# Lesson video

In progress...

Hi, my name is Mrs, Ben.

And for this lesson, I will be your teacher, does maths fascinate you? Maths has fascinated the world for years and years and years, we find maths in our everyday life, from the way things are built to the money that we spend, to patterns and things that we like the look of, all of that comes down to maths.

Well, we're going to be exploring some maths in this lesson and the maths we'll explore is addition.

Let's begin by looking at the lesson agenda.

Throughout the lesson, we will use a range of addition strategies.

We will be deciding on the most efficient strategy, and then we will have chance to practise some addition strategies.

At the end of the lesson, there'll be an independent task for you to have a go out by yourself.

I know you'll be keen to find out how you got to answer, so I will go through the answers with you.

There are just a few things that you might need for this lesson, something to write with, so a pencil or a pen, something to write on, so some paper and possibly something to represent dienes.

So in this picture, you can see I've got 10 spaghetti sticks to represent one hundreds.

One spaghetti stick represents a 10 and one piece of fusilli twisty pasta to represent the ones.

If you don't have those things to hand, just pause the video here whilst you go and get them.

Remember, try to work somewhere quiet, where you won't be disturbed.

As we are looking at different addition strategies in this lesson, I thought it would be helpful to remind you of a few that you've probably come across before.

Our first one is partitioning and recombining.

This is where we might take a number such as a two digit number, and we will pull it apart.

We will partition it into tens and ones, and then we will add tens and ones separately and then recombine the two parts together.

But we'll be able to use known facts.

For example, if I know that three plus three is six, I know that 30 plus 30 is equal to 60.

We can adjust near multiples of 10.

So if I'm adding nine or 19, I might adjust my calculation to use 10 or 20 and then take one off later on in the calculation.

So I might round up say I have 37.

I might round it up to 40, and then near adjust later by subtracting the three extra that I've added on originally.

If you've forgotten any of these strategies, don't worry, because I'm going to be showing them to you in this lesson.

So job number one is to grab your pencil and your paper and have it go to the calculation that you can see on the screen, do it in whichever way that you prefer to work it out.

But pause the video here whilst you have ago.

Okay, then let's have a look together.

So there's going to be many, many different ways of working this out.

And the fascinating thing for me is that everybody watching this video will have a different method, or you probably will have a different method.

So there's not a right way or a wrong way.

However, there will be a way that is the most efficient, so efficient means to do something the best way.

It's not going the long way round or the slowest way around.

It will be quick and effective.

So as I go through some different strategies with you, I'd like you to think about which is the most efficient strategy.

Let's take a look at my friend, Anna.

Anna used the partitioning and recombining strategy.

She said, I added the tens.

So her calculation looks like this.

Say the calculations on the screen with me, 40 plus 30 is equal to 70.

Four plus nine is equal to 13 and 70 plus 13 is equal to 83.

So she's found that the total is 44 plus 39.

Did you get that total too? So let's just remind ourselves here.

What should you do in this part where it's in green? Okay, so she's partitioned, our two digit numbers and she's added the tens together.

When she partitioned our two digit numbers.

The second step was to add the ones together.

And lastly, she has recombined them together.

So that is quite a good strategy, isn't it? Do you think that's efficient? Perhaps it is the most efficient strategy, but let's have a look and see what Junaid did.

So this is Junaid and Junaid's strategy was to adjust near multiples of 10.

So he's noticed that 39 is very close to number 40.

So this is what he says.

"I know that 39 is one less than 40.

I am going to take one from 44 to add to 39.

This leaves me with the expression 43 plus 40.

So let's watch and see how Junaid manages this.

So we can see that the hole is going to go here.

And we have two parts, these two parts are represented with dienes and this group shows 39 and this group represents 44.

Does it matter that it's 39 plus 44 and not 44 plus 39? Absolutely not because when we have addons or we're doing addition, we can use a lot of commutativity, change the order of the addons and the sum will still stay the same.

Okay then, so he has taken a one from here, the 30 to the 44 and he has added it onto 39.

So what do you notice here about the yellow blocks? Well, they can be regrouped into one 10.

So we now have one 10 here just because he moved one block from the one, one, sorry, from the set of 44.

He then says that this calculation is a lot easier to solve.

I can count on in multiples of 10 or using known facts.

So he says, I know that four plus four is equal to eight.

So 40 plus 40 is equal to 80.

So 43 plus 40 is equal to 83.

So he has actually combined those let's see over here.

And the total is 83.

So we mentioned two different strategies there.

He's adjusting near multiples of 10 because he took that one and added it to the 39 to make a whole group of four, a whole group of 10.

So that full set was now 40.

And he changed his calculation.

When it came to do the calculations in the end, he's actually used a known fact as well.

It's used his knowledge of double four, meaning equal to eight.

So we can be very flexible in the way that we use our maths but whose methods did you prefer? Which strategy was most efficient? We had Anna that was partitioning and recombining and Junaid who was adjusting near multiples of 10 and using known facts.

What do you think? My personal opinion is it is easier to add multiples of 10 and adjust for this question.

Partitioning and recombining is a very safe method that's absolutely fine, but by just taking one from 44 and adding it to 39 was much more quick, it was much faster because we had then had 43 plus 40, which then gave us a total of 83, much faster to calculate, I think.

Okay, pause the video here whilst you have a go at 25 plus 38, when you're ready, come back to me.

But I want you to think carefully about the strategy that you use.

Try to use the most efficient one off you go.

Okay, how did you get on? Which method did you do? Which strategy did you use? Partitioning and recombining? Adjusting near multiples of 10? Rounding and adjusting perhaps.

Well, let's have a look to see how our friends managed this one.

So here is Anna again, and she likes this partitioning and recombining strategy.

So she said, "I added the tens, I added the ones I recombined and added them together." Can you guess what's going to come up on the screen next? That's right.

Say it with me 20 plus 30 is equal to 50.

Five plus eight is equal to 13, 50 plus 13 is equal to 63.

So we know that 25 plus 38 is equal to 63 and we've used a partitioning and recombining strategy, do have a more efficient strategy to use? Let's see what Junaid uses.

So Junaid likes adjusting near multiples of 10 strategy.

He said, I know that 38 is two less than 40, I am going to take two from 25 to add to the 38.

This leaves me with the expression 23 plus 40.

Now maybe you want to have a go at drawing out some dienes here just before I go ahead in the video, just to see if you can work out what it is that Junaid has actually done.

Okay, here is our part, whole model again, and we've put in our information.

So in this set, what value is it? We've got 38 there and the value of this part is 25.

Good, so look carefully.

Junaid has taken two from here and he has added it onto the 38.

So now he has gotten another group of 10 that's right? So we can regroup that and we now have 10.

So what's our new calculation.

Well, we have 40 here plus 23.

So 40 plus 23 is quite quick to calculate and we know that it is 63 all together.

So he says, "This calculation is a lot easier to solve.

I can count on multiples of 10 or using known facts." This time he said, "I know that two plus four is equal to six.

So 20 plus 40 is equal to 60.

So 23 plus 40 is equal to 63.

It's really important that I show you the dienes here.

So you can get the picture of what is happening in the maths in your minds because if I'd have just shown you this part here, it's really abstract, doesn't really have much meaning.

So you have to be able to build the pictures of the dienes of the tens and the ones in your mind.

We've got one more strategy to look at this time.

This is an Zara and Zara's strategy is to round and adjust.

She likes rounding and adjusting, so let's see what she says.

I can round 38 to 40.

I can calculate 40 plus 25, which equals 65.

I can take away the two I added on when rounding.

So let's have a look at the dienes here that Zara has used.

The value of this part is 38.

And the value of this part is 25.

So she said, she's going to round to 38.

So let's watch what happens.

So she is added two more and you can see this represented in this image here with two pink blocks.

We've got a number line here as well.

That's going to help support our thinking as because we're using rounding.

So she knows that 40 plus 25 is equal 65.

That's quite easy for her to calculate now.

Nice numbers very nice when we used multiples of 10.

So 40 plus 25 is equal to 65.

What does she need to do then? She's added on two extra because she made 38, two one's bigger or two one's greater.

She's made it over here, didn't she? So she's changed this to a 10 by adding two on.

So that can't stay there.

She can't just change the value completely of 38.

So she's going to have to take two off.

So here we go.

We can see it here on the number line as well.

So 65 subtract two is equal to 63.

Now that we've looked at three different strategies for solving that calculation of 25 plus 38, which strategy do you think was the most efficient? Anna used the partitioning and recombining strategy.

Junaid adjusted near multiples of 10 and Zara use the rounding and adjusting strategy.

I personally think it was easier to round and adjust for this question.

I found it easier to do 25 plus 40 and rounded up 38 to 40, and then take off two from my answer.

So here's a little reminder of the different strategies that we could use.

There are more strategies to use, but these are just four that we've looked at in this lesson.

I'd like you to think about how you would do 66 plus 67.

Would you partition and recombine? Would you use known facts, would you adjust now multiples of 10 or would you round and adjust? Again there is no set way of doing this, but there are strategies that are more efficient than others.

So pause the video here whilst you have a go and then I'll show you my answers.

So on the screen, you can see a bit of a mind map of my ideas.

Let's go from the top.

So up here I thought, right, well, 66 plus 67, 67 is quite close to 70.

So I'm going to do what Junaid did earlier and I'm going to move three over from the 66 up to the 67.

So now I have 70, my new calculation is 63 plus 70, which I know is equal to 133.

Let's have a look at this rounding and adjusting strategy.

So I rounded 67 up to 70, and then I started with the largest number.

So I calculated 70 plus 66.

Remember I can change the order of addings because of the law of communitativity, so we can change the order of the addons and the sum will still remain the same.

So I calculated that 70 plus 66 is equal to 36.

I did this by partitioning, really.

I added to 30 to get me to a hundred, and then I added the rest to get me to 136 So on the screen, you can see a bit of a mind map of my ideas.

I'll go through them with you now you may have some similar strategies.

You might have some different strategies.

So up here, I used Junaid's strategy and I moved three over from 66 and put them onto 67.

So I've altered the values of the two addons.

My new calculation is 63 plus 70 equals 133.

I tried this strategy next rounding and adjusting.

So I changed the order of the add ons and started with 67 and then rounded it up to 70.

I know I've added three extra on, I needed to add 66.

So I added a chunk of 30 to get me to a hundred.

And then there were 36 left to make sure I added 66 in total.

So I added another 36 and that got me to 136, but I rounded it, didn't I? Now I added three on.

So those three that I added on extra have to come off.

So I then need to take away three or subtract three, 136 subtract three is equal to 133.

Let's go over here.

I'm going to go through the partitioning strategy next.

So I pulled apart the tens and the ones, and then added them together.

60 plus 60 is equal to 120, six plus seven is equal to 13.

So 120 plus 13 is equal to 133.

I quite enjoyed that strategy.

That was nice and easy for me to look at and to think about.

I also thought about the fact that I know and I thought, hmm, it's actually near doubles, isn't it? They're 66 plus 67, the difference is only one.

So I've used this to add to the two together.

So I thought, right, well, I'll make it double.

So 66 plus 66 equals 132.

By doubling, I actually took one off the 67.

So I need to add that extra one back on.

So 132 plus one is equal to 133.

Did you use any of the same strategies as I did? Which strategy did you use? And was it efficient? Of my strategies, which one do you think is the most efficient? I liked the partitioning strategy, but I think to do it mentally for a mental calculation, near doubles might've been the most efficient strategy.

Junaid said, if I take any two digit number, reverse the digits and add the two numbers together, the total will be an even number.

For example, 82 with the digit reversed is 28, 82 plus 28 equals 110, which is an even number.

Is it always, sometimes or never true? Don't forget to justify your reasoning with some proof.

When you ready come back to me and we will go through some answers together.

So I decided that this statement was sometimes true.

What did you work out? Here is my proof.

What did you work out? Here is my proof.

The first thing I wanted to do for the investigation was to try some different digits, both digits, the eight and the two that we saw in Junaid's example are even numbers.

Well, I wondered what would happen if we had odd numbers.

So in this example, both digits are odd.

I chose 19 reverse the digits, so I did 91.

The strategy I use to find the answer to this one, or the total was the adjusting near multiples of 10.

So I've actually taken one from the 91 and moved it over to the 19 to make 20.

So I now have 20 plus 90, which is very easy to calculate, it's 110.

So I found that it gave me an even total, just like Junaid had said, so two add numbers, reversed will give me an even total.

So to investigate further, I thought, well, what if I have the digits where one is an even number and one digit is an odd number.

So my even number is two, my odd number is seven, reverse the digits and added them together.

So I rounded 27 to 30 this time by adding three extra on.

30 plus 72 is equal to 102 because I added three extra on when I rounded, I've got to take them back off.

102 subtract three is equal to 99.

So I noticed that my total was an odd number.

So it's not always true what you need had said.

So we're trying to enter the way I put my odd digit first followed by an even number.

So I chose 36 and reverse the digits to plus 63.

And I used a partitioning and recombining strategy to add them together.

30 plus 60 equals 90, six plus three equals nine 90, plus nine is equal to 99.

And again, I found that the total was odd.

So that helped me conclude that Junaid statement was sometimes true.