Lesson video

Hi, my name is Mr. Peters, and welcome to today's lesson.

In this lesson, we're gonna be thinking about the reasons why we can change a two-factor calculation into a three-factor calculation.

This is a really useful lesson again as we can apply this to our everyday mathematics and make our work more efficient and, ultimately, a lot easier.

Should we get started? So by the end of this lesson today, you should be able to explain the reasons for why you might change a two-factor calculation into a three-factor calculation.

Throughout this lesson today, we've got a number of keywords we're gonna be referring to.

Let me say them first and you can have a go afterwards.

The first one is array.

Your turn.

The second one is factor.

Your turn.

The third one is associative.

Your turn.

And the last one is efficient.

Your turn.

So an array is when items are arranged in either rows or columns.

Factors are whole numbers that exactly divide another whole number.

The associative law states it doesn't matter how you group or pair values in a calculation, the result would remain the same, and this can be applied in both addition and multiplication.

And then finally, working efficiently means finding a way to solve a problem, which is quick whilst also allows you to maintain accuracy.

Look out for these words throughout our lesson as we go.

This lesson today is broken down into two cycles.

The first cycle, we'll think about identifying three factors in an array, and the second cycle, we'll be changing two-factor calculations into three-factor calculations.

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

So here we've got an array.

In this array, we've used a number of dots to represent 16 multiplied by five.

Can you see where the 16 and the five are represented within our array? Take a moment to have a think.

Yeah, that's right.

We can see here that we've got one row here of 16 dots, haven't we? And if we continue that down, we have now got five rows of 16 dots.

So in each row or in each group, the 16 represents the number of dots in each row or group, and the five would represent the number of rows or number of groups that there are altogether.

Watch what happens now.

What do you notice about the array this time? Take a moment to have a think about how you could explain what you've just seen.

So what did you notice? Well, Sofia's saying that a number of counters or dots in each array has stayed the same.

There's not any more or there aren't any less.

However, each row or each group of 16 counters has now been divided into two rows of eight counters, hasn't it? So we can say that another way of writing 16 is eight multiplied by two.

Can you identify now where the eight multiplied by two is in our array? That's right, the eight multiplied by two represents a number in each group, doesn't it? We now also have a five which represents a number of groups we've got altogether.

So we can say that there are five groups of eight multiplied by two, or we could say there is eight multiplied by two five times.

So we can say that 16 multiplied by five is equal to eight, multiplied by two, multiplied by five.

Can you see what Sofia is saying and how our array has changed from representing two factors to now representing three factors? Well done if you can.

Having a look at our array in a bit more detail now, we can say that the eight and the two are associated, aren't they? They're in the same group.

There are five groups of eight multiplied by two.

So we can represent this using the associative law.

We can say that the eight multiplied by the two are associated, and we group that pair of factors together using the brackets.

The eight multiplied by two is what is in each group at the moment.

And then we know we can multiply that by the five for the number of groups and that will give us the total amount, which we know is 80.

We could, however, associate the two and the five in our calculation.

Have a look at our array now.

I've taken away the groups, however, what happens if I do this? What would you notice now? That's right, each group now has five rows and two in each row.

So we can represent each group as two multiplied by five or five multiplied by two.

And then we would have eight groups altogether, wouldn't we? So we could say that there are eight groups of two multiplied by five, or there is two multiplied by five eight times.

When we think about how we represent this as an equation, we can write eight multiplied by two multiplied by five.

But this time, in the groups, we associated the two and the five together, didn't we? So now we've got eight multiplied by two multiplied by five with the two and the five in brackets.

And again, this will give us exactly the same product, won't it? This calculation would become eight multiplied by 10, which would then become 80, wouldn't it? Here's another example, six multiplied by 14.

Have a look carefully what's happening.

That's right.

Where's the six and the 14 in our calculation? Well, the six represents a number of groups this time, doesn't it? And the 14 represents the amount of counters in each group.

Let's all think about how we can adapt these two factors into three factors now.

Have a look carefully.

What's happened to each row in each array? That's right, each row has been divided into two rows within the group.

So one row of 14 has now been divided into two rows of seven within the group.

So another way we can say 14 then is seven multiplied by two, isn't it? The seven multiplied by two represents the amount of counters in each group, and the six represents the number of groups.

So we can say that there are six groups of seven multiplied by two, or there is seven multiplied by two six times.

Again, we could rewrite our equation.

Six multiplied by 14 is therefore equal to six multiplied by seven, multiplied by two.

So our array was originally represented by two factors, and this new array is now representing three factors.

So having a think about this in more detail again, what numbers are associated with each other? That's right.

In a group, there are seven counters in two rows, aren't there? So we can say the seven and the two are associated with each other.

And we can write this as six multiplied by seven, multiplied by two in brackets.

We know that seven multiplied by two is equal to 14.

So now we can do six multiply by 14, and that would give us 84.

Now the six multiplied by 14 was a bit of a tricky calculation there.

I wonder if we can make that slightly easier.

Well, let's have a look at our array now.

What'd you notice happened to each of the groups this time? That's right.

The groups has now changed to two groups.

We had six groups in the last example, and now we've only got two groups, haven't we? And in each group what numbers are associated? That's right.

The six and the seven are associated because there are seven counters in each row and there are six rows in each group.

So we could represent this as two groups of six times seven or six times seven, two times.

Again, as a calculation, we can write six multiplied by seven in brackets because those are the factors that are associated.

And then we can multiply that by two.

Using our multiplication facts, I know that six times seven is 42 and then times in that by two, that gives us 84.

Sofia's saying that was a lot easier than doing six times 14, wasn't it? And I tend to agree with her.

You're right, Sofia, I think.

Okay, time for you to check your understanding now.

Tick the correct expression that represents the array.

Is it A, B, or C? That's right, it's B, isn't it? We can see that we've got five groups.

So the five represents the five groups.

And then in each group, we've got two rows of seven.

So we associate the two and the seven together by using the brackets.

Well done if you've got that.

Okay, this time we'd like to circle the array to show six multiplied by two, multiplied by five with the two and the five associated with each other.

Okay, hopefully you got that.

It would look a little bit like this, wouldn't it? We can now see that we've got six groups and then in each group we've got two rows of five counters.

So the two and the five are associated in each group.

Well done if you've got that.

Okay, time for you to check your understanding now.

I'd like you to have a think about how you can divide up each of the arrays to show the factors.

In the first one, you need to show the two factors and then the second and the third array, you need to show how it can represent three factors as well.

You need to do that for example A, and you also need to do that for example B.

Good luck with that, and I'll see you when you get back shortly.

Okay, welcome back.

Let's see how you got on.

So, for the first array, we can see that we've grouped this into seven groups with 15 in each group.

In the second example, we've associated the three and the five.

So we needed to create groups which have three rows with five in each group.

We can see we were able to do that for the majority of the ones going down vertically on the left hand side.

However, to fit the two on the other side wasn't as easy.

And at the moment, they looked like one column of 15, don't they? So we could draw some little dotted lines in here to show each of the rows and where five is represented in each one of those rows, couldn't we? And then finally for the last example, the five and the seven are associated.

So, and that means in each group we need to have five rows of seven counters.

And again, we've shown that we've got three groups and then we've got five rows of seven in each group.

Well done if you've got that one.

Okay, so having a look at the first array, then we can see that we've got five groups of 14 in each row, haven't we? In the second array, we can see that we need to have the five in our calculations.

So we're gonna let the five represent the number in each row and in each group, we're gonna have two rows.

So we're gonna do five multiplied by two to represent each group.

And then we're gonna have to multiply that by seven.

And we would associate the five and the two together in each group.

So that represents two rows of five counters.

And then the seven would represent the seven groups.

And then for the last example, let's sort of think about what would be associated here.

So again, we've got two rows and we've got seven in each row.

And there are five groups altogether.

So we can record this as five multiplied by two, multiplied by seven.

And then again here we can see that the two and the seven are associated with each other as they are the number of counters in each row, in each group.

So we've got five groups of two times seven.

Which calculation would be easiest to work out the total amount do you think? That's right.

I think it would be the middle one, wouldn't it? We could associate the five and the two together to make 10 and then multiply that by the seven to make 70 altogether.

Well done if you managed to get all of those.

Okay, onto part two of our lesson then, changing two-factor calculations to three-factor calculations.

Okay, have a look here then.

We've got 24 multiplied by five.

This could be quite a tricky calculation to do mentally at first and it's not in our times tables.

So how could we go around tackling this quite easily? Well, at the moment, our calculation has two factors and Laura is suggesting that we could work out the factors of 24 in order to help us make this calculation slightly easier.

"Shall we draw it as an array?" she's asking.

Sofia says, "Let's not bother drawing an array.

I think it'll take too long, and it'd be rather inefficient.

So I think we could do it without one." So let's have a look at our two factors to start off with.

Which one of these would be best represented as the product of two factors? Well, Sofia thinks the 24 would be the best.

If we find the factor pairs for 24, we might have lots of different options that we can use in order to make this calculation easier.

Let's see what we can come up with.

So we know that one and 24 are factors of 24.

We know that two and 12 are factors of 24.

We know that three and eight are factors of 24, and we know that four and six are factors of 24.

So these pairs of numbers multiplied together, don't they, to make 24.

Thinking carefully now, which one of these pairs would make our calculation the easiest? Laura suggesting that we look at 12 multiplied by two.

I wonder why she's thinking that.

Ah, that's right, because 12 multiplied by two, multiplied by five would then become the calculation, wouldn't it? And if we do 12 multiplied by two, multiplied by five, we can then apply the associative law.

Let's have a look.

12 multiplied by two, multiplied by five.

And then if we put the brackets around the two and the five, that's a nice, comfortable number to be working with, isn't it? 'Cause we know that two multiplied by five is 10, isn't it? So our calculation then becomes 12 multiplied by 10, which altogether is 120, isn't it? That's made our calculation a lot easier rather than doing 24 multiplied by five, isn't it? Sofia suggesting a different strategy.

She said she would've used the six and the four.

Let's see how that would've worked.

She said her calculation would've become six multiplied by four, multiplied by five, and then she would've associated the four and the five with each other.

So her calculation becomes six multiplied by four, multiplied by five, with the four and the five in brackets, that means we're gonna calculate the four and the five first, which gives us 20.

So it becomes six multiplied by 20, which would be again equal to 120 because we know that six multiplied by two is 12.

So six multiplied by 20 is 12O.

Two both very equally valid ways of tackling this problem.

Well done both of you.

Here's another example.

Looking at our factors again.

One of them is five and the other one is 36.

So which factor are we gonna convert into the product of two other factors? Well, having five as a factor is really useful.

'Cause if we can get a two, then we could make it into a 10, couldn't we? So let's think about the factors of 36 that we can think of.

Well, we could have one and 36, we could have two and 18, we could have three and 12, or we could have four and nine.

So which pair would you use this time to make your calculation easier? Laura is saying she'd use the two and the 18 again, just like we mentioned before.

That's because five multiplied by two multiplied by 18 is a lot easier to work out.

Five multiplied by two would be equal to 10.

So we now associate the five and the two together.

So that means five multiplied by two gives us 10, and then we multiply that by the 18, which gives us 180, doesn't it? Again, that makes our calculation a lot easier rather than the five times by 36.

Sofia is saying that we could have also used the four and the nine, couldn't we? We could have converted our calculation into five times four times nine and associated the five and the four together.

Associating the five and four together is nice because that always gives us 20.

And then we know that two times nine is 18, so 20 times by nine is 180.

Again, two great examples of calculations that are better off being broken down into three factors to solve.

Okay, time for you to check your understanding again.

Can you tick the factor pairs of 18? I'll give you a moment to think about that.

That's right.

It's B and C, isn't it? Two multiplied by nine is equal to 18, and three multiplied by six is equal to 18.

And have a look here.

Can you choose the most efficient calculation for 32 multiplied by five? Take a moment to have a think.

That's right.

It'd be A, wouldn't it? Because five is already a factor, it's really helpful to try and find two to be another factor.

So if the factor you are working with, with the five, is actually an even number, we know that we can always break it down into a factor of two and another factor.

So we can say here that 16 multiplied by two, multiplied by five would be the most efficient way because we can associate the two in the five together.

That gives us a 10, and then multiply that by 16 to give us 160.

Well done if you've got that.

Laura is now asking, how could we change calculation B and C to make them more efficient? Well, I'd say for B, we could associate the four and the five together rather than the eight and the four.

'Cause the four and the five will give us a 20.

And we know that two times eight is 16, so 20 times eight is 160.

And then for C, how would you do that? Well, actually, using the associative law might not be best here.

We might need to use the commutative law when you might need to change the order of the factors.

If we change the 16 and the five around, our calculation can then become two multiplied by five, multiplied by 16, and then we can associate the two and the five together at the front.

So two multiplied by five becomes 10, and then we multiply that by the 16 for the 160.

Okay, onto our final task for today then.

What I'd like you to do here is change these two-factor calculations into three-factor calculations.

An example has been given to you for the first one.

And then once you've gone through them, I want to ask yourself, what was it you were noticing, particularly in column B? And then for the question two, Andeep is calculating five multiplied by 26.

He says he has simplified his calculation to this.

Will he get the correct answer? Can you explain your thinking? Good luck with those two tasks, and I'll see you again shortly.

Okay, let's go through these together then.

So 32 multiplied by 50.

It would be best if we broke down 32 into two factors.

So that could be 16 and two.

Then we can associate the two and the 50 together to make a hundred and therefore give us our 1,600 when we multiply that by 16.

For the next one, we've got 20 and 15.

Well, if we break down the 15 into three and five as factors, then we can associate the 20 and the five together to make 100 and then multiply that by three, that would give us 300.

And the next one's 75 multiplied by four.

Well, if we break down the 75 into three lots of 25, we can write our equation as three multiplied by 25 multiplied by four.

We know that four lots of 25 is 100, and then we can multiply that by three again.

So that would give us 300 again.

Well done if you've got those ones.

Okay, and then working through the right-hand side, 18 multiplied by five.

So if we break down 18, so it becomes the product of two factors, we could do that into two and nine, couldn't we? And then we can associate the two and the five again 'cause we're working with a five.

So two multiplied by five is 10, and then multiply that by the nine, that gives us a 90.

Hmm, 180 multiply by five.

If we break down the 180 into its factors, again, we could break this into 90 and two, couldn't we? The two multiply by the five gives us a 10 again, and then we could multiply that by the 90.

So our answer is 900.

How does that relate to our last one? That's right, we could break 1,800 down into factors of 900 and two couldn't we? Again, multiply the two and the five together, that gives us 10 and then multiply all of that by 900, which would be equal to 9,000.

Let's have a look at the next row, 20 times by 25.

Well, we could break down either factors really here, but if we break down the 25, that will give us five and five as factors, won't it? So our calculation becomes 20 multiplied by five, multiplied by five, and then we can associate the 20 and the five together.

That would give us 100 and then multiply that by five and that would give us 500.

We could do 200 multiplied by 25 now.

Well, let's break that down into two factors.

So we keep breaking down 25 into five and five.

That means we can now associate the 200 and the five together.

That's gonna give us 1000, isn't it? And then multiply that by five, so our answer will be 5,000.

And then finally we've got 2000 multiplied by 25.

So again, break down the 25 into five and five as factors.

We can multiply 2000 by five, which gives us 10,000 and then we multiply that by five, which gives us 50,000, doesn't it? What did you notice then about the calculations in B? That's right, we were using our times table knowledge to still find the factors for numbers where they were 10 times the size or 100 times the size.

If I know the factors of 18, then I can also find out the factors of 180 or 1,800 by making one of those factors 10 times or 100 times the size.

Well done if you spotted that.

Okay, and on to Andeep's question here.

He says by breaking down 26 into factors of 20 and six, he can calculate this a lot more easily.

Did you agree with him? Sofia doesn't agree with him.

She says that he's accidentally partitioned 26 into 20 and six, and 20 and six aren't factors of 26, are they? 20 and six would add together to make 26, but they don't multiply together to make 26.

And we are looking for two numbers that multiply together to make the 26.

So he might be better off using two and 13 as Laura is suggesting as a factor pair.

Then he can multiply the five and the two together to make 10 and then multiply that with 13 to make 130.

Well done if you managed to get that one as well.

Okay, that's it for our learning today.

Hopefully you're feeling a lot more confident about how you can apply the commutative or the associative law to calculations where we did originally have two factors and then break them down into three factors instead.

So to summarise our learning.

When multiplying two numbers together, one of the factors could be expressed as a pair of factors and this may help us to make the calculation more efficient.

You can represent expressing two factors as three factors using an array, and then the associative law is particularly useful for helping us to find a pair of factors that could be multiplied together to make the calculation more easy and efficient.

Thanks for learning with me again today.

I've really enjoyed it.

I hope you have, too.

Take care and I'll see you again soon.