Lesson video

Hello, my name is Mr. Tazzyman.

Today I'm gonna be teaching you a lesson from the unit that is all about multiplying and dividing by two-digit numbers.

There might be a few procedures to follow today, but it's also important that you understand why we do each step of the procedure as well.

Okay, I hope you're sitting comfortably, 'cause we're ready to start learning.

Here's the outcome for today's lesson then.

By the end, we want you to be able to say, "I can explain when it is efficient to use factorising or long multiplication to multiply by two-digit numbers." These are the keywords or phrases that you're going to expect to hear during the lesson.

We've got associative law and efficient or efficiently.

I'm gonna say these and I want you to repeat them back to me.

I'll say, "My turn," say the word or phrase and then I'll say, "Your turn." Ready, my turn.

Associative law.

You turn.

My turn, efficient or efficiently.

Your turn.

Brilliant.

Now let's have a look and see what they mean.

The associative law states that it doesn't matter how you group or pair values, i.

e.

which we calculate first.

The result is still the same.

It applies for addition and multiplication.

Working efficiently means finding a way to solve a problem quickly whilst also maintaining accuracy.

Here's the outline then for today's lesson.

We're gonna start by thinking about factorising or long multiplication, which do you choose? And for the second part, we're gonna look at choosing the most efficient strategy.

Sam and Andeep are gonna join us on our journey again.

Hi, Sam, hi, Andeep.

They're gonna be discussing some of the maths and that will help you to learn.

Ready to start? Let's go.

Sam and Andeep are figuring out how many Olympic badges will be produced if there were 16 badges for each of the 324 events at the Olympics.

Here's Sam's method and here's Andeep's method.

What do you notice? Which is more efficient? So you can see that Sam has used a long multiplication method here, whereas Andeep has decided to use his understanding of the associative law by factorising 16 into eight and two and using a short multiplication method to work out two partial products.

Both methods use multiplication to find the product.

Sam's method uses long multiplication and Andeep's method used factorising to find the products.

You could say Sam's method is more efficient as there as less instances of regrouping.

For the marathon event, there are 243 athletes participating and each athlete needs 35 kits prepared for them.

What's the total number of kits required? What do you notice, which is more efficient? Here's Sam's method and here's Andeep's method.

Both methods use multiplication to find the product.

Sam's method uses long multiplication and Andeep's method uses factorising to find the products.

You could say Sam's method is more efficient as there are less instances of regrouping.

Sam's method also multiplies by five and three tens using three times tables facts.

She may be more confident with these than the seven times table.

Okay, let's check your understanding then.

Which method would you use to solve the problem and why? Justify your thinking to your partner.

We've got the multiplication expression 125 multiplied by 46.

If you were going to find the product, would you use, A, factorising or B, long multiplication? Pause the video and have a think.

Welcome back.

The factors of 46 are one, 46, two and 23.

It is not more efficient to multiply 125 by 23 and then by two, so long multiplication is the more efficient choice.

You can see that B has been ticked there.

Sam and Andeep are organising t-shirts for the Olympics.

Each of the 24 racks holds 30 t-shirts.

How many t-shirts are there in total? We've got the expression written down there, 30 multiplied by 24.

How many ways can the equation be solved using factors? I wonder.

Both of those are composite numbers, so we know there might be lots of different ways.

Here they are.

We've got 30 multiplied by 24 is equal to three multiplied by 10 multiplied by 24, 15 multiplied by two multiplied by 24, six multiplied by five multiplied by 24.

In those first three examples, you can see that the factor of 30 has been factorised to give two other factors.

In the next ones I'm about to read out, it was 24 that was factorised instead.

So 30 multiplied by 24 is equal to two multiplied by 12 multiplied by 30 or three multiplied by eight multiplied by 30 or six multiplied four multiplied by 30.

There are so many ways we can calculate the total by factorising one of the factors.

Which would be most efficient? Have a look at all those multiplication expressions, the ones that we calculated and worked out.

Which of those would be easiest? Which would be most efficient? Mm, what do you think? Could you explain why as well? Andeep says, "I prefer this equation, because I can multiply by three, then scale by 10." That makes sense.

Or this one, because I can double 30, then use six multiplied by 12 to find 60 times 12.

Okay, let's check your understanding again.

How many ways can the equation be solved using factors? Pause the video and see if you can work that out.

Welcome back.

Here's what you may have got.

19 multiplied by 30 is equal to two multiplied by 15, multiplied by 19 or three multiplied by 10 multiplied by 19 or five multiplied six multiplied by 19.

19 itself is a prime number, so that only has two factors.

That meant that that couldn't be factorised, but 30 was factorised in each of these examples.

Okay, let's check further.

Which of these equations is the most efficient way to calculate the product? You got 19 multiplied by 30, which we've just thought about and we've got our three different options.

Which do you think would be most efficient way? Is it two multiplied by 15 multiplied by 19 or three multiplied by 10 multiplied by 19 or five multiplied by six multiplied by 19? Have a think and pause the video to decide.

Welcome back.

We ticked B.

You may have said B because it will be more efficient to multiply by three then by 10 and this could be done using a mental strategy.

Okay, it's time for your first practise task then.

You will need a set of zero to nine cards.

In pairs, take turns to generate a three by two digit multiplication equation.

Use both the factorising and long multiplication method to calculate the product.

Which method did you prefer and why? So you might say something like this, "I preferred using the long multiplication strategy, because" or "I preferred using the factorising strategy, because." Here are the cards.

Pause the video here, have a go at that and I'll be back shortly with some feedback.

Welcome back.

You might have got something like this.

Sam and Andeep drew up 243 multiplied by 18.

Sam used long multiplication whereas Andeep used factorization.

Andeep said, "I preferred using the factorising method, because it was easy for me to double then multiply by nine." So what Andeep had done is, he used the associative law to factorise 18 into two and nine.

He worked out firstly 243 multiplied by two, then he worked out the product of that multiplied by nine to get the overall product.

Let's move on to the second part of the lesson then, choosing the most efficient strategy.

The swimming pool venue is 110 metres long and 69 metres wide.

What is the area of its floor? What is known, what's unknown? Well, the pool venue is 110 metres long and it is 69 metres wide.

110 and 69 are the factors.

What is unknown is the total are of the floor, that's the product.

So we've got the multiplication expression, 110 multiplied by 69 and that is equal to, at the moment, an unknown, something we need to calculate.

Which method is the most efficient to use and why? Andeep says, "Using long multiplication can be more efficient than factorising when multiplying a three-digit number by a two-digit number." Here's the long multiplication.

He gives the product of 7590 and there's some regrouping involving, both in the multiplication parts to find the partial products, but also in the addition of the partial products as well.

"I'm not sure I needed this layout for this one though," says Andeep.

"I could've partitioned 110 into 100 and 10 and multiplied those by 69.

10 multiplied by 69 is equal to 690.

100 multiplied by 69 is equal to 6900.

6900 plus 690 is equal to 7590.

That was more efficient for me," says Andeep.

The Olympic event is selling tickets.

Each ticket costs 12 pounds.

How much do 25 tickets cost? What's known, what's unknown? Well, we know 12 pounds and we know 25 tickets.

So 12 multiplied by 25, they are our factors.

What's unknown is the total cost of the tickets, that's the product.

Andeep has a go at factorising to do that and he factorises 12 into four multiplied by three.

"I found this more efficient than using long multiplication," says Andeep, "I could multiply 25 by four mentally." Is there another strategy for this example? Well, you could use partitioning in this way.

25 multiplied by 10 is equal to 250, 25 multiplied by two is equal to 50, 250 plus 50 is equal to 300.

So in this method, the 12 pounds has actually been partitioned rather than factorised.

It's been partitioned into 10 and two.

"I can also partition either factor.

12 is equal to 10 plus two, so I can calculate mentally this way too," says Andeep who summarised what we've just talked about.

Okay, it's time to check your understanding then.

Which method is most efficient to use and why? Justify your thinking to your partner.

The multiplication expression we've got is 219 multiplied by 38.

Do you think that would be more efficiently solved using factorising or long multiplication? Pause the video and have a go.

Welcome back.

We've ticked long multiplication.

The factors of 38 are one, two, 19 and 38.

It is likely more efficient to use long multiplication than multiply by 19.

Not many people know their 19 times table very well.

Okay, let's move on.

Andeep organised five boxes of brochures.

In each box there are three packs of 12 brochures.

Sam arranges the same number of brochures in three boxes.

How many packs could there by in each of Sam's boxes? So what's known and what's unknown? Well, we know there are five boxes with three packs of 12 brochures.

These are our factors.

What's unknown is the number of packs and brochures.

So we've put the multiplication expression in that we know and we've put the factor in that we know in the second multiplication expression.

We've got 12 multiplied by five multiplied by three is equal to three multiplied by something, which is the number of packs, multiplied by something, which is the number of brochures.

We can use the law of commutativity and association here.

You can reorder the numbers in the equation, because it will give you the same product.

If Sam now has three boxes, it could mean that Sam has five packs of 12 brochures.

We can even change the order of some of those different factors.

As Andeep says, you can reorder the numbers in the equation because it will give you the same product.

You can use factorising to find more combinations for this question.

Five multiplied by three, well, you know that that's equal to 15.

Now let's factorise 12.

We know the factor pairs of 12 are one and 12, two and six, three and four.

So two multiplied by six multiplied by 15 could be another way of arranging the same number.

Let's check your understanding then.

How many other ways are there of arranging the same number of brochures? Andeep says, "What other factors make 12?" Can you apply the laws of commutativity or association?" Pause the video and have a go.

Welcome back.

Here's what you may have got.

Six multiplied by two multiplied by 15 or three multiplied by four multiplied by 15 and of course using commutativity, you could also have got four multiplied by three multiplied by 15.

These arrangements will give the same number of brochures.

All right, it's time for the second practise task then.

Number one, solve each question using the most efficient strategy.

A, a local park is 113 metres long and 64 metres wide.

What is the total area of the park? For B, Sam arranges eight packs of 23 visitor badges.

Andeep says that he only need to buy six packs of 34 visitor badges to have the same number of badges.

Is Andeep correct? Why, why not? For C, there are 237 athletes, each require 34 bottles throughout the event.

How many bottles is this altogether? And finally, for D, Andeep organises seven boxes of brochures.

In each box there are six packs of 13 brochures.

Sam arranges the same number of brochures in six boxes.

How many packs could there be in each of Sam's boxes? Pause the video here and have a good go of those worded problems. Good luck.

Welcome back.

Let's do some marking then.

For 1A, the total area is 7232 metres squared, because 113 multiplied by 64 is equal to 7232.

For B, Andeep is incorrect.

Sam's total number of visitor badges is eight multiplied by 23 which equals 184.

Andeep's total number of visitor badges is six multiplied by 34 which equals 204, so Andeep has more.

For C, there were 8058 bottles needed altogether, because 237 multiplied by 34 is equal to 8058.

And for D, the answer was seven packs.

In a sense, all that's happened, is that through commutativity, the factors have swapped positions.

In the context, what's happened is, Andeep has organised six packs into seven boxes, whereas Sam has arranged seven packs into six boxes.

There are the same number of brochures for each of them.

We've reached the end of the lesson, so it's time to summarise.

We understand that multiplication can be simplified by factorising one of the factors.

We also understand that if the multiplication will involve regrouping then a written method may be more efficient.

We know that if the multiplication will not involve regrouping, then a mental or informal method may be more efficient.

My name is Mr. Tazzyman.

I hope you enjoyed learning today, I did.

Maybe I'll see you again soon.

Bye for now.