Lesson video

Hello, I am Miss Miah, and I'm so excited to be a part of your learning journey today.

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

Today you'll be able to divide a two-digit by a one-digit number using representations with regrouping and remainders.

Your keywords are on the screen.

I'd like you to repeat them after me.

Remainder.

Regroup, regrouping.

Good job, let's move on.

A remainder is an amount left over after division.

This happens when the first number does not divide exactly by the other.

You can also say that you will get a remainder if the dividend is not a multiple of the divisor.

The process of unitising and exchanging between place values is known as regrouping.

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

One 10 can be regrouped for 10 ones.

So for this first lesson cycle, you are going to be understanding all about remainders.

What I want you to think about is a time when you had to share something, maybe there wasn't enough.

Was there any leftover? In this lesson you'll meet Andeep and Izzy.

Andeep has seven cakes and wants to divide them between three of his friends.

Three ones is one each, that's three.

Six ones is two each, that's six.

There is one cupcake leftover.

This is known as a remainder.

So in order to write this down, it would be two remainder one.

So what did you notice? If the number of ones is not a multiple of the divisor, there will be a remainder as part of the quotient.

So in other words, there will always be something leftover if the ones tens or hundreds is not a multiple of the divisor.

Right, let's have a look at it another way.

76 marbles are shared between three children.

How many marbles does each child get? So I've got the part-whole model here, 76, and I've got my table.

So we're going to begin by partitioning 76 into tens and ones.

So we've got 70 and six.

We're going to divide each part by the divisor.

So seven tens divided by three is two tens.

And we can see that that's been represented using the place value counters.

Now we need to regroup the one 10 for 10 ones, and that's because seven tens is not a multiple of three.

So we've got one 10 remaining, and that is the 10 now that we are going to regroup.

Altogether, we have got 16 ones.

So what did you notice? We had 70 and six before.

By regrouping the one 10 for 10 ones, you've partitioned 76 in a new way.

So now we've got 60 and 16.

You can represent this as a part-whole model.

So now we can divide the ones.

We can use skip counting.

So we've got three, six, nine, 12, 15.

So 15 is five ones each.

There is one marble left over.

So 16 divided by three is five and remainder one.

So lastly, we're going to add our partial quotients.

So two tens and five ones is equivalent to 25 and a remainder of one.

Each child will get 25 marbles each, and there will be one marble leftover.

Over to you, I want you to calculate how many cupcakes each child get and if there's any leftover.

So each child will get two cupcakes with one leftover.

Moving on.

26 sweets are shared between five children.

How many sweets does each child get? So our division equation is 26 divided by five.

Andeep thinks that we can't solve this calculation because the divisor is greater than the two tens in the dividend.

What do you think? How would you approach this question? Izzy says that you can solve it, but you will need to regroup.

So we're going to begin by partitioning 26 into tens and ones.

So we've got 20 and six.

Now, there are not enough tens to make a group of five tens, so you have to regroup two tens into 20 ones.

So we can regroup the two tens for 20 ones.

So have a look at your part-whole model.

Instead of having 20 and six, we now have zero tens and 26 ones.

Now we're going to divide the ones by skip counting in fives.

So five, 10, 15, 20, 25.

Five ones each is 25.

There is one leftover, a remainder of one.

So 26 divided by five is five ones and remainder one.

So now we need to add the partial quotients, and that is five ones and a remainder of one, which is five remainder one.

So each child will get five sweets, and there'll be one sweet leftover.

Andeep and Izzy are working out 36 divided by five.

Here's Andeep's calculation, and here's Izzy's calculation.

Have a look at both.

Who is correct, and what mistake has the other person made? Have a think.

Well, I can see Andeep has used the partitioning strategy and Izzy's used her place value counts to help her.

Izzy is incorrect because there are not enough tens to make a group of five tens.

So what Izzy should have done is regrouped the three tens first into 30 ones, and then divided by five through skip counting.

But she hasn't.

She's just taken the three tens and she's tried to share them as they are, but we know that's not going to work.

We need to regroup.

Over to you, I'd like you to fill in the blanks.

You can pause the video here and when you're ready, click Play so we can move on.

So how did you do? Each child gets three cupcakes with two leftover.

So the answer is three remainder two.

Right, onto our task.

For task A question 1, you are going to be filling in the blanks using the representation on the left-hand side to help you.

For task A question 2, you're also going to be filling in the blanks using the representation.

This time you've got place value counters to help you.

You can pause the video here.

Off you go, good luck.

So how did you do? So we can see that the dividend must be 17 because there are 17 cupcakes altogether, so the total number of cakes.

The four represents the number of children the cakes are shared between, so that is our divisor.

The four represents the equal number of cakes that each child gets.

And the one represents the number of cakes leftover.

If you got that, well done, you can give yourself a tick.

Good job.

Moving on.

The 95 represents the total number of marbles.

The three represents the number of children the marbles are shared between.

In other words, the divisor.

The 31 represents the equal number of marbles that each child will get, so this is known as the quotient.

And the two represents the number of marbles that are leftover.

So this is also the quotient as well because we then write this as 31 remainder two.

If you've got all of that correct, well done, give yourself a tick.

Moving on to the second part of our lesson cycle.

So this time we're going to use everything that we've learned in our first lesson cycle and apply it.

39 pebbles are shared equally between two children.

How many pebbles does each child get? So what we're going to first work out is the division equation.

By looking at this question, I know that there is going to be a remainder.

I know that 39 is an odd number, and if we're dividing by two, we're going to end up with something leftover because two is an even number.

So our division equation is 39 divided by two.

39 is our dividend and our divisor is two.

We know how many pebbles there are, what we don't know is how many pebbles each child will get.

So that would form our quotient.

So how many pebbles does each child get? We begin by partitioning 39.

So Izzy is saying that she's going to think carefully about how she will partition her dividend.

She could partition 39 like this, 30 and nine, or like this, 38 and one.

Or even like this, 20 and 19.

Now she thinks that partitioning like this, so into 30 and nine, so parts of 30 and nine would be easier for her because she can use her times tables facts to help her divide.

So now what we do is divide each part by our divisor of two.

30 divided by two is 15, nine divided by two is four remainder one.

Now all we have to do is add our partial quotients.

15 add four and a remainder of one is 19 remainder of one.

So that means each child will get 19 pebbles with one leftover.

Your turn.

Your division equation is 46 divided by five.

You can pause the video here to fill in the blanks.

Off you go.

So how did you do? Well, what you should have got is 45 as the part that you're first dividing by, which is nine, and then you have a remainder of one because one is not divisible by five.

Nine add remainder one is nine remainder one.

Let's move on.

Izzy says she can work out if there will be a remainder for any of these dividends by partitioning.

So she's calculating 47 divided by five.

She knows that five groups of nine ones are 45 ones, that can make one part.

And before we carry on, using your times tables facts can help you to partition your dividend effectively.

So this is because if we find the highest multiple of our divisor, we can then use that as one part because we know the division facts to that if we know our multiplication facts to that.

And then what's left over is our second part.

So 45 add something gives me 47.

Two is the other part.

Now two is not divisible by five, so the remainder here is two.

The quotient is nine remainder two.

Over to you.

Use the part-whole model to check whether there will be any remainders.

You can pause the video here.

Off you go.

How did you do? So what we should have got is 48 divided by four is 12.

One is not divisible by four, so that's a remainder of one.

12 add your remainder of one is 12 remainder one.

Now, you can also record division equations in another way.

Sometimes we calculate this type of strategy in our head without even knowing.

I used to do this a lot, but I just didn't really know how to record it.

So here we are.

Let's explore this now.

So we've got two different methods here.

What's the same and what's different? Well, our division equation is 39 divided by two.

And here we've got the part-whole method, so the partitioning method, being used for our division.

And then we've got an informal written method on our right-hand side.

Let's look at what's the same first.

The dividend and divisor remain the same.

We're still calculating the quotient, so that remains the same as well.

The layout is what is different.

The regrouping happens at different stages.

So for this part of the lesson, I'm going to have a turn first and I want you to see how I record my calculation, and then I would like you to have a go.

So pay attention.

79 divided by six.

We're going to divide each part by the divisor.

So we're going to divide our tens first.

Seven tens divided by six is one 10 remainder one 10.

One 10 and nine ones is 19 ones.

So this is the part where I divide my ones now.

I've got 19 ones altogether, I'm going to divide that by six.

That gives me three remainder one.

I now have to add my partial quotients.

So one 10, add three and a remainder of one is 13 remainder of one.

So the quotient is 13 remainder one.

Over to you.

I'd like you to use the informal written method to calculate 43 divided by three.

Please pause the video here and have a go.

So how did you do? So you should have started off by dividing your tens.

So four tens divided by three is one 10 remainder one 10.

Then you're going to regroup.

So one 10 can be regrouped as 10 ones, or you could simply just add your one 10 to your three ones at this point and get 13 ones.

You then divide your ones.

13 ones divided by three is four remainder one.

One 10 add four and a remainder of one is equal to 14 remainder one.

If you managed to get that, well done.

Give yourself a tick.

So your divisor is four.

By partitioning each dividend, can you identify what the remainder will be? For example, here we've got 37.

Now remember, we're dividing by four.

I know that four multiplied by nine is 36.

36 can therefore be one part.

That means the other part is one, because 36 add one is 37.

Now, because one is not divisible by four, that is my remainder.

That means I have a remainder of one.

So look at your other dividends.

So you've got 38, 39, and 40.

I'd like you to divide them by four and identify what the remainder will be.

What do you notice? You can pause the video here.

So how did you do? Well, let's look at 38.

The highest multiple of four in 38 is 36.

The missing part is two.

Two is not divisible by four, so that's a remainder of two.

Then if we go onto 39, our remainder is three.

And again, our highest multiple of four is 36.

And then if we look at 40, 40 is a multiple of four, so there are no remainders.

So what you should have noticed is that as the dividend increased by one, so did the remainder up until the next multiple of the divisor.

On to your final tasks for this lesson cycle.

For task B question 1, you're going to be solving the problems using informal methods.

You can choose which method to use as long as it's efficient for you.

I want you to avoid using short division at this point.

So, 37 marbles are shared between three children.

How many marbles does each child get? B, 58 glow sticks are shared equally between four children.

How many glow sticks does each child get? And C, 87 cupcakes are packed into boxes of seven.

How many boxes can be filled? Are there any cupcakes leftover.

A top tip, figure out what division equation you need first to calculate this equation.

For question two, you are going to tick the two division equations, which have an answer of four remainder two.

And then you're going to explain your reasoning to your partner, or you can jot this down on your task sheet.

You might want to draw a part-whole model to justify your answer.

You can pause the video here.

Off you go, good luck.

So how did you do? So for question 1A, the quotient that you should have got was 12 remainder one.

So each child would've got 12 marbles.

The division equation that you should have got for this question was 37 divided by three.

For question 1B, the division equation that you should have got was 58 divided by four.

And the quotient that you should have got was 14 remainder two.

If you got those two questions correct, well done.

Give yourself a tick.

For 1C, the question that you should have got was 12 remainder three.

And the division equation that would've helped you get there was 87 divided by seven.

So ultimately, there are 12 boxes and three cupcakes leftover.

Let's move on and have a look at this a bit more deeply.

So the question here was to tick the two division equations which have an answer of four remainder two.

And then you would've had to explain your reasoning.

So A was 16 divided by four.

Now, you know that 16 is a multiple of four, so there's not going to be a remainder, so we can eliminate that straight away.

Now if we have a look at 17, we know that 16 divided by four is four, and 17 is one more than 16, so one will be our missing part.

But one is not divisible by four, so that's a remainder of one.

We still can't tick it because that's four remainder one and not four remainder two.

Now if we look at 18 divided by four, at this point we didn't even need to calculate because we already know that 17 is one less than 18, and that's given us a remainder of one.

18 is one more than 17, so surely that would've given us a remainder of two.

Because our multiple of four that we can have is 16, our missing part is two.

Two is not divisible by four, so our quotient would've been four remainder two.

So if you gave that a tick, good job.

Now, having a look at D, I know straight away that this division equation will not have a remainder because 20 is a multiple of five, whereas 22 is two greater than 20, which means that the two ones that are formal part are a remainder.

So the quotient would be four remainder two.

You can now partition the dividend into tens, which helps when dividing a two-digit by a one-digit number.

You also understand that if the dividend is not a multiple of the divisor, there will be a remainder.

You understand that if the remainder is a number of tens, you can regroup the tens for ones and continue to divide.

And lastly, you understand that if the number of ones is not a multiple of the divisor, there will be a remainder as part of the quotient.

I really hope you enjoyed this lesson, and I hope to see you in the next one.