Lesson video

Hi, my name's Mr. Peters.

Thank you for joining me today for this lesson.

In this lesson, we're gonna be thinking about how we can apply our understanding of the properties of numbers, as well as how we can use the commutative and associative laws to help simplify our calculations and make them a lot easier for us to work out mentally.

This is a really useful lesson and will help you to work quickly and efficiently with numbers mentally.

If you're ready to get started, let's get going.

So by the end of this lesson today, you should be able to say that I can use the properties of numbers, as well as the commutative or associative laws to help me simplify calculations.

Throughout this lesson today, we've got three key words we're gonna be referring to.

I'll have a go at saying them first, and then you can repeat them after me.

The first one is factor, your turn.

The second one is commutative, your turn.

And the last one is associative, your turn.

Thinking about what these mean in a bit more detail, factors are whole numbers that exactly divide another whole number.

The commutative law states that you can write the values of a calculation in a different order and it won't change the calculation.

The result of the calculation will remain the same.

It applies to both addition and multiplication.

Meanwhile, the associative law is slightly different.

The associative law states it doesn't matter how you group or pair the values in a calculation, for example, which order you decide to calculate the pairs in, the result will remain the same.

And again, this can apply to both addition and multiplication.

Throughout this lesson today, we've got two cycles.

The first cycle, we're thinking about manipulating one factor, whereas the second cycle, we'll be thinking about manipulating two factors.

If you're ready, let's get started.

Throughout our lesson today, we've also got Sam and Jacob, who'll be joining us, sharing their thinking, as well as any questions that they've got along the way.

Okay, so Sam and Jacob are actually spending some time looking at different two digit times two digit calculations.

Sam's having a look at this one here, it's 20 multiplied by 25, and he thinks there's gonna be lots of different ways that we could calculate this.

Let's have a look.

Jacob thinks using our factors would definitely be a useful way to help us calculate this.

He says that he knows 20 can be composed of two factors.

He knows 20 could be composed of one multiplied by 20, it could be composed of two multiplied by 10, or it could be composed of four multiplied by five.

We can now change this two factor calculation into a three factor calculation, as you can hopefully see underneath now.

So we just need to choose which pair of factors we should apply into our three factor calculation now.

Well, Jacob has said that he's noticed that we're going to be multiplying by 25.

Then he's thinking, hmm, 25 is quite a nice number if I can find another number to multiply it with.

So he's looking to find a number that he can multiply with the 25 to make this calculation easier to calculate.

There we go.

He spotted two multiplied by 10.

He's saying he could multiply the 25 with the two, first of all, and then he could multiply all of that by the remaining 10.

At the moment, we can't do that.

If you look at our calculation, we've got the two and the 10 next to each other and the 25 is to the right hand side of the calculation.

We're going to need to rearrange the factors in our calculation, and we can do that using something which Sam has referred to as the commutative law.

So now we've rearranged our factors.

We've got 10 multiplied by two multiplied by 25.

That's right, Jacob.

Now we've got those two factors next to each other.

We can associate those two factors together.

To do this, we can draw some brackets around those two factors, which means that we're going to multiply these two factors first.

So we can now multiply these two factors.

Two multiplied by 25 is equal to 50, isn't it? And we are now gonna be multiplying that 50 by the remaining factor that was 10.

So our calculation is now 10 multiplied by 50.

Could you have a go at figuring that out? That's right, 10 multiplied by 50 is equal to 500, isn't it, Sam? Well done.

Sam's saying, I saw that you used the two and the 10 as factors, but actually I prefer to use the four and the five as factors.

Let's have a look at our calculation now.

Hmm, I wonder why Sam preferred to use four and the five as factors of 20.

Have a think for yourself quickly.

So let's see what Sam's thinking was then.

Well, he's realised that, again, 25 can be multiplied by four, can't it? That will give us 100.

That's a nice easy number to multiply with.

So at the moment, our calculation reads four multiplied by five multiplied by 25.

We're gonna have to apply the commutative law again to rearrange the factors.

And now that we have rearranged or reordered the factors, we can associate, can't we, the four and the 25 together using the brackets.

Now we can multiply these two numbers.

Four multiplied by 25 is equal to 100, and then we can apply, and then we can rewrite our calculation as five multiplied by 100, which we know, again, is equal to 500.

Nice work, you two.

Two different ways of solving this calculation so far.

I love your thinking.

Sam found another way as well, he thinks.

We could have also manipulated the other factor.

So far, we've been manipulating the 20, haven't we? And using that 20 to become the product of two factors.

We're now going to use the 25 to become the product of two factors.

Let's have a look.

That's right, Jacob.

25 could be decomposed to five multiplied by five, couldn't it? So there we go.

We can now see the factors of 25.

We know that one pair of factors could have been one multiplied by 25, but again, that's not really gonna help us in this calculation, whereas decomposing five into five multiplied by five is gonna help us with this calculation.

Hmm, why is that gonna help us? Take a moment to have a think.

Well, let's have a look at how this is recorded as a three factor expression now then.

We can record it as 20 multiplied by five multiplied by five, can't we? And you're right, this is now gonna make our life easier, isn't it? Because we can now associate the 20 and the five with each other.

We know that 20 multiplied by five is equal to 100.

So once again, we've now got 100 multiplied by five, and that is equal to 500 again.

Three different ways of solving this calculation, all of which have been the result of our factor knowledge of numbers.

We've changed the calculation now slightly and we're gonna look at a different calculation.

I wonder, maybe take a moment for yourself, how do you think you might tackle this calculation? Let's see how Sam and Jacob tackled it.

Jacob is saying that he knows 24 can be composed of a different range of factor pairs.

Let's have a look.

We could compose 24 of one and 24.

It could be two and 12, it could be three and eight, or it could be four and six.

Once again, we can write now this as an expression of three factors being multiplied with each other.

And Jacob's saying, hmm, 35's an interesting number.

It's not gonna allow us to make a number that could be 100 this time as it's not a factor of 100, but it might be helpful to take this 35 and turn it into a multiple of 10.

So what could we multiply this 35 by to make it into a multiple of 10? That's right, we could multiply it by two, couldn't we? Two lots of 35 would be equal to 70.

So if we chose the two and the 12 as our factor pair here, then, we've now got two multiplied by 12 multiplied by 35.

We can reorder the factors, can't we, by using the commutative law.

So we've now got 12 multiplied by two multiplied by 35.

And of course, now we can associate the two and the 35 together.

We've now got 12 multiplied by 70.

Hmm, I already think this is a bit of an easier calculation to tackle than 24 multiplied by 35.

That's because as Sam's pointing out, we know 12 multiplied by seven is 84.

So 12 multiplied by 70 is gonna be 10 times larger.

So it's gonna be 840.

That's definitely allowed us to use timetable facts to help us solve that calculation quickly and easily, hasn't it? And now we're gonna think about how we could have tackled it using the other factor as well.

Jacob is saying 35 can be made up of different factors.

Let's see, it could be one and 35, it could be five and seven.

Hmm, we know the one and 35 isn't gonna help us, so we can use the five and seven.

I wonder how that would help us.

Well, let's rewrite our expression now into three factors.

24 multiplied by five multiplied by seven.

Sam's saying that he knows 25 multiplied by five is 125.

He knows that four 25s are 100, so five 25s are 125, but we don't want five 25s, do we? We want five 24s, or we want 24 lots of five.

So instead of having 25 lots of five, we could have 24 lots of five.

That means we just need to take off one lots of five, don't we? So if we know that 25 lots of five is 125, then we can say that 24 lots of five is 120.

So our calculation now becomes 120 multiplied by seven.

And again, we can use our times table facts to help us here, can't we? We know that 12 multiplied by seven is 84, so 120 multiplied by seven would be 840.

Well done if you managed to spot that as we went along.

Okay, time for you to check your understanding now.

I'm gonna have a go first, and then you're gonna have a go on the other side.

We're going to tackle the same calculation, however, you're gonna decompose a different factor to the one that I am.

So I'm going to do 22 multiplied by 15, and I'm going to decompose the 15.

We know that 15 could be composed of one and 15, or it could be composed of three lots of five, couldn't it? So we're going to use the three and the five here.

We're gonna write our expression like this, 22 multiplied by three multiplied by five.

Hmm, what might be easiest to work out here? Well, I know that two multiplied by five is 10, and 20 multiplied by five is 100.

So I think the multiplying the 22 by five would be a lot better than multiplying by the three.

So 22 multiplied by five, well, that gives me 110.

And now all I need to do is multiply that by three.

110 multiplied by three is equal to 330.

There we go.

Your turn now, could you have a go? You need to try and get 330.

So you know you've got it right if you've got that.

Have a go at decomposing the other factor.

Okay, let's see how you got on.

We know 22 could be composed of one and 22 or two and 11.

So let's use the two and 11, and let's write this as a three factor calculation now.

We've got two multiplied by 11 multiplied by 15 here.

Hmm, would it be easier if we multiplied the two by the 11 this time? That would just take us back to our original calculation of 22 multiplied by 15.

So we don't wanna do that, do we? But we could reorder the factors here using the commutative law.

Two multiplied by 15 would give us 30.

And then we've got 30 multiplied by 11, haven't we? We know that three times 11 is 33, so 30 times 11, again, would be 10 times the size, so that becomes 330.

Well done if you managed to get that.

Okay, onto our first task for today then.

What I'd like you to do is have a go at solving each of these equations by manipulating one of the factors.

Good luck, and I'll see you back here shortly.

Okay, the first one, 25 multiplied by 16.

You may have chosen a different strategy to the one that I'm going to show you, and that's absolutely fine.

Let's see how you got on.

I've chosen here to decompose the 16.

I know 16 is the product of four multiplied by four.

So taking one of those factors of four and multiplying that by 25 will allow me to get to 100 nice and easily, won't it? So let's rewrite our calculation as 25 multiplied by four multiplied by four.

Associate the 25 and the four and multiply them together.

That gives us 100.

100 multiplied by four is equal to 400.

The second one, 15 times by 42.

Here, I'm going to decompose the 42.

Let's have a look at the factor options we could have chosen.

Lots of different options here.

I'm looking for a factor from that list that would be useful to multiply with 15.

Well, I've actually gone for six and seven here.

The reason for that is because we can rewrite our calculation as 15 multiplied by six multiplied by seven.

I'm gonna associate the 15 with the six.

And I know that 15 multiplied by six is equal to 90.

So now I've got 90 multiplied by seven.

I know that nine multiplied by seven is 63.

So 90 multiplied by seven is 630.

And the third one, 18 multiplied by 33.

This time, I chose to go for the 18.

I decomposed 18 into these different factors here.

I wonder which pair you chose if you did it like this.

I chose to use the three and the six as the factor pair.

We rewrite our calculation of three multiplied by six multiplied by 33.

Now, here, I had to think a little bit carefully.

I decided to associate the three and the 33 together.

The reason for that is that I know three multiplied by 33 is 99, and 99 is nearly 100.

So now I've got six multiplied by 99.

I know that six multiplied by 100 is 600.

So six multiplied by 99 is the same as saying 99 lots of six.

If 100 lots of six is 600, then 99 lots of six is just taking off one lot of six.

So instead of 600, it would become 594.

There we go.

Hopefully, you managed to get those three answers as well using a strategy which you felt was convenient for you.

Okay, well, now we're gonna move on to our second cycle, manipulating two factors.

Okay, so in this cycle, as we said, we're gonna be breaking down each of the factors in the calculation and seeing how we can be even more flexible with those.

So let's start off with breaking down our 16 into two factors.

We know that these are the factor options for 16.

It could be one and 16, two and eight or four and four.

And then let's also break down the factors of 36.

So we've got one and 36, two and 18, three and 12 or four and nine.

Now, we could use any two pairs of these factors here and we could combine them into a four factor expression like we're going to now underneath.

So which pairs should we use to make our calculation easier, do you think? Take a moment for yourself.

Well, let's use Sam's strategy.

Sam's suggesting we use the two and the eight, as well as the four and the nine.

Hmm, I wonder why he's chosen those.

Let's place them in here.

You can see that they've been reordered as well.

Sam and Jacob have put them in a different order in the calculation.

Let's see why they chose to do that.

So they've decided to multiply the nine and the eight together to start off with.

So we can apply the associative law here to multiply these two together first.

Nine multiplied by eight is equal to 72.

So our calculation now becomes 72 multiplied by four multiplied by two.

Hmm, what would they do next? So as Jake has pointed out, we can now associate the 72 with the four and multiply these two together.

This would give us 288.

And now that we've got 288, we can just double that by multiplying that by two.

288 doubled is equal to 576.

You may feel that made your calculation slightly easier.

I certainly felt that multiplying 72 by the four and then multiplying that by two was a lot easier than trying to multiply 16 by 36.

Okay, and let's have a look at one more example as well then, shall we? We've got 54 multiplied by 24 this time.

Let's find the factors that we could have for 54, and let's list the factors we could have for 24.

Hmm, which pair of factors should we use this time? Sam's suggesting we use the six and the nine and the two and the 12.

Let's write this as a four factor expression now.

And as Jacob has said, we're going to use the commutative law to put them in an order which he thinks is gonna make our calculation easier.

So he suggests that we start off with by multiplying the 12 and the nine with each other.

12 multiplied by nine we know is 108.

So now our calculation reads 108 multiplied by six multiplied by two.

We can now associate the 108 with the six.

So 108 multiplied by six, that's an easier calculation, isn't it? We know that 100 times by six is 600, and six eights are 48.

So altogether, that's gonna give us 648.

And then to finish off, we just need to do 648 multiplied by two.

648 multiplied by two, I can do that mentally, that's 1,296.

There we go, two calculations made easier by using our understanding of our factors, and as well as applying the commutative and the associative laws.

Jacob says he enjoys trying to find lots of different ways that he can tackle a calculation using his factors, and that's a really great way to feel, Jacob.

Well done, you.

Okay, time for you to check your understanding now.

True or false, 28 multiplied by 32 is equal to eight multiplied by seven multiplied by four multiplied by four.

Take a moment to have a think for yourself.

Yep, and the answer's true, isn't it? Have a look at the justifications.

Which one of these helps you? That's right, it's B, isn't it? It is true.

The factors have just been reordered using the commutative law, hasn't it? We know that the factors of 28 could be four and seven and the factors of 32 could be four and eight, and they've just been placed in a different order.

Well done if you've got that.

And here's another check for understanding.

Can you choose a pair of factors that would help make this calculation easier? I'll give you a moment to have a think.

That's right, it's A, isn't it? Two and 12, why would that make it easier? Often, having one of the factors as a two means that all we need to do at one point is do some doubling, which will certainly make our calculation easier if we're confident with doubling numbers.

However, while some people may feel that this made it easier, you may have felt that B or C made it easier, and that's okay too.

Whatever makes you feel more confident and in the end allows you to be more efficient with your calculating.

Okay, and our final practise for today then.

What I'd like you to do here is solve these expressions again, but this time breaking down both of the factors so they are the product of two other factors.

And then for task B, what I'd like you to do is tackle this little problem here.

I'd like to use the digits zero to nine, and you can only use them once.

And what you need to do is create a calculation which is two digits multiplied by two digits which would be equivalent to a four factor expression.

I wonder what's the largest number you could possibly make? Good luck with that, and I'll see you back here shortly.

Okay, let's see how you got on then.

Here are the list of factors you could have had for 48 and for 15.

Again, you might have chosen a different pair of factors to the pair that I'm going to choose here to complete this calculation.

And that's fine, as long as you got to the same answer at the end and you felt it was efficient for you.

So this time, we're going to use the four and the 12 and the three and the five.

That gives us a four factor expression that looks like this.

We can then associate the 12 and the five with each other and multiply them together.

That gives us 60.

So now we've got 60 multiplied by four multiplied by three.

This is a lot easier because 60 multiplied by four is 240, as we know, because six times four is 24, and that's just 10 times larger.

And now we've just got to do 240 multiplied by three, and that gives us 720.

Well done if you've got that.

For the second one, here's the list of factors you could have chosen from.

For this one, we've chosen eight and nine and two and six.

Here, they are listed as a four factor expression.

This time, however, we decided to multiply the nine and the six with each other by associating these two factors in the centre of our expression.

We know that nine multiplied by six is equal to 54.

So our calculation now becomes eight multiplied by 54 multiplied by two.

We're then going to associate the 54 and the two with each other.

That gives us 108, and now that leaves our calculation as eight multiplied by 108.

Again, I know that 100 multiplied by eight is 800, and eight eights are sixty four.

So our answer will be 864.

And the last one, 66 multiplied by 18.

Here are the list of factors you could have chosen from.

We're gonna choose two and 33 and three and six for this one.

Here it is, written as a four factor expression again.

Notice this time to speed us up even more so, I'm going to associate both of these pairs together.

So we've got two multiplied by six now, and we've got 33 multiplied by three.

Two multiplied by six is 12, and 33 multiplied by three is 99.

Ah, and look again, we've got a factor that's nearly 100 again, haven't we? But this time, it's 99.

We know that 100 lots of 12 is 1,200, so 99 lots of 12 would be 12 less, and that gives us 1,188.

Well done if you managed to get all three of those and you found an efficient way to tackle them all.

And onto our second task then.

Hopefully, you might have found your own way to tackle this.

There might have been lots of different solutions that you came up with, and that is equally fine.

Here's a solution that we came up with.

We used 20 multiplied by 63, and we said that was equal to five multiplied by four multiplied by nine multiplied by seven.

The five and the four factors came from the 20, and the seven and the nine came from the 63.

I wonder, did you have a strategy which enabled you to fill in the boxes using the digits zero to nine only once.

Notice how our product here was 1,260.

That was the largest number at this moment that we found.

But we're not saying that is the largest number that it could possibly be.

You might have found a larger one.

If so, well done.

Jacob's saying you could also extend this challenge to try and find the smallest product you possibly could as well.

If you've got some extra time on your hands, you might like to investigate that for yourselves.

Okay, that's the end of our lesson for today.

I really enjoyed that lesson.

Hopefully, you did too.

To summarise what we've learned, we can think about our learning is this.

Calculations of large factors can have one or two of their factors broken down into their respective factors.

Finding a factor that can multiply another number to make a multiple of 10 can often make calculations easier.

And finally, knowing your times tables will allow you to find factors more quickly, and as a result, allow you to be more flexible with your calculations.

That's the end of our learning for today.

Thanks for joining me.

Take care, and I'll see you again soon.