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I can use expanded multiplication to record regrouping from the 10s to the 100s.

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Licence

This content is © Oak National Academy Limited (2023), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

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Key learning points

If the product of the single digit and the tens number is greater than 100 then you regroup ten tens into one hundred.
If the product of the single digit and the tens number is greater than 100 the partial product will be a 3-digit number.
There are similarities between the representations of multiplication.

Common misconception

Pupils may not align or regroup correctly leading to an inaccurate calculation.

Model accurately what happens when we regroup from 10s to 100s. Why did this happen? How do we record this?

Keywords

Expanded multiplication - Expanded multiplication is a way of recording the steps focusing on partitioning one or more factors and showing partial products.
Regroup - The process of unitising and exchanging between place values is known as regrouping.

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(no audio) <v ->Hello, I'm Miss Mia, and I'm so excited</v> to be a part of your learning journey today.

I hope you enjoy this lesson as much as I do.

Today you will use expanded multiplication to record regrouping from the 10s to the 100s.

Your keywords for today are, expanded multiplication, and regroup or regrouping.

You may have come across these words before.

If you have, try and remember what these words mean.

Expanded multiplication is a way of recording the steps of a calculation, focusing on partitioning one or more factors and showing partial products.

The process of unitizing and exchanging between place value is known as regrouping.

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

One 10 can be regrouped for 10 ones.

So today we'll be looking at multiplying a two-digit number by a one-digit number using expanded multiplication.

In this case, there will be regrouping happening from our 10s to 100s.

And our first lesson cycle is about understanding expanded multiplication.

In this lesson, you will meet Aisha and Jun.

So let's begin.

You may have seen something like this.

This is an example of expanded multiplication and we can see that we're multiplying 41 by two.

In this example, the partial product in the ones is a one-digit number.

You may have also seen this.

In this case we're multiplying 16 by four.

In this example, the partial product in the ones digit is a two-digit number.

However, partial products may not always be a one-digit or two-digit number.

Have a look at this example.

We're multiplying 33 by four.

We've already multiplied our ones.

Now when moving on to our 10s.

We end up with 12 10s.

You'll be exploring what to do when this happens.

There are four rows each with 33 chairs.

How many chairs are there altogether? Now we can see that the array has displayed this calculation.

What I want you to think about is, what multiplication equation is needed? If you got 33 multiplied by four or four multiplied by 33, you are correct.

You can have them in any order because multiplication is commutative.

So we're going to start by partitioning 33 into three 10s and three ones.

We are going to be using expanded multiplication.

Aisha's going to use the base 10 blocks to help her.

Now the larger factor goes at the top because this makes it easier to calculate our equation.

Now when it comes to the base 10 blocks, we need four groups of 33, and we're also multiplying by four, so the smaller factor goes at the bottom.

So we'll do that and we will have our four groups of 33 there represented by our base 10 blocks.

We're going to multiply the ones first.

Three multiplied by four ones is equal to 12 ones.

There is one 10 in 12, so we regroup and place the one in the 10s column.

There are two ones in 12, so we put the two in the ones column.

Make sure you align these digits correctly.

This will make it so much easier when we regroup our partial products and it will lead to better accuracy.

Next we're going to focus on our 10s.

Four multiplied by three 10s is 12 10s.

There are more than 10 10s in 12 10s.

We must regroup.

We'll now place the 100 in the 100s column, place the two 10s underneath aligned correctly under the 10s column, and place zero as your placeholder.

Now Aisha's wondering, \"Why do we have a three-digit partial product?\" Jun's answer to that is because 12 10s is 120.

This time we are adding a three-digit partial product to a two-digit partial product.

We're going to be recombining our partial products.

This part is similar to column edition, so we're going to be starting off by looking at our ones first.

Two ones add zero ones is two ones.

You're going to place the two in the ones column.

One 10 add two 10s is three 10s.

Place the three in the 10s column.

And lastly, 100 add zero 100s is 100.

Place the one in the 100s column.

And you can see that all these digits have been aligned correctly.

So the product is 132.

There will be 132 chairs.

Aisha is calculating 32 multiplied by four.

You can see her calculation over here.

She's got 20 as her answer.

Is Aisha's calculation correct? I want you to explain your thinking to your partner.

So how did you do? Well, Aisha did not regroup.

Aisha did not regroup her 10s correctly.

Moving on, there are four rows each with 26 chairs.

How many chairs altogether? Aisha believes that she will not need to regroup in the 10s column and she believes that her product will be a two-digit number.

Do you agree? And what is the multiplication needed? I'm going to give you a moment to think.

So let's calculate this together.

The equation required is either 26 multiplied by four or four multiplied by 26.

Now the larger factor goes at the top because this makes it easier to calculate our equation.

Which means the smaller factor goes into our ones column and it needs to be aligned correctly.

You're going to multiply the ones first.

Four multiplied by six ones is 24 ones.

20 ones can be regrouped for two 10s.

A two is written in the 10s column.

There are four ones in 24, so we put the four in the ones column.

Next we move on to multiplying in our 10s.

So four multiplied by two 10s is eight 10s.

We place the eight in the 10s column.

There are zero ones in eight 10s, so we place zero as our placeholder in the ones column.

Four add zero ones is four ones.

We then move on to our 10s column.

Eight 10s add two 10s is 10 10s.

10 10s equals a hundred.

We must regroup.

So a hundred has zero 10s, so you put zero as your placeholder in the 10s column and you place the one in the 100s column.

What do you notice? I'll give you a moment to think.

Sometimes it does not look like we need to regroup our 10s to 100s, but when we sum partial products, we do.

Over to you.

12 10s can be regrouped as 100s and 10s.

You can pause the video here.

Off you go.

How did you do? If you got 10 and two 10s, you are correct.

Back to you.

The 100 is placed in the column.

You can pause the video here.

Off you go.

How did you do? So you should have got 100s column.

This is because 12 10s is regrouped as 100, two 10s and zero ones.

So the 100 needs to go in the 100s column.

Okay, onto the first task of this lesson cycle.

For this question, you will fill in the gaps by calculating the partial products.

You can pause the video here.

How did you do? So for question one, we were multiplying 53 by four.

Our first partial product was already filled out for us, but our second partial product was not.

So instead of looking at the ones column first, we needed to look at the 10s column.

Four multiplied by five 10s is 20 10s.

20 10s is regrouped as 200.

So you place the two in the 100s column, zero as a placeholder in the 10s column and the ones column.

Now let's move on to part two.

Here we were calculating both partial products and the product.

So we will start off by looking at our ones column.

Six multiplied by one is six ones.

We finished that.

We can now move on to multiplying in our 10s column.

Six multiplied by six 10s is 36 10s.

36 10s can be regrouped as 360.

So the three in 300 is placed in the 100s column, and the six in six 10s is placed in the 10s column.

We write zero as our placeholder in the ones column.

We then recombine both partial products to get 366 as our product.

If you've got both of those correct, good job.

Now we're going to be moving on to using expanded multiplication to solve problems when there are instances of regrouping from our 10s to 100s.

Let's go.

Tickets at a football stadium cost 41 pounds each.

How much did Jacob and Aisha spend individually? So Jacob's bought three tickets and Aisha's bought five.

So what multiplication equations are needed to solve this problem? Have a think.

Right, so if you've got three multiplied by 41 or 41 multiplied by three, and for the second equation, if you've got five multiplied by 41 or 41 multiplied by five, you are correct.

Here we've got three tickets bought at 41 pounds and in the second equation we've got five tickets bought at 41 pounds.

So I will talk you through my expanded calculation.

I placed my larger factor at the top, which is 41, and I placed my smallest factor, three, at the bottom.

I started by multiplying my ones.

Three multiplied by one is three.

If you've got that, good job.

Then I started multiplying in my 10s.

What is three multiply by four 10s? If you've got 12 10s well done.

Now 12 10s can be regrouped as 100 and two 10s.

Here we can see that it's been aligned correctly.

Now when you recombine both partial products, you end up with a product of 123.

Your turn.

I would like you to use expanded multiplication to calculate five multiplied by 41.

You can pause the video here.

How did you do? So I will talk you through this calculation.

You would've placed your larger factor at the top, which is 41, and you would've placed five at the bottom in the ones column.

You then would've started multiplying in your ones.

So five multiplied by one one gives you five ones.

You place that underneath in your ones column.

You then would've multiplied five by four 10s, and that would've given you 20 10s.

Now 20 10s can be regrouped as 200.

You place the two in the 100s column and you put zero as your placeholder in the 10s column and the ones column.

You then recombine those partial products, quite similar to column edition, and you should have got your product as 205.

So all together Lucas spent 123 pounds and Aisha spent 205 pounds.

If you got that, good job.

Back to you.

So the missing partial product for four multiplied by 41 is either 16, 160 or 106.

And I'd like you to explain how you know to your partner.

You could pause the video here.

How did you do? So let's have a look at the ones column.

Four multiplied by one one is four ones, and that's already been filled out for us.

So we would've had to look at our 10s column.

Four multiplied by four 10s is 16 10s, but our partial product is a three-digit number.

So 16 10s must be regrouped.

16 10s regrouped is 100, 6 10s and zero ones, which is 160.

So if you've got 160, you are correct.

Well done.

Let's move on.

Find the missing digits.

Now, I used to really struggle with this type of question, but what I realised was that if you break down the question and start by looking at separate columns first, it should make it easier.

So what I want you to begin by thinking about is what is known and what is unknown.

I'm going to give you a moment to think.

Let's start with what is known.

Our product is known, our partial products are known.

What's missing or what is unknown is our smallest factor, the factor that we are multiplying by, and part of our larger factor.

Begin by looking at our ones column.

The missing factor must be a one-digit number that multiplies by two to make eight.

So let's use our two times tables for help us.

So something multiplied by two ones will give us eight ones.

I know that four multiplied by two ones will give us eight ones.

So the missing factor is four.

Now we can look at our 10s column.

In order to support us so our partial product here is 280 or otherwise equivalent to 28 10s, four multiplied by something gives us 28 10s.

I know that 280 is four multiplied by something 10s.

I can use my times tables to help me.

Four multiplied by seven 10s is 28 10s.

That's because if you know that four multiplied by seven is 28, four multiplied by seven 10s must be 28 10s.

So that means the missing digit is seven.

Okay, back to you.

The missing partial product for 38 multiplied by five is? And your options are 530, 105, or 150.

You can pause the video here.

Okay, how did you do? Let's have a look at our ones first.

Five multiplied by eight ones gives us 40 ones, and that's been recorded.

Our first partial product is known, but we don't know our second partial product, so we must look at multiplying our 10s.

So if I know that five multiplied by three is 15, I also know that five multiplied by three 10s is 15 10s.

If we regroup 15 10s, that is 100, fives 10s and zero ones, which is equivalent to 150, which means 150 is our answer.

Back to you.

So Aisha believes the product of the single digit and 10s number will always need regrouping.

Is this always true, sometimes true or never true? You could pause the video here.

Okay, what did you think? So let's start off with some examples.

21 multiplied by two does not require regrouping when multiplying in the 10s column.

51 multiplied by two does require regrouping when multiplying in the 10s column.

And we know that because two multiplied by five 10s will give us 10 10s, and we do need to regroup that.

So that means her statement is sometimes true.

Back to you.

Draw a line to match each multiplication with the correct regrouping.

So you've got the equations 31 multiplied by four, 33 multiplied by two, and 14 multiplied by three.

Now the instances of regrouping you are matching to are zero regroup, regrouping in ones to 10s, regrouping in 10s to 100s.

You can pause the video here to have a go.

So how did you do? If you got 31 multiplied by four as having regrouping in the 10s to a 100s, you are correct.

33 multiplied by two results in no regrouping.

And 14 multiplied by three, we can see that there will be regrouping in the ones to 10s because three multiplied by four will give us 12 ones, and 12 ones is greater than 10 ones so we must regroup.

Okay, onto our final three tasks.

So the first task, you'll be completing these multiplication equations using expanded multiplication.

Your equations are 51 multiplied by five, 84 multiplied by two, and 73 multiplied by three.

For the second question, I would like you to find the missing numbers.

And for the third question, using expanded multiplication, how close can you get to the target of 364? Now, Aisha's already used 364 multiplied by one, so you can't use that.

Can you think of any more? You can pause the video here to have a go.

How did you do? So let's look at 51 multiplied by five.

The partial products recorded here are five and 150.

The product itself is 155.

If you got that correct, give yourself a tick.

Let's look at the second question.

So the larger factor at the top should have been 84 and we're multiplying by two.

The partial products you should have got were eight ones and 160 as your second partial product.

After recombining the partial products, you should have got 168 as your product.

If you got that, give yourself a tick.

And for the final equation, you should have got nine as your first partial product and then 210 as your second partial product.

Summed together, these partial products give us 219 as our product.

If you got that, give yourself a tick.

Let's move on.

So for both questions, please check that you've got the correct digit in the missing gaps.

If you managed to find all the missing numbers, well done.

And for question three, now, I love these types of questions because they really do make you think methodically.

After multiply one by 364 to get my product.

Another example answer that you could have had was 182 multiplied by two.

182 is the larger factor, and then we multiply that by two, and then you would've had three partial products here.

So you would've multiplied in your ones first to get four ones.

Then you would've multiplied in your 10s, followed by your 100s.

All three partial products summed would've given you 364.

Now it's important to note that we have three partial products here.

We are multiplying by three-digit number.

Some of the alternative answers that you could have had were four multiplied by 91 and seven multiplied by 52.

If you managed to get all of those variations, good job.

Well done.

You made it to the end of the lesson.

To summarise our learning today, we multiplied a two-digit number by a one-digit number using expanded multiplication with regrouping happening in our 10s to 100s.

You now understand how to use expanded multiplication, especially when regrouping 10s to 100s.

You can use expanded multiplication to regroup 10s to 100s, and you can use this to solve problems. I hope you enjoyed this lesson as much as I did.

(no audio).

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It is greater than 50. 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The arrow is pointing to the __________.", "image": "https://www.thenational.academy/_next/image?url=https%3A%2F%2Foaknationalacademy-res.cloudinary.com%2Fimage%2Fupload%2Fv1698137184%2Fndkve0asckfrma0shigc.png&w=640&q=75", "imageKey": "778da13f-5096-b602-bd74-4a4f74069bcc/question/72a15627571fd83abdadaa0a4d6d04b5.png" }, "questionType": "Multiple choice", "correct": [ { "text": "partial products" } ], "choices": [ { "text": "products" }, { "text": "factors" }, { "text": "partial products" }, { "text": "recombined amount" } ], "id": "2239c2e5-cd02-2a0e-d291-1a19adab73e6", "lessonId": "580ac7f4-14a0-e936-8b63-28815aefa02b" }, { "type": "Question", "question": { "text": "12 tens can be regrouped as ______ hundred and 2 tens." }, "questionType": "Short answer", "correct": [ { "text": "1" }, { "text": "One" }, { "text": "one" } ], "id": "a11a53fd-64f1-f927-280d-ebdb0d7b76c5", "lessonId": "580ac7f4-14a0-e936-8b63-28815aefa02b" }, { "type": "Question", "question": { "text": "The digits '3' and '6' in the number 365 represent: " }, "questionType": "Multiple choice", "correct": [ { "text": "3 hundreds and 6 tens" } ], "choices": [ { "text": "3 tens and one 6" }, { "text": "3 ones and 6 tens" }, { "text": "3 hundreds and 6 ones" }, { "text": "3 hundreds and 6 tens" } ], "id": "1df3848b-2cd6-e486-0a2a-141d6171aded", "lessonId": "580ac7f4-14a0-e936-8b63-28815aefa02b" }, { "type": "Question", "question": { "text": "Match each multiplication with the correct regrouping." }, "questionType": "Matching", "pairs": [ { "left": { "text": "23 × 2" }, "right": { "text": "0 regroup" } }, { "left": { "text": "24 × 3" }, "right": { "text": "Regrouping in 1s to 10s" } }, { "left": { "text": "42 × 3" }, "right": { "text": "Regrouping in 10s to 100s" } } ], "id": "3abc9742-7ea1-fcbc-c75d-9bdd620b1aa4", "lessonId": "580ac7f4-14a0-e936-8b63-28815aefa02b" }, { "type": "Question", "question": { "text": "Choose the correct partial products for this expanded method.", "image": "https://www.thenational.academy/_next/image?url=https%3A%2F%2Foaknationalacademy-res.cloudinary.com%2Fimage%2Fupload%2Fv1698137198%2Frrnps0gdepgfa2lvmc6g.png&w=640&q=75", "imageKey": "778da13f-5096-b602-bd74-4a4f74069bcc/question/020b7745ee39679570772774e2f9d8c8.png" }, "questionType": "Multiple choice", "correct": [ { "text": "6 + 420" } ], "choices": [ { "text": "6 + 42" }, { "text": "6 + 420" }, { "text": "6 + 402" }, { "text": "402 + 60" } ], "id": "817a0719-ea22-1c2b-8d23-ca040e161608", "lessonId": "580ac7f4-14a0-e936-8b63-28815aefa02b" }, { "type": "Question", "question": { "text": "Aisha has 47 packets of sweets. Each packet has 8 sweets. How many sweets does she have altogether?" }, "questionType": "Short answer", "correct": [ { "text": "376" }, { "text": "Three hundred and seventy six" }, { "text": "three hundred and seventy six" }, { "text": "three hundred and seventy-six" } ], "id": "fb41b70f-dc7f-4e1e-cad0-9c83390fe480", "lessonId": "580ac7f4-14a0-e936-8b63-28815aefa02b" } ], "lessonId": "580ac7f4-14a0-e936-8b63-28815aefa02b", "meta": { "gradeStageId": "ks2", "subject": "Maths", "grade": "Year 5", "topic": null, "unit": "Multiplication by partitioning leading to short multiplication (2 by 1-digit)", "lessonCount": "10", "age": [ 9, 10 ], "gradeId": "age_9_10", "gradeStage": "Key Stage 2", "lesson": "Multiply a 2-digit number by a 1-digit number using expanded multiplication (regrouping 10s to 100s)" } } ] }