Lesson video

Hi, my name is Mr. Peters.

Thanks for joining me today for this lesson.

In this lesson today, we're gonna be using our knowledge and understanding of factors and how we can use this knowledge of factors to 100 in particular to help solve calculations more efficiently.

If you're ready to get started, let's get going.

So by the end of this lesson today, you should be able to say that I can explain how to use factor pairs of 100 to solve calculations efficiently.

Throughout this lesson, we've got three key words we're gonna be referring to.

I'll have a go at saying them first and then you repeat them after me.

The first one is factor.

Your turn.

The second word is commutative.

Your turn.

And the last one is associative.

Your turn.

Let's have think about what these mean then.

Factors are whole numbers, which exactly divide another whole number.

The commutative law states that you can write the values of a calculation in a different order without changing the calculation.

The result will stay the same.

This applies to both addition and multiplication.

The associative law states it doesn't matter how you group of pair values i.

e, it doesn't matter which group or pair you calculate first, the result will always again, stay the same.

This once again applies to both addition and multiplication.

Okay, so this lesson is broken down into two cycles.

The first cycle will be thinking about factors of 100 and the second cycle, we'll be thinking about simplifying equations using factors of 100.

If you're ready, let's get going for the first cycle.

Throughout the session today, we're gonna be joined by Andeep and Sophia who again, as always, will be sharing their thinking and any questions they have throughout the lesson.

So let's start here with Andeep and Sophia exploring the factors of 100.

You can see here that we've got one long row of tiles which represent 100 tiles.

And we're gonna work through this and find the factors of 100 and represent this on our factor bug.

So this one long row of tiles represents one row and 100 columns.

So we know that one and 100 are factors of 100.

We can represent this as the antennae on the factor bug, can't we? We know that every number has one and itself as a factor, and we represent these on the antennae of the factor bug.

We know that this can also be represented as a multiplication and division equation.

We know that 100 is equal to one times by 100, and we know that 100 divided into one row is equal to 100.

Sophia's saying that she knows that 100 is an even number, so two can also be a factor.

Let's have a look at that.

She knows that 100 equal to two, multiplied by 50, and that 100 divided by two is equal to 50.

Therefore, if you look at our array now, we will change our array to show two rows and 50 columns.

Let's add those factors onto our factor bug now.

"Hmm, I wonder, can 100 be divided by three?" You're right, Andeep.

It can be divided by three, but it can't be divided by three equally, where the factors would be positive whole numbers.

So whilst we know that 100 is equal to three multiplied by 33 and a third, or 100 divided by three is equal to 33 and a third, the 33 and a third isn't a positive whole number, therefore it cannot be a factor.

Sophia's pointed out that often where two is a factor, four could also be a factor.

Let's have a look at that now.

Yep, 100 can be divided into four equal rows of 25.

And again, we can add that to our factor by 'cause four and 25 is factors and this can be recorded as an equation here.

100 is equal to four multiplied by 25.

100 is the product, four and 25 are the factors.

And as a division equation, 100 divided by four is equal to 25.

We know that the dividend represents the product of 100 and we know that the divisor and the quotient, the four and the 25 represent the factors as well.

Okay, I wonder if we can find any more factors now.

Well, Sophia's saying she knows that 100 ends in a zero and any number that ends in a zero or a five is a multiple of five, isn't it? So that must mean we can make it into five rows.

So we can see here we've got five equal rows and each row has 20 in it.

So we know that five and 20 can also be factors of 100.

We know that 100 is equal to five multiplied by 20, and we know that 100 divided by five is equal to 20.

And finally we know that all multiples of 10 are also multiples of five.

And this multiple of five ends in a zero, so therefore it must be a multiple of 10.

So we can divide it into 10 equal rows.

So that would be 10 rows and 10 columns.

And we can see that on the top left there.

Hmm, what'd you notice about that? That's right.

It makes a square, doesn't it? So it's a square number.

When we record the factors of a square number on our factor bug, we only need to record the factor once.

So we're gonna do that as the tail on our factor bug.

Because the factor is repeated, we don't need to write it down twice.

So we found a great deal of factors for 100, but we need to know when to stop looking for any more factors, don't we? So by linking factors to division, this can help us 'cause there's a rule that we have that supports us with this.

Let's have a look at our division equations here.

Remembering the terminology of a division equation.

We've got the dividend divided by the divisor is equal to the quotient.

100 divided by one is equal to 100.

One divided by two is equal to 50.

100 divided by four is equal to 25, and 100 divided by five is equal to 20.

And then finally 100 divided by 10 is equal to 10.

Take a moment to have a think.

What do you notice is happening to the quotient each time and what's happening to the divisor each time? That's right.

You might have noticed that the quotient is decreasing in size and the divisor is increasing in size, isn't it? And we have a rule here that we know that when the divisors becomes greater than the quotient or the quotient becomes smaller than the divisors, we know that we've exhausted the number of factors that we can find and there's no more to find.

So we can see here, that the quotient 100 is larger than the divisor, and it's still larger here.

And the quotient of 25 is still larger here than the divisor of four.

Once again, the quotient is still larger here.

we've got 20 and the divisors is five.

However, now we've got the quotient which is 10, and the divisor that is 10.

If we try to find any more after this, we know that the divisor would then overtake the size of the quotient, and would therefore end up with us repeating factors.

So at this point here, we know there's no more factors to find.

Let's think about the factors of 100 then, and what this tells us about the number itself.

Take a moment to have a think for yourself.

Remembering, what the factors can tell us about the type of number that it is.

That's right.

There are more than two factors, so therefore we know it's not a prime number, it must be a composite number.

And it also has an odd number of factors, doesn't it? Which means, it must be a square number.

We know it's only square numbers that have an odd number of factors and we already figured that out previously when we looked at the shape of the square of 100 with the tiles.

So we can say that 100 is a composite number, and we know that a square number is a special type of composite number.

Okay, time for you to check your understanding now.

tick the factor pairs of 100.

Take a moment to have a think.

That's right.

It's A, C and D, isn't it? Two and 50 are a factor, pair of 100.

Five and 20 are a factor pair of 100.

And four and 25 are also a factor pair of 100.

We can then also go on to apply our understanding of factor pairs to 100, in real life contexts.

Let's have a look here.

Here you can see an electronic pad and at the top of the pad you can see the battery.

And Sophia is saying that she often wants to know how much battery is left on her pad, to know how much longer she might have left to play on it before she needs to charge it.

Have a look here on the right hand side.

We've now got a battery which has been divided into sections to show how much charge is left in it.

We know that when the battery is full, we can say that it is 100% charged.

Having a look at this battery, we can see that there are four intervals.

We know that four and 25 are factors of 100, therefore each interval is 25%.

So we can see that the battery has 25% charge left in it, doesn't it? Hmm, it's getting towards the end of its battery life, Sophia, you might need to charge it again soon.

Again, we could look at this in the context of money as well.

Have a look here.

We've got five lots of 20 Ps here, haven't we? Andeep is saying that he found these five 20 Ps in his wallet, how much money does he have altogether? Well, we know that five and 20 are factors of 100, therefore, we could say that we have 100 pence or we know that 100 pence is equivalent to one pound, isn't it? Okay, another quick check for understanding now, have a look at this battery.

It's been divided in some equal parts to show how much percentage battery life is left in it.

Take a moment to have a think, how much charge is in it? That's right.

We know that a full battery has 100% charge in it, doesn't it? And our battery has 10 intervals in it, and there is one interval which doesn't have any battery life in it.

So that one interval represents 10%.

That means we've lost 10% battery or there is 90% battery leftover, isn't there? Okay, time for you to check your understanding now.

What I'd like you to do here is fill in the missing numbers for each of these examples here, using your factor pairs to tell me what the temperature is on each of the thermometers.

And then for question two.

Years, four, five, and six are raising money towards their school charity.

Each class is looking to raise 350 pounds.

What's the total percentage of money they've raised so far? Again, I'd like to use your fact pairs of 100, to help you answer how much money they have raised so far.

Good luck with those two tasks, and I'll see you back here shortly.

Okay, let's go through these then.

So we can see the first thermometer is divided into 10 intervals.

And if the thermometer was at the top, that would be 100 degrees.

Therefore, each interval is worth 10 degrees and the thermometer is matching the first line of 10 degrees.

We know that 10 and 10 are factors of 100.

So the temperature is 10 degrees.

The second example, well again, our hole would be 100 degrees and this time the intervals there are five of them.

So that means each interval is worth 20.

So we know that five and 20 are factors of 100.

So the temperature is 20 degrees.

And the last one, this time there are four intervals.

We know that the factors of 100 are four and 25, therefore, each interval is worth 25 degrees, isn't it? We can see that three intervals are shaded, or actually one interval isn't shaded and therefore that's 75 degrees.

Okay, and for task two, we can see that 100% would be the full amount of money of 350 pounds.

This has been divided into five equal parts for the first class.

So each one of those segments is worth 20%.

We've got four of those segments, so that's 80%.

So the first class has raised 80% towards their target.

The second class, their total is divided into 10 equal parts.

And the red line goes up to the end of the nine parts leaving one section, which represents the money that's left to be raised.

We know that each part is worth 10%, so 100% minus 10% would be equal to 90%.

And then the last one year six has been divided into four equal parts.

We know the factors are four and 25 that make 100, and two of those parts have been raised so far.

So that would be 50% of the total amounts, 'cause two lots of 25 is equal to 50.

Well done if you managed to get all of those.

Okay, now we're looking at starting cycle two here.

Simplifying equations using factors of 100.

Okay, so let's see how we can use these factor pairs to 100 to help us manage calculations more easily.

Have a look at this calculation here.

Got 25 multiplied by 20.

We can use our knowledge that five and 20 are factors of 100 to help us here.

I can see that one of the factors is 20, therefore if I can find a factor of five to multiply with it, we'll be able to multiply by 100.

So let's have a look at 25 then.

Well, I know that 25 can be broken down into factors of five and five.

We can now say that 25 multiplied by 20 is equivalent to five, multiplied by five, multiplied by 20.

We can then apply the associative law to pair two of the factors together.

If we pair the five and the 20 together, as I said previously, we'd be able to multiply these two together and this would be equal to 100, wouldn't it? So five, multiply by 100, that gives us 500 altogether.

That was a really quick and easy way of tackling that problem, wasn't it? Which looked quite tricky at first.

Let's have a look at another example.

This time we've got 36 multiplied by 50.

Take a moment to have a think.

Which factor do you think will break down into other factors and which factor do you think will keep to try and multiply it for something that would be equivalent to 100? Well, let's have a look, shall we? Again, we can use our knowledge that two and 50 are factors of 100, can't we? So if we keep the 50 where it is and break down the factors of 36, we need to break it down so that one of the factors is two, of course, because we want to use that two to multiply it with the 50, don't we? I know that two and 18 are factors of 36.

So we can say that 36 multiplied by 50 is equal to two multiplied by 18, multiplied by 15.

We can now apply the commutative law to reorder the factors, can't we? We know that it doesn't matter which order the factors are placed in, so we've now reordered it.

So we've got two multiplied by 50, multiplied by 18.

Can you see how the two and the 50 are next to each other now? That means we can now associate these two together.

So using the associative law, we can multiply this pair of factors together, first.

Two multiplied by 50 is equal to 100, and then we just need to multiply that by the 18, don't we? And we know that 18 lots of 100 is equal to 1,800.

Sophia is feeling a lot more confident now.

She says this is definitely gonna help her to make her calculations more efficient when working mentally in her head.

Here's one more example to look at.

Have a look at the factors carefully.

Which one of these factors would we keep and try and use that and multiply something with it to make something that would be equal to 100? Andeep's suggesting that Sophia has a go at this one.

Well, Sophia's noticed that she can use her factor pairs of 100 of four and 25, can't she? We can see we've already got the 25, so we need to find a four that will multiply with it and that would be equal to 100, wouldn't it? So we can break down the 48 into factors of four and 12 this time, that works out perfectly.

So we can say that 25 multiplied by 48 is equivalent to 25 multiplied by four, multiplied by 12.

Again, look at our equation now.

Have we got two factors that we could associate together that would be equal to 100? We do, don't we? We could associate the four and the 25 together, that would be equal to 100, and then 100 multiplied by 12 would be equal to 1200 or 1,200 wouldn't it? Well done Sophia, you got there, didn't you? Fantastic.

Okay, time for you to check your understanding again, now.

Can you match the equation to the factor pair that would be best to help solve the calculation more quickly? Okay, let's see how you got on.

Well, for the first one you can see one of the factors is 25, can't you? So therefore using four and 25 would help us.

So we'd need to break down that 32 to find a factor of four, wouldn't we? 65 multiplied by 20, one of the factors is 20.

So it would benefit us to use five and 20 as a factor pair, wouldn't it? So again, we need to break down that 65, so we've got a factor which is five.

And 50, multiply by 74.

Of course we know that two and 50 would be a factor pair of 100 that would help us here.

We'd need to break down that 74 to find a factor that is two.

Hmm, what goes with two to make 74? That's right.

36 isn't it? And another quick check for understanding, I'm gonna have a go first and then you can have a go at your own example.

Are you ready? We're gonna look at 20 multiply by 65.

I've already realised that there's a 20 here, so I want to get a five here that's gonna multiply with that to make 100.

So we're gonna decompose our 65 into factors of five and 13.

We can now rewrite our calculation as three factors.

We've got twenty multiplied by five, multiply by 13.

We then associate the 20 and the five with each other and multiply those together, that will make 100.

And now we've got 100, multiply by 13 is equal to 1,300.

Right? Your turn now, what I'd like you to do is have a go at tackling 80 multiply by 25.

Take a moment to have a think.

Okay, let's go through this together then.

Well, we can see we've got a 25 as one of the factors, so we need to break down the other factor to find a factor.

That would be four, because we know the factor pairs would be four and 25.

That would be equivalent to 100.

So 80 can be decomposed into factors of 20 and four.

We can now rewrite our calculation as 20 multiplied by four, multiplied by 25.

If we associate the four with the 25, and then multiply those two together, that would give us our 100.

So we've got 20 multiplied by 100 and that is equal to 2000, isn't it? Well done if you managed to get that.

Okay, and time for you to have a go at practising this now.

I've given you a number of calculations here and what I'd like to do is fill in the missing numbers as on the left hand side, you break down a two factor calculation to a three-factor calculation and then finally to the product.

And then finally, on the right hand side, I've taken out the middle step.

So you can go from working out with the two factors and thinking about what that would be as a product by breaking down one of those factors into another two factors.

And then for task two, what I'd like you to do is find three factors that are multiplied together to make 800, and within that you must use a factor pair that makes 100.

See how many different solutions you can come up with.

Good luck with those two tasks and I'll see you back here shortly.

Okay, let's go through these two then.

16 multiplied by 25 is actually equal to four multiplied by four, multiplied by 25.

That's because we've decomposed the 16 into four and four as factors.

We can now associate the four and 25 together to make 100.

So the answer would be 400 altogether, because we multiply that by the remaining four.

And for the second one we've got 32 multiplied by 25.

I can see already that I've got 25 as a factor, and I've also been given four as a factor amongst the three factors in the next stage of the calculation.

So what I'm gonna do here is decompose our 32 into eight and four as factors, and then I'm going to multiply those by 25.

We can associate the four and 25 together, multiplying those to make 100, and then we can multiply that by the remaining eight, which would be 800.

And again, for the last one, we've got 84 multiplied by 25 here, again, I can see that I've got a 25, so I'm looking to get a four as a factor from the 84.

84 can be decomposed into 21 and four, and then we can multiply that by 25.

Associate the four and the 25 together, that would give us 100 once we've multiplied them and then multiply that by 21.

So it gives us 2,100.

Well done if you've got those ones.

Okay, and the right hand side, then.

We've got 16 multiplied by 50.

Hmm, well I can see I've got the 50, so I need to decompose the 16 and it would benefit me if one of those factors was a two, wouldn't it? So I know that 16 can be decomposed into two and eight as factors.

Multiply the 50 and the two together, that would give me 100 and then multiply that by the eight.

That would give us 800.

Second example, I've got 32 times by 50.

Again, I can see I've got a 50, so I'm gonna break down my 32, and I'm gonna break that down as two and 16 this time because I need the two to multiply with the 50.

Two multiply by 50 would be equal to 100.

And then I multiply that by the 16, that gives us 1,600.

And then the last one, 84 multiplied by 50.

Well again, I've got a 50 here, so I need a two to be multiplying with the 50.

That would be equal to 100.

So we can decompose the 84 into two and 42, multiply the two and the 50 together, that would give us 100 and then multiply that by the 42 that remains, that would give us 4,200.

Have a look at all the products to both of our answers in both columns.

Did you notice anything? That's right.

If you look carefully, the factors on the left hand side were exactly the same except for we had 25s, didn't we? And on the factors on the right hand side, they were 50s, so we actually just doubled one of the factors in each of our calculations, and that's exactly what Sophia's picked up.

She's noticed that the products on the right hand side, were actually double those on the left hand side and that's because one of the factors has been doubled.

Therefore, we know that if you double the size of a factor, you also need to double the size of the product.

So actually we could have just calculated all the ones on the right hand side just by using the ones that we've learned already on the left hand side.

Well done, if you spotted that.

And then for task two then, let's use our factor pairs of 100 to help us with this calculation.

You could have recorded this as one multiplier by 100, multiplied by eight is equal to 800.

We know that one and 100 are a factor pair of 100.

We could have done two multiplied by 50 multiplied by eight, that would be equal to 800 again.

We could have done four multiplied by 25 multiplied by eight, that would be equal to 800.

And we could have done five multiplied by 20, multiplied by eight would be equal to 800.

Were there any others we could have done? Yeah, that's right.

We could have also done 10 multiplied by 10, multiplied by eight, which would be equal to 800.

Can you see how I worked through that systematically? Starting off with one of my factors being one, and then the next factor being two and finding the pairs for each of those as we went along.

Well done if you managed to come up with those for yourself too.

Sophia's wondering whether she could have extended her learning, by instead of finding factor pairs of 100 to make this calculation, using factor pairs of 200 instead.

Maybe if you've got some spare time, you could go on and do that for yourself.

Okay, that's the end of our learning for today.

To summarise what we've learned today, we know that knowledge of factor pairs of 100 can be applied in our everyday lives.

We know that you can try to make calculations easier by looking for factor pairs of 100 to help you calculate these.

And we also know that we can apply the commutative and associative laws to help us make our calculations easier when using factor pairs of 100.

Thanks for joining me today.

That was a really good lesson.

I really enjoyed myself and hopefully you did too.

Take care.

I'll see you again soon.