, that's the bit in the middle.

We know from the question that the HCF is 15 and the prime factors of 15 are three and five, so we know that three and five need to go in the intersection.

In here, we are going to need to place the other prime factors of 105.

Well, 105 is three multiplied by five multiplied by seven, and we've already got the three and the five, so we need to put the additional factor of seven into the prime factors of 105.

We're also told that the LCM is 420, and we know that the LCM is the product of all numbers in the Venn diagram.

So far the product of the factors in our Venn diagram is 105.

The LCM is 420.

So I do 420 divided by 105, which leaves me four, which is the product of two and two.

And so, therefore, these are the two remaining prime factors of x and they go here.

Therefore, x must be the product of these prime factors.

The ones in the loop for x.

X is two squared multiplied by three multiplied by five, so x is 60.

Let's have a go at another one.

Same problem, but different numbers this time.

So we're going to solve it with a Venn diagram.

We know that HCF is the intersection, and we know that the HCF is four, so we're going to put two squared in here.

You could put two and a two, I've decided to write it in its exponent form.

The other prime factors of 12 go here.

We've already got four.

Four multiplied by what makes 12? That's three.

Then the LCM is the product of all of the numbers in the Venn diagram.

So far we've got 12, so 120 divided by 12 is 10, and 10 is the product of two and five.

So two and five are going to go here, and now we can work out what x is by finding the product of all of the prime factors in x and that is in here.

So it's going to be two cubed multiplied by five, which is 40.

X is an integer.

LCM of x and 20 is 60, and the HCF of x and 20 is five.

Which of the following is the correct first step to solve this problem? And that's B.

The five is the highest common factor, it needs to go in the intersection.

Which is the next step, the correct second step, to solve this problem? And that is C.

We need to make sure that the prime factors of 20 make 20.

So five we already had, we need to multiply that by four, which is two squared.

And what would be the final correct step? And the final correct step is B.

We've already got 20.

The product of the numbers that were in our Venn diagram to start with were 20, but we know the product of all of them needs to be 60.

20 multiplied by what makes 60? That's three.

So three must have been the missing prime factor.

Just to finish it off then, now I'd like you to tell me what integer x is.

And it's 15 because we find the product of the prime factors that are in the x part of the Venn diagram.

Now, Task C, I'd like you to have a go at this one first, so pause the video and then come back when you're ready.

Question number two.

Question number three.

It's a lot more challenging this one, so pause the video, but you've got all the skills you need to be successful and then I'll be waiting when you come back to check your answers.

Super work, well done.

Question number one, you had two and seven in the middle, and then three and two and five giving us the x was 140.

Two, two and three squared in the middle, three and five prime factors of 270 giving us the missing prime factor of the x was two squared, that means the x is two cubed multiplied by three squared, which is 72.

And let's take a look at the answer to question three.

So we've got five is the highest common factor of A, B, and C, and then we can fill in our other prime factors, and then we can find the values of A and B.

A was two and B was also two.

That was really challenging that last one.

How did you get on with that? Brilliant.

Well done.

Now let's summarise the learning from today's lesson.

The HCF and LCM of two or more integers can be found by listing the factors and multiples.

The HCF and LCM of two or more integers can be found by using their prime factorizations.

When using a Venn diagram, the HCF is found by finding the product of the intersection, and the Venn diagram, to find the LCM, you find the LCM by finding the product of all of the prime factors in the diagram.

Well done during today's lesson.

You've done fantastically well.

Look forward to seeing you again really soon, and take care of yourself and goodbye.