Lesson video

Hiya, my name is Ms. Lambell, really pleased that you've decided to pop along and do some maths with me today.

Welcome to today's lesson.

The title of today's lesson is HCF or LCM, and it's in the unit, Properties of Number.

By the end of this lesson, you'll be able to use prime factorization of two or more positive integers to efficiently find either the highest common factor or lowest common multiple, depending on the context of the question.

Now, lots of people find it really difficult to decide whether they need to be finding the highest common factor or lowest common multiple.

So you might find today's lesson challenging, but stick with it, you'll be fine by the end of the lesson.

A couple of things that we need to just refresh our memory on, and that's our abbreviation.

So our abbreviation for highest common factor is HCF, and our abbreviation for lowest common multiple is LCM.

Today's lesson is going to be split into three learning cycles.

The first one will be highest common factor in context, the second one, the lowest common multiple in the context, and then finally, whether to use the HCF or the LCM based on the context.

So we'll look at just HCF, and then just LCM, and then this will be the one that's a little bit more challenging, we need to be able to recognise from a context of a question whether we need to be finding the HCF or the LCM.

So we're gonna start with HCF in a context.

Group A has 30 students and group B has 42 students.

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

Here we would need to be finding the HCF, the highest common factor.

So here are our factors of 30.

So remember we start with one and the number itself.

So one multiplied by 30, and then we're gonna do them systematically and in pairs.

So two multiplied by 15, three multiplied by 10, and five multiplied by six, and now are factors of 42.

So one and 42, two and 21, three and 14, six and seven.

The highest number that appears in both of those lists, so the HCF of 30 and 42 is six.

The largest teams they could get into would be six because remember we said they needed the same number in each team.

Here's Andeep.

Now Andeep has two lengths of ribbon.

We can see the lengths of those ribbon are 315 centimetres and 240 centimetres.

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

What is the longest length of ribbon he can make? Now I think you'll agree that if we were to list the factor pairs of 315 and 240, that that wouldn't be the most efficient way of doing it.

But we know that we have a method of using the prime factorization of numbers to help us define the highest common factor or the lowest common multiple, and in this case, the highest common factor.

So here are our two products of primes, and here's our Venn diagram.

So remember we start with the smallest prime number that is in one of our products.

So that's two, and it's two to the power of four.

And we can see there are no factors of two in 315.

So it just needs to go in the 240.

Right, moving on to three, we've got a three squared and a three.

So three is common to both, but 315 needs the other factor of three.

And then we have five, and that's in both.

So it's going to go into the intersection, that middle part of our diagram, and then we have the number seven, which is just in 315.

So that's gonna go just in 315.

Can you remember how we find the highest common factor from a Venn diagram? I'm sure you can.

What we do is we multiply together, we find the product of the common factors.

So here our common factors are three and five, and the product of three and five is 15.

The longest length of ribbon that Andeep is able to make is 15 centimetres.

That's the product of the intersection.

Let's check our understanding up till now.

Sophia is working out the highest common factor of 35 and 45.

Without doing any calculations, I want you to decide how you know Sophia must be wrong.

So remember, no calculations, just looking at what she's got to say.

She says, "I've drawn a Venn diagram and the HCF of 35 and 45 is 315".

How do you know that Sophia is not right? Hopefully, you said something like, a factor of a number cannot be greater than the number itself.

So factors, remember the highest factor of a number is that particular number.

So 315 can't possibly be a factor of 35 or 45.

We're now ready to have a go at some questions independently, but as always, remember if you do think you need a bit more confidence than you maybe would want to go back and re-watch the examples.

There are a few questions in this task and each of them are on their own separate slide.

So what I'm gonna do is read the question with you and then you'll pause the video and then come back when you've got the answer.

But I'll read the question first.

Aisha's little brother is sorting his 30 cars and 50 bricks into equal rows.

What is the maximum number in each row if there are no rows with both cars and bricks? So now pause the video, work out your answer, come back when you're ready.

Let's move on then to question number two.

Jacob has 24 white, 42 red, and 48 yellow flowers.

What is the greatest number of identical bunches he can make that uses all of the flowers? So keyword here, identical bunches, and we need to use all of the flowers.

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

Question number three, Lucas is making a model.

He needs to cut his wire into the longest equal lengths possible.

One piece is 420 centimetres and the other is 120 centimetres.

How long can he make each piece? So this is very similar to the question we had Andeep with the ribbon.

Pause the video and have a go.

Brilliant, well done for coming back and persevering with these questions.

Superb.

Let's have a look at the answers.

So one, the answer.

So we've written 30 as a product of its prime factors and 50 as a product of its prime factors.

We could then use a Venn diagram if we wanted to, but we can see here that the highest common factor is two multiplied by five, which is 10.

Like I said, you could have drawn a Venn diagram.

Moving on to question number two.

So 24 written as a product of prime factors, 42 and 48 as well.

So here our highest common factor is the product of two and three, which is six.

And finally question number three, again, writing them as product of prime factors.

The highest common factor is two squared multiplied by three, multiplied by five, and that's 60.

Well done if you've got all of those right.

If you didn't, maybe have a look and see where you went wrong, but I'm sure you did get them right.

Let's now move on to our second learning cycle.

So our second learning cycle is now thinking about lowest common multiple in a context.

So we've just looked at highest common factor and we're now going to look at lowest common multiple.

Alex, Sam, and Laura are making daisy chains.

Alex's daisies are all three centimetres long.

Sam's are all five centimetres long, and Laura's are all six centimetres long.

They want to make them all the same length.

How long will the chains be? Maybe pause the video and have a think about it.

Let's have a look then.

So let's first put two of Alex's daisies together.

So now the length of Alex's chain is six.

Oh, that matches with Laura, but it doesn't match with Sam.

So we'll add another daisy on for Sam, which is 10 centimetres.

Now Alex and Laura still match, but Sam doesn't.

So let's catch Alex up.

We'll add a third one onto Alex's chain.

That's nine centimetres and then we'll add another one on, that's 12 centimetres.

Let's go and we'll catch Laura up.

So Laura's gonna add on her second flower or a second daisy, and now hers is 12 centimetres.

So again, we've got Alex and Laura.

They've each got a chain which is 12 centimetres long, but Sam hasn't.

So we are gonna add in another daisy for Sam.

So Sam's chain now is 15 centimetres long.

So let's catch Alex up.

So oh, Alex's is now 15 centimetres long.

So Alex and Sam now match, but Laura doesn't.

Or does she when we add a third daisy into her chain? No, that takes us too far.

She's now got a chain which is 18 centimetres.

So let's go back to Alex.

So Alex now adds another one.

His is now 18 centimetres.

Oh, and that matches Laura, but Sam's doesn't match again.

So let's add on another daisy for Sam.

Sam's chain is now 20 centimetres long.

That hasn't really helped.

So we'll now go back and we'll add another daisy on for Alex and let's add another daisy on for Laura.

We've now got all three of them are different.

I'm going to go back to Alex and add another daisy onto his chain.

Now Alex and Laura's of the same length, but Sam's isn't.

Will there ever be a time when Sam's is the same as Alex and Laura's? It's not looking like there's going to be.

Let's add on another one for Sam then.

So Sam's is now 25 centimetres.

So still doesn't match Alex and Laura's.

Let's add another daisy on for Laura.

Laura's is now 30 centimetres.

Add another one for Alex, 27, and another one 30, right? So we've got Alex and Laura, theirs are the same again.

Is Sam gonna match this time? Thank goodness for that he does.

Sam's daisy chain is now the same length as Alex's and Laura's.

The shortest daisy chain they would be able to make is 30 centimetres.

We're now going to look at a different problem.

Izzy is splitting her stickers into smaller groups.

She can make groups of 10, 12, or 28 with none left over.

So if she were to sort them out, she could make powers of 10, powers of 12, or powers of 28.

We want to work out what is the smallest number of stickers that she has.

Now here we will be looking to find the lowest common multiple because we will be making groups of stickers.

So we'd be increasing the number of stickers each time.

Here are those three integers written as a product of their prime factors because we now know that we can use those to help us efficiently find the lowest common multiple of two or more integers.

And here's our Venn diagram.

So just as we've done previously this lesson, we're going to place our prime factors into the Venn diagram, starting remember with the lowest one, which is two.

So we can see that two appears in all of them.

The one of the two's appears as in all three.

So that's going to go into the middle.

And then 12 and 28 also share a second factor of two.

So that's going to go in the 12 and 28 intersection.

We're then going to take each of the other primes.

But as you notice here, there are no other sharing of prime integers, sorry, prime factors.

So we are going to just put them in the correct place.

So we're going to put the three in, 12, the five in the 10, and the 28 in the seven.

Now have a think.

Can you remember what we need to do to find the lowest common multiple? What was it we did to find the lowest common multiple? So the smallest number of stickers she can have is 420.

How did I find that? So remember to find the lowest common multiple of two or more integers, we multiply together all of the prime factors in our Venn diagram.

Now I'd like you to have a go at this check for understanding.

Which Venn diagram and answer is correct for the lowest common multiple of two cubed multiplied by three squared multiplied by five squared and two squared multiplied by three cubed multiplied by five? So which one gives is the correct Venn diagram, but also gives the correct answer finding the LCM? Pause the video, have a go at this, and then come back when you're ready.

Superb.

Well done.

Let's have a check in then and see which one is the correct one.

So the correct answer was C.

So if we have a look here, we can see that filling in our Venn diagrams, that they both share a factor of two squared.

So it couldn't have been Venn diagram A.

So we've now narrowed that down.

And if actually we look, the two Venn diagrams are exactly the same for B and C.

So we are then just looking to find which one has found the correct lowest common multiple.

So here remember we need to multiply together all of the prime factors in the Venn diagram, which is two cubed, three cubed, and five squared, giving us 5,400.

Common mistake here is B.

What this person has done, they filled in that Venn diagram absolutely perfectly and well done to them.

But what they haven't, what they have done, sorry, is they have found the highest common factor and not the lowest common multiple.

You're now ready to have a go at some of these independently.

Here we are going to have one question on each slide.

So I'm gonna read the question with you and then you are gonna pause the video and have a go.

But I'll read the question first with you.

So Jun goes swimming every four days and he goes to the cinema every 15 days.

Which is the first day that he does both? Pause the video, have a go at this question, remembering to use a Venn diagram if you want to, and then come back when you are ready.

So question number two, Jacob is baking.

He can bake 12 cupcakes and 20 biscuits at a time.

He needs to make the same number of each.

What is the smallest amount he can make? So pause the video and then come back when you're ready.

And question number three.

So there's lots of information here, but remember, some of it, we'll be able to sort of ignore.

Let's pick out the key bits of information.

So I'm gonna read it through and then I'll pick out the key bits of information to help you.

So Sam was making sandwiches for a party.

Bread comes in loaves of 24 slices, cheese and packs of 15, and ham in packs of 40 slices.

Each sandwich has two slices of bread and one slice of cheese and ham.

What's the fewest of each he needs to buy to ensure there is no wastage? So key things to remember here is that he's going to use two slices of bread for each sandwich and one slice of cheese and one slice of ham.

We want to know the fewest of each that he needs to buy.

Good luck with this one.

It's a little bit more challenging the previous two, but I have every faith that you'll be able to do it.

Pause the video and have a go.

Great work.

Let's have a check of those answers now then.

So Jun goes swimming every four days and he goes to the cinema every 15 days.

Jun's very lucky as far as I'm concerned.

What is the first day that he does both? So write in firstly, write in our two numbers as a product of their prime factors.

And then we are remembering to put that into a Venn diagram if we need to.

I've not shown the Venn diagram here, but the correct answer was 60.

So on the 60th day, he would go swimming and to the cinema.

Very busy, Jun.

Question number two.

So Jacob and his cupcakes and his biscuits.

I hope he's making those for some particular event, not just to eat himself.

So here again, write them as prime factors.

And then the lowest common multiple was 60.

So the lowest that he can make of each is 60.

Well done if you got that right.

And then question number three, like I said, this one was a little bit more challenging.

Each sandwich uses two slices of bread.

So first thing we wanted to do was to think about how many sandwiches could we make from one loaf of bread.

So one loaf of bread, there are 24 slices, meaning we can make 12 sandwiches.

So actually instead of writing the prime factorization of 24, I've done 12 because I know that I need to use two pieces of bread for each sandwich.

So written out the prime factors and then put those into the Venn diagram, or maybe you can do it without, that's absolutely fine either way.

I actually prefer the Venn diagram.

And the lowest column multiple is 120.

So he needs to buy 120 of each item.

He may also have gone to work out how many loaves of bread that was and how many packs of cheese and how many packs of ham.

Well done if you pushed your learning on and you had a go at that.

We are now ready to move on to our final learning cycle.

So we now know how to find the highest common factor and the lowest common multiple in a context.

But you've been in a learning cycle that told you it was HCF or LCM.

So here, I'm going to be wanting you to work out for yourself whether you need to be finding the highest common factor or the lowest common multiple.

Let's take a look.

So Oak Academy are having a fete.

They're going to sell hot dogs.

And sausages come in packs of 12 and rolls come in packs of nine.

They don't want any leftover, so they don't want some extra sausages left over or some extra rolls left over.

So the need to buy the same number of each.

So in this problem, would we be finding the highest common factor or the lowest common multiple? So like I said, what is the smallest number of hotdogs they can make? Here is our Venn diagram.

I've filled it in for you.

So we've got 12 as prime factorization and nine, and I filled them in.

What would we need to find, HCF or LCM? So key thing to think about here is they would be buying multiple packs, they'd be buying multiple packs of sausages and multiple packs of rolls.

So that then gives us a bit of a clue as to we will be finding the lowest common multiple.

Can you remember how we did that? Was it just the prime factors that were common or was it all of them? Well done, you're right.

Yes, it's all of them.

So the lowest common multiple is two, multiplied by two, multiplied by three, multiplied by three, which is 36.

So the smallest number of hotdogs that they can make is 36, but they'll probably make multiples of that to make sure that they've got plenty for the fete.

Another problem now then, we've got two schools are going on a trip.

They need to get the students across a lake in boats.

Without mixing the schools and making sure the groups are the same size, what's the smallest number of crossings needed? So here, have a think.

Again, we'll be finding the HCF or the LCM.

So there were 120 students at school A and 156 at school B.

There's our Venn diagram.

So we'll be finding that HCF or the LCM.

Brilliant, well done.

It's the HCF.

So key thing here is on the previous example, we were buying multiple packs.

Here we're splitting the students into smaller groups.

Okay, so we are finding factors.

How do we find the highest common factor? Yep, you've got it, well done.

We multiply together the common factors, factors that are in the intersection.

We find the product of those.

So two squared multiplied by three, which is 12.

We are ready now to have a go at this check for understanding.

Andeep is making bracelets.

He wants each one to be the same.

He has 24 red beads, 32 yellow beads, and 50 white beads.

He says, "To work this out, I need to find the LCM".

Pause the video and have a think about whether you agree or disagree with Andeep.

Okay, so hopefully you came up with no, he needs to work out the highest common factor.

So it's very much like the previous question we were doing.

The LCM would mean increasing the number of beads, wouldn't it? We would be buying multiple packs of 24 red beads, multiple packs of the yellow beads.

And he needs to be splitting the beads down into smaller groups to make each of the bracelets.

So those are the two key differences between them to hopefully help you find, am I'm looking for the HCF or the LCM.

Right, ready to go now.

Last task.

Well done for sticking with it.

You've done fantastically so far.

Let's finish off on a high note.

Question number one.

First thing you're going to do is you're going to decide if you would be finding the HCF or LCM in each of those following.

So I'm not going to read all of them out.

I'm gonna get you in a moment to pause the video and then you can read them through at your own pace.

Here all I'm asking you to do is identify whether you need to find the HCF or the LCM.

You don't have to work out the answers.

When you're done, you can come back.

Pause the video, good luck.

Now I'm going to make it a little bit more challenging because this time, I want you to do both things.

I want you to identify whether you need to be finding the highest common factor or lowest common multiple, but I also then want you to go on and to find that highest common factor or lowest common multiple.

So pause the video and then work through each of the problems. When you are ready, come back and we'll check those answers.

That was quick.

Well done.

Let's have a check in then and see how we've got on with those.

I'm pretty certainly, you've probably got them all right 'cause you've done so well so far already this lesson.

1A, we needed to find the LCM, so we're going to be each time increasing the number of steps.

So that's gives us the clue that it's the lowest common multiple.

1B was the highest common factor.

So this time, Aisha is going to be splitting the people down into groups.

So we're gonna be looking for a factor.

C was lowest common multiple.

Again, we've got granddad who's baking a cake and he needs to make sure that he's got the smallest number of pieces.

So he'll be cutting them into multiple number of pieces.

And then moving on to question two, part A, the answer was six.

Part B, it was 30, and then part C, so this was a little bit more challenging again.

I tried to make that last question a bit more challenging for you 'cause you are ready for it.

So here we needed to find the lowest common multiple, which was 264 minutes.

But then I needed to convert that into hours.

So I knew that four hours was 240 minutes, so I had 24 minutes left.

If they left at 9:00 AM and then four hours, 24 was the next time that they were both there, we know that they leave the same time again at 13:24.

There you might have 1:24, but remember you would need the PM to show that it was in the afternoon.

Now let's summarise the learning that we've done this lesson.

So in real life situations, we can find ourselves needing to find both the highest common factor and the lowest common multiple.

So we looked at those separately, but ultimately, we need to be able to recognise that for ourselves.

So we then looked at different context in real world situations that rely on being able to identify the highest common factor or lowest common multiple is required to solve the problem.

You've done fantastically well today.

There was some really challenging stuff in there.

Well done for sticking with it.

Thank you so much for joining me.

I've really enjoyed it.