Lesson video

Hello, I'm Miss Mia 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.

In this lesson, you'll be solving problems involving the 3, 6 and 9 times tables.

I'm super excited because now you are going to be using your knowledge of the 3, 6 and 9 times tables to solve problems. Your keywords are on the screen now and I'd like you to repeat them after me.

Divisible/divisibility.

Multiple.

Now, you may have seen these keywords in previous lessons and now we're just going to have a quick recap of what these words mean.

So divisibility is when division of a number results in another whole number.

A number is divisible by another if it can be shared exactly with no remainder.

A multiple is the result of multiplying a number by another whole number.

So, this lesson is all about our 3, 6 and 9 times tables.

We've got 2 lesson cycles here and our first lesson cycle is all about thinking of a number.

Our second lesson cycle then moves on to solving problems to do with our 3, 6 and 9 times tables.

So are you ready? Let's begin.

In this lesson, we are going to have Andeep and Izzy, who are going to be helping us with our mathematical thinking.

Now, Izzy is playing a maths game with Andeep.

"I'm thinking of a number that is greater than 20 and less than 30.

It is a multiple of 3 and a multiple of 9, but not a multiple of 6.

What number can this be?" "So I can write that as 20 is less than 30." So, he's now identified what those numbers are.

So it can be 21, 22, 23, 24, 25, 26, 27, 28, 29.

But what I believe in is the process of elimination.

So using our divisibility rules, we can now eliminate those numbers to try and figure out what number Izzy is thinking about.

So that's what Andeep says.

He's going to use his divisibility rules.

So for a number to be a multiple of 6, it has to be an even number and divisible by 3.

Now, knowing this, we can use the process of elimination to get rid of the numbers that do not fall under these conditions, so we can now eliminate the even numbers.

And it's interesting he said even numbers because we can now identify what those even numbers are and eliminate them, so then we know what numbers we are working with, and these are going to be odd numbers.

So let's begin.

So he can now eliminate 21 because although it is a multiple of 3, it is not a multiple of 9.

We can eliminate 22 because it's an even number.

We can eliminate 23 because 23 is not a multiple of 9.

We can eliminate 24 because it's an even number.

We can eliminate 25 because it's an odd number and it is not a multiple of 3.

We can eliminate 26 because it's an even number.

Now, we'll leave 27 because it is a multiple of 3 and the digit sum for 27 is 9.

So we'll leave that there for now.

We'll eliminate 28, digit sum for 29 is 11, which is not a multiple of 3, so it is not divisible by 3.

Which leaves us with 27.

"The number that you're thinking about must be 27." "Yes, that's correct." Well done if you managed to figure out that it was 27.

Over to you, can you find Andeep's missing number? I'd like you to justify your thinking to your partner.

So he says that the number that he's thinking about is greater than 30, but less than 40, is a multiple of 3 and 6.

You can pause the video here.

Remember to use your divisibility rules to help you.

Off you go, good luck.

So, what number did you get? The number that Andeep must be thinking about is 36, and that's because the digit sum is 9 and which means it's a multiple of 3 and it is an even number, which means it is divisible by 6.

So 36 was the missing number.

Well done if you got that correct.

No, Izzy plays another round with Andeep.

This time the numbers are larger.

"I'm thinking of a number that is greater than 400 and less than 410.

It is a multiple of 3 and a multiple of 6.

What numbers could this be?" "So I can write that as 400 is less than 410" And the numbers that he's identified that fall fall into this are the numbers that you see on the screen here.

So from 401 all the way up to 410.

Now, Andeep says he can use the divisibility rules to help him.

He says for a number to be a multiple of 3, the digit sum has to be a multiple of 3.

It also has to be an even number for it to be divisible by 6.

We can eliminate our odd numbers straight away.

Now, he then says that the digit sum for 405 is 9.

That's a multiple of 3, but it's an odd number so that will not work.

So he can now eliminate 405.

He says that knowing 405 is a multiple of 3, he can eliminate 403, 404, 406 and 407 because the next multiple of 3 has to be 308.

So the digit sum for 408 is 12.

That's a multiple of 3.

It's also an even number, meaning it is divisible by 6.

The digit sum for 402 is 4 plus zero plus 2, which is equal to 6.

That's a multiple of 3.

It's also an even number, meaning it is divisible by 6.

Yes, that's correct.

Over to you.

Can you find and Andeep's missing number? And again, I'd like you to justify your thinking to your partner.

Now, Andeep says, "It is greater than 350 but less than 360.

It is a multiple of 3 and 6." Remember to use your divisibility rules to help you.

You can pause the video here and click play when you're ready to rejoin us.

So how did you do? Well, the number Andeep is thinking about is 354, and that is because the digit sum is 12.

It is a multiple of 3 and it is an even number, which means it is divisible by both 3 and 6.

Izzy now thinks of a 4 digit number.

She says, "I'm thinking of a number that is greater than 1,275 and less than 1,280.

It is a multiple of 9 and it is a multiple of 6.

What number can this be?" So Andeep then identifies that he can write this as 1,275 is less than 1,280.

And here are the numbers on the screen.

So he's going to use his divisibility rules again and he's going to use the process of elimination to help him.

So he remembers that for a number to be a multiple of 9, it's digit sum also has to be a multiple of 9.

Now it has to be an even number as that's the only way it can be divisible by 6.

So, he says he can begin by eliminating the odd numbers.

So we can eliminate 1,275 1,277 and 1,279.

The digit sum of 1,278 is 1 plus 2 plus 7 plus 8.

And we know that is equal to 80, is a multiple of 9 and is an even number.

So we can circle that as a potential number that Izzy is thinking about.

Then if we look at 1,276, the digit sum for 1,276 would be 1 plus 2, which is 3.

3 plus 7 is 10.

10 plus 6 is 16.

16 is not a multiple of 9, so that won't work.

And the digit sum for 1,280 is 11 and 11 is not a multiple of 9, so that won't work either.

So, that means the number that Izzy is thinking about is 1,278.

Izzy says, "Yes, that's correct." Over to you.

Can you find Andeep's missing number? I'd like you to justify your thinking to your partner.

Now, he says, "It is greater than 3,365 but less than 3,370.

It is a multiple of 3, 6 and 9." You can pause the video here and click play when you're ready to rejoin us.

So, which number did you get? If you got 3,366, you are correct, and that is because the digit sum is 18 and it is an even number.

Onto the main task for this lesson cycle.

Question 1.

Izzy is thinking of a number that is greater than 300 and less than 310.

It is a multiple of 3, 6 and 9.

What number can this be? Question 2, Andeep is thinking of a number.

It is greater than 500 and less than 520.

It is divisible by 3, 6 and 9.

What numbers could he be thinking about? Question 3, Andeep is thinking of a number.

It is greater than 2,345 and less than 2,350.

It is divisible by 3, 6 and 9.

What numbers could he be thinking about? Question 4.

Using your knowledge of the divisibility rules between 3s, 6s and 9s, circle the numbers that are multiples of 3, 6, and 9.

You can pause the video here, click play when you're ready to rejoin us.

Off you go, good luck.

So what did you get? Now, for question 1, the number is 306, and that's because 306 is an even number, which means it is divisible by 6 and the digit sum is 9.

And because it's a multiple of 3, it means it is also a multiple of 6.

Now, for question 2, 504 was the correct answer, and that's because the digit sum for 504 is 9.

Now 9 is a multiple of 3 and 9 and 504 is an even number so it is divisible by 6.

Question 3.

So, Andeep's number was 2,349, and that's because the digit sum for 2,349 is 18.

18 is a multiple of 3 and a multiple of 9.

Well done if you got that correct.

For question 4, the way you could have answered this was by remembering your divisibility rules.

So, all multiples of 9 are multiples of 3, and even multiples of 3 are multiples of 6.

These are the numbers that you should have circled.

Well done if you circled them.

Well done if you managed to circle all of the ones that are on the screen correctly.

Let's move on.

For this lesson cycle, we are now going to be solving problems. Are you ready? Let's get started.

Andeep and Izzy are playing a game during PE using hoops and bean bags.

The aim of this game is to score as many points by throwing the bean bags aiming for the middle.

Izzy and Andeep start with 3 bean bags each.

Izzy goes first.

Each number represents the amount of points.

So she's thrown 1 bean bag into zone 3.

Then she's thrown her second bean bag into zone 6, and she's managed to get the bullseye, she's scored 9 points, so zone 9.

Izzy says she scored 18 because 3 add 6 add 9 is equal to 18.

"I'm going to have another go." So let's see what she scores this time.

Right, she scored 6 points, and another 6 points, and 3 points.

So this time she knows she scored 15 because 2 times 6 is equal to 12 and then 12 add 3 is 15.

Over to you.

How many points did Izzy score? You can pause the video here and click play when you're ready to rejoin us.

So, how many points did Izzy score? Izzy scored 18 points because 3 times 6 is equal to 18.

And notice here how we've used multiplication to help us, and that's because it's far more efficient to use our multiplication knowledge than to use repeated addition.

It can take a bit longer.

Now, this time Andeep starts with 5 bean bags.

Let's see what he scores.

6 points, another 6 points and another 6 points, and again, and then 3 points.

Unfortunately he did not get 9 points this time round.

So Andeep says he scored 27, and he knows that because 6 add 6 add 6 add 6 add 3 is equal to 27.

"There must be a quicker strategy.

I can use grouping to help me." So, we know that 4 times 6 is equal to 24.

So we've got 4 bean bags in zone 6, so that's 4 times 6, which is equal to 24.

And then we've got 1 bean bag in zone 3.

So 1 times 3 is equal to 3.

So 24 add 3 is 27.

Now Andeep has 5 bean bags.

He makes a prediction.

"The highest I would've been able to score is 45 points." Do you agree? Well, "It's true because the highest score is 9, and I have 5 bean bags so that's 9 times 5 which is equal to 45." Over to you.

If Andeep starts with 7 bean bags, what is the highest amount of points he can score? So he's got 7 bean bags there.

The points zone remains the same.

I'd like you to justify your thinking to your partner.

You can pause the video here.

So, what did you get? Well, the highest amount of points is 9, and Andeep has 7 bean bags.

So 9 times 7 is equal to 63, which means the highest amount of points Andeep can score with 7 bean bags is 63.

Now Andeep has 6 bean bags.

He makes another prediction.

"The lowest amount of points I would've been able to score is 18 points." Do you agree? Well, it's true because the lowest score is 3 and Andeep has 6 bean bags, so that's 6 times 3, which is equal to 18 points.

Over to you.

If Andeep starts with 4 bean bags, what is the lowest amount of points he can score? You can pause the video here and click play when you're ready to rejoin us.

So what did you get? Well, the lowest amount of points that Andeep can score is 3, and Andeep has 4 bean bags so that means 4 times 3 is equal to 12.

The lowest amount of points Andeep could have scored was 12.

Well done if you got that correct, fantastic job.

Now Andeep throws 8 bean bags.

How can he efficiently calculate the total score? Now, we can see that he's got quite a few points across the different zones.

So he says that he can figure this out using his times table knowledge.

In other words, grouping.

So we're going to start off with our outer zones first, which is 3 points.

Now he's got 3 bean bags in the point 3 zone, so that's 3 times 3 which is equal to 9.

Then he's got 4 bean bags in the 6 point zone, which gives us 4 times 6.

That's equal to 24.

And lastly, 1 bean bag in the central point, he's managed to 9 points once, so that's 1 times 9 which is equal to 9.

So what we would do after that is add those together to get our final score.

Or another way that we could have recorded this was 3 times 3, add 4 times 6, add 1 times 9, which is equal to 42.

Which way would you have preferred to calculated this? You may have used repeated addition or you may have used grouping to help you.

I would argue grouping is far more efficient.

So Andeep's total score is 42 points.

Now, Izzy scored 18 points using 3 bean bags.

Andeep says he can figure out what ways Izzy might have done this.

So, "She may have thrown all 3 bean bags into the 6 zone.

That's 6 times 3, which is 18." That's one way she could have scored 18 points.

"Or she may have also thrown one bag each in each of the zones." So that's 1 bag in the 9 zone, 1 bag in the 6 zone, and 1 bean bag in the 3 point zone, which is also equal to 18.

So there are 2 combinations of ways she could have scored 18 points.

So Izzy scored 27 points.

How many ways could she have scored this? Now for this question, think about how many bean bags you'll use.

You can change that amount.

You can pause the video here and click play when you're ready to rejoin us.

So what did you get? Well, she may have thrown 3 bean bags into the 9 point zone.

So that's 9 times 3, which is equal to 27.

Or she may have thrown 9 bean bags into zone 3, that's 3 times 9, which is equal to 27.

Or she may have thrown 3 bean bags into zone 6 and one in zone 9.

So that is 3 times 6 add 1 times 9, which is equal to 27.

Well done if you manage to get some or most of those combinations.

Onto the main task for your lesson cycle.

So for question 1, you're going to play the game.

You could use small cubes or paperclips if you don't have a large space and bean bags.

Start with 5 bean bags and calculate efficiently who scored the most and who scored the least.

Now do remember if you are using small cubes, just be careful 'cause you don't want to flick it too hard or it could lead to an injury.

Then have a go with 8 bean bags each.

How many ways can you score 27 points? For question 3, Izzy has 5 bean bags.

What is the maximum score she could have scored? What is the minimum score that she could have scored? For question 4, Andeep has 7 bean bags.

What is the maximum score he could have scored and what is the minimum score he could have scored? Now, question 5, Izzy scored 30 points, what could she have scored? Question 6, how could you score 60 points? Can you find more than one way? And question 7, if Andeep has 12 bean bags and scored the minimum score and Izzy has 6 bean bags and scored the maximum amount of points, who would score more? Explain how you know.

You can pause the video here.

Off you go, good luck, have fun, and then click play when you're ready to rejoin us.

So how did you do? For question 1, you may have got something like this.

5 bean bags each, so 2 times 3, add 3 times 6 is 24 points or 1 times 9, add 4 times 6 is 33 points.

For question 2, you may have got 3 times 9, add 1 times 3 is equal to 30, or 5 times 6 is equal to 30, or 4 times 3 add 3 times 6, which would be equal to 30.

So there were many combinations there.

For question 3, now, the maximum score Izzy could have got with 5 bean bags.

Well we know that 9 is the maximum points that you could score.

So 9 times 5 is equal to 45, and the minimum score was 15, and that's because the lowest points that you could get were 3.

And if there's 5 bean bags, that's 3 5 times, which is 15.

Now, Andeep has 7 bean bags.

In order to calculate the maximum score, we know that the highest amount of points that he could have scored again was 9.

So that's 7 9 times, which is 63.

And lastly, the minimum score would've been 21, and that's because 3 7 times is 21.

And lastly, Izzy scored 36 points.

What could she have scored? Well, here are some solutions.

She could have scored 6 6 times or 4 bean bags in the 9 point zone 4 times.

6 points 3 times as well as 9 points 2 times to get 36 points altogether.

Now, for question 6, here are some more solutions in ways that 60 points could have been scored.

So 10 bean bags in the 6 point zone would've given 60 points.

10 bean bags in the 3 point zone add 5 bean bags in the 6 point zone would've given 60 points, or 20 bean bags in the 3 point zone would've also given 60 points.

Now for question 7, Andeep's points would've been 36 because 12 times 3 is 36, and Izzy's points would've been 54 because 6 times 9 is 54, so Izzy would've scored more.

Well done, you've made it to the end of this lesson.

I really hope you had fun.

So let's summarise our learning.

Today you solved problems involving the 3, 6 and 9 times tables.

You should now understand that knowing the 3, 6 and 9 times tables can help you to solve problems. You also can use the relationship between the 3s, 6s and 9s to help you solve problems. Thank you for joining me in this lesson.

I really hope you enjoyed learning about your 3s, 6s, and 9 times tables.

Bye.