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 2-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.

I can explain how to use long multiplication to multiply a 3-digit by a 2-digit number.

These are some of the key words that you are gonna expect to hear in the lesson, partial product and regroup.

I'll say them again, but I want you to repeat them back to me.

I'll say, "My turn," say the words and then I'll say, "Your turn," and you repeat back what I've said.

Ready? My turn.

Partial product, your turn.

My turn, regroup.

Your turn.

It's important that we also understand what these actually mean.

Here's what partial product means.

Any of the multiplication results we get leading up to an overall multiplication result is a partial product.

You can see that below in these representations as well.

We've got the multiplication.

16 multiplied by 4 is equal to 64.

16 as a factor has been partitioned into 10 and 6.

The multiplication of each of those partitioned parts of 16 are the partial products, and you can see those in the jottings on the right.

4 multiplied by 10 is equal to 40, 4 multiplied by 6 is equal to 24.

Those have then been added together to give the overall product.

Here's what regroup means.

Regrouping is the process of unitizing and exchanging between place values.

For example, 10 ones can be regrouped for one ten.

One ten can be regrouped for 10 ones.

This is the outline for the lesson then.

We're gonna start by regrouping in long multiplication and then we're gonna look at solving problems. Sam and Andeep are gonna be with us today helping out.

They're experts in long multiplication because they've done a lot of learning so far.

They're gonna respond to some of the math prompts and discuss some of the learning.

So listen to them really well, because it's really helpful.

Hi Sam, hi Andeep.

You ready to go? Are you ready to go? Alright, let's learn.

153 tickets are bought for an Olympic event.

If each ticket costs 22 pounds, how much was spent in total? Well, let's start with what's known and what's unknown.

Hmm.

Well, what's known is that there are 153 tickets bought and that each ticket is 22 pounds.

These are our factors.

What is unknown? The total amount spent.

The product is missing.

So there, we can write a multiplication equation of 153 multiplied by 22, and that's equal to an unknown.

Something we need to calculate.

It's going to involve some long multiplication.

So long multiplication can be used with a 3-digit number.

Andeep says, "I'm multiplying a 3-digit number by a 2-digit number.

He's written the 3-digit number first as the largest factor.

We've got 153 multiplied by 22.

Then there's space for two partial products and the sum of those partial products at the bottom.

I put the 3-digit number first.

First, multiply the ones, 153 x 2.

3 ones x 2 = 6 ones.

So six goes in the ones column.

5 tens x 2 = 10 tens, or 100.

I can regroup 10 tens for 100.

I have a zero tens and 100, which I put at the top.

So you can see zero has gone into the tens column and then there's an exchange 100, which has been put at the top above the two factors as a reminder.

1 hundred x 2 = 2 hundreds plus 1 hundred.

I have added the regroups 100 here.

We've got our first partial product.

It's 306.

Now you might be tempted to just walk away at this point and think, "I've done some multiplication, so I've finished." But of course, we know that there's two partial products that we need to get.

Then we need to add them together.

So Andeep has calculated the first partial product, then multiplied by the tens, 153 x 20.

A placeholder is used because we are multiplying by a multiple of 10.

We know that all multiples of 10 have a zero in the ones column.

3 ones x 20 = 60 ones, or 6 tens.

So six goes into the tens column.

5 tens x 20 = 100 tens, or 1000, which is regrouped.

So we have a zero in the hundreds column and you can see again, we've put a one above the two factors at the top to remind us that there's a regroup 1,000 we need to add on when we make our next calculation.

1 hundred x 20 = 20 hundreds, or 2,000 plus the regroups 1,000.

The second partial product is 3,060.

What do we need to do next? Ah, yes, we need to sum the partial products.

We end up with 3,366.

So there are 3,366 seats altogether.

What did you notice about the calculation? What do we have to do? Well, Andeep says, "One of my partial products had 4 digits because we are multiplying a 3-digit number.

I had to regroup from the hundreds to the thousands." Let's your understanding so far.

What is the multiplication equation needed to solve the problem? I'll read the problem to you now.

Sam and Andeep visit the Olympic village, which houses 340 apartments.

If each apartment hosts 78 athletes, how many athletes can the village accommodate? What do you know? What's the unknown? Pause the video and come up with a multiplication equation to solve.

Welcome back.

The multiplication equation would've been 340 x 78 is equal to an unknown.

You know there are 340 apartments.

That's one factor.

Each apartment hosts 78 athletes.

That's the other factor.

The product is missing.

Table tennis events use a lot of table tennis balls.

I can imagine.

During the Olympics, they use 42 boxes.

Each box contains 165 table tennis balls.

How many balls do they use? You can see we've got our question template again.

What's known, what's unknown? Well, let's see.

There are 42 groups, or boxes, of 165.

What's unknown? The total number of table tennis balls.

So what will our equation look like? 165 x 42 is equal to an unknown.

Long multiplication can be used to calculate the product and Andeep says, "Remember to put the larger number at the top." So we start with 165 and we write down 42 as the next factor.

First, multiply the ones, 165 x 2.

Well, we've got some regrouping that's occurred here.

Then multiply the tens, 165 x 40.

We put down a placeholder because we are now using multiples of 10 and we end up with 6,600 as our second partial product.

There's also been some regrouping.

Look at the top.

You can see.

Now some of the partial products, 6,930.

There are 6,930 table tennis balls altogether.

What did you notice? We kind of talked it through as we went, but what did you see? Well, Andeep says, "You had to regroup from the hundreds to the thousands." Sam has used long multiplication to calculate the number of seats there are at the tennis event.

321 x 24.

Here is Sam's calculation.

Is Sam correct? Why or why not? Sam has regrouped incorrectly.

This one is incorrect.

This is how it should have looked and you can play a bit of a spot the difference here.

If you look closely at that first partial product, you can see that Sam managed to reverse the digits when doing 3 hundreds multiplied by 4 ones.

You should have ended up with 1,200.

Sam wrote 2,100 instead.

Let's check your understanding then.

Sam, Andeep and Aisha are solving 232 x 42.

Who is correct? Justify your thinking to your partner.

You've got two calculations to look at there.

Sam's calculation and and deeps calculation.

Pause the video here and decide who's correct.

Welcome back and deep is correct.

He has regrouped correctly.

If you need to have a closer look at that, then pause the video now.

Here's your first practise task then.

For 1 a, b, and c, you need to fill in the gaps.

So all we're looking at here is if you are setting out a long modification correctly.

And then for d and f, you need to fill in the gaps and complete the long multiplication.

Pause the video here and have a go at that.

Good luck.

Welcome back.

Here are the completed gaps.

For a, it was 328 above 47.

For b, you had 123 above 42, and for c, you had 246 above 25.

For d, If you'd worked through all of the long multiplication, you should end up with the product of 7,889.

For e, it was 7,686, and for f, it was 7,824.

If you didn't get those, please do take a moment now to go back and carefully check through by comparing the answers with your own jottings.

Often, there are mistakes that can be made at the regrouping stages or sometimes the final addition step.

Pause the video now to do that.

Okay, it's time for the second part of the lesson then, solving problems. Sam and Andeep are playing a game.

They have six cards.

Andeep shuffles them and picks out one card at a time to make a multiplication equation.

He's got 123 x 44.

Andeep uses long multiplication to find the product.

He starts with the greatest factor on top and he puts 44 in.

First, multiplied by the ones, 123 x 4.

We've got regrouping there and we've ended up with the partial product to 492.

Then multiplied by the tens, 123 x 40.

Remember that placeholder because now we're using multiples of 10 and the partial product is 4,920.

Now sum the partial products.

We end up with 5,412.

Andeep's score is 5,412.

Now Sam has a go.

Andeep had 123 x 44 and a score of 5,412.

Do you think Sam will score higher or lower? 421 x 23.

First multiply the ones, 421 x 3.

1,263 is the first partial product.

Then multiplied by the tens, 421 x 20.

In goes the placeholder and then we have a second partial product of 8,420.

Then recombine the products.

You get 9,683.

Sam's score is 9,683.

Time to check your understanding then.

Andeep has another go.

He uses long modification to find the answer.

Is he correct? Why or why not? Pause the video and have a go.

Welcome back.

Andeep is incorrect because he forgot to use zero as a placeholder.

The second partial product should be 4,880 and I have to say, it's a very common error of that.

Please do remember to use that placeholder because we're switching from using ones to using tens.

The highest score wins.

The rules change: players can now rearrange their digits.

Andeep picks the following cards and arranges them like this.

214 x 32.

What advice would you give to Andeep to get the highest score possible? So how could he rearrange those cards? Could he swap any over to try and make a higher product? Hmm.

Andeep says, "I think I need the largest value digits in the 100s and 10s positions." He rearranges the numbers.

This time he's got 432 x 32.

"I think this is the highest possible score.

400 multiplied by 30 is 12,000 so my answer will be greater than 12,000." Really good reasoning.

Okay, let's check your understanding then.

Look at Sam's cards and arrange them to find the highest possible score.

Sam got a four, a one, a four, a three, and a two.

Pause the video here and have a go at finding a way of getting the highest possible score.

Welcome back.

Here are two solutions that you might have considered.

431 x 42 or 432 x 41.

Notice that there's a change in the ones column and that's because it doesn't actually matter which way around they go.

The products will still be two.

Both products here would be greater than 16,000.

Andeep makes a prediction.

"If I want the smallest product, I need to make the 3-digit number as small as possible." Do you agree? Justify your thinking to your partner.

Hmm.

Andeep also needs to think about the size of the second factor.

Andeep thinks about the value of both factors.

The products will be greater than 3,600 because 12 x 3 = 36 so 120 x 30 = 3,600.

What if I used a 2 in the 10s digit of the 2-digit number?" Hmm.

Well, the products will be greater than 2,600 because 13 x 2 = 26.

So 130 and 30 x 20 = 2,600.

Having 23 groups will result in a product smaller than having 34 groups.

Okay, let's check your understanding again then.

Arrange the card so you will find the smallest possible product.

Pause the video here and have a go at that.

Welcome back.

Did you arrange them like this? 134 x 22, keeping the hundreds and tens digits small results in a smaller product.

Okay, it is time for the second practise task then.

Continue playing the card game with a partner.

Arrange the cards (1, 2, 2, 3, 4, 4) in long multiplication layout.

Number one, round one.

Each player picks 5 cards randomly from the deck of 6 cards.

Lay them out in order as they appear.

Calculate the product for each term.

Take 3 turns.

For number two, we've got round two.

Highest score challenge.

Players pick another set of 5 cards from the deck.

Arrange the cards to achieve the highest possible score using long multiplication.

Players have 1 turn to arrange and calculate the score.

For number three, round three.

Players pick 5 cards from the deck.

Arrange the cards to achieve the lowest possible score using long multiplication.

Players have 1 turn to arrange and calculate the score.

Okay, pause the video and have a go at three rounds.

I'll be back later on with some feedback.

Enjoy and good luck.

Welcome back.

Here's how round one went for Andeep and Sam.

Andeep picked 4, 4, 3, 2, and 1 and ended up with an overall product of 9,303.

For number two, Andeep got 4, 2, 2, 2 and 1.

He rearranged to find what he thinks is the highest product, which is 13,472 by arranging the factors as 421 x 32.

He made sure the 100s number and 10s number in the factors were larger digits.

For number three, he rearranged to find what he thinks is the lowest product, 2,852.

He got 124 x 23 using those digit cards.

He made sure the 100s number and 10s number in the factors were smaller digits.

Okay, here's a summary then of today's lesson.

Long multiplication is a written strategy that can be used to multiply a 3-digit number by a 2-digit number.

Partial products are combined to make the product.

We are also able to regroup in long multiplication, including from hundreds to thousands.

My name's Mr. Tazzyman.

Hope you enjoyed today's lesson.

Maybe I'll see you again soon.

Bye for now.