**Division, the Bus Stop Method, and Long Division** *KS2 Revision*

## What is Division?

#### Things to Remember:

• Division is splitting up a number or object into pieces that are the same size.

• Division is the opposite of multiplication.

o We can use our knowledge of times tables to find answers.

• We can count up in steps with the number we divide by.

• \div is our symbol for division.

In class your teacher might have said something like “I am going to split you up into groups of 5.” This is division! Division is just splitting things up into equal pieces. Here your teacher has split the class into groups with 5 students in. We have a fancy symbol for division, and it looks like this: \div. Let’s look at two ways of doing a division with small numbers.

### Counting Up

A class has 20 students and are split into groups of 5. How many groups are there?

In maths, this question would just look like:

20\div5=\_\_\_

Because we are dividing by 5, we know that each group will have 5 students in, so we just need to count up in 5s until we get to the number we are dividing, which is 20.

1\times5=5

2\times5=10

3\times5=15

4\times5=20

So, we can see that we need four 5s to make 20, so the answer is:

20\div5=4

Times Tables

It is important to know that division is the opposite of multiplication, so if you know your times tables you will be able to do these divisions pretty easily.

What is 24\div 8?

If we think about this as “How many 8s go into 24?”, then it is just a times table question. We know that three 8s make 24, so the answer must be 3.

24\div8=3

And we can check this with our method of counting up:

1\times8=8

2\times8=16

3\times8=24

We were correct, we needed 3 of them!

Now that we know what division is, we need to look at REMAINDERS.

Calculate 17\div5.

So, this question is asking us “How many 5s go into 16?” But we know that 16 isn’t in the 5 times table, so this can’t work. Let’s try counting up instead and see if that can help us.

1\times5=5

2\times5=10

3\times5=15

4\times5=20

This hasn’t given us a perfect answer, but we can see that our answer will be somewhere between 3 and 4. We want to take the smaller number, which is 3, and then find out how much is left. This is our remainder, which we find by taking our starting number (16) and subtracting the value of the 5s we wanted (15).

16-15=1

So, our remainder is 1. We have two ways to use this:

Leave as a remainder

16\div5=3\text{ remainder }1

Turn into a fraction

To turn a remainder into a fraction, we just write it as the remainder over what we divided by

16\div5=3\frac{1}{5}

Note: You can actually use these two methods for bigger numbers, but it can get quite tricky!

## Bust Stop Method (Short Division)

**Things to Remember:**

• The bus stop method is useful when we are dealing with big numbers or times tables we don’t know.

• The number we are dividing by goes on the outside (left).

• The number we are dividing goes on the inside (right).

• Having a strong knowledge of multiplication and times tables will help here.

The bust stop method is useful for when we are dealing with big number or tricky times tables that we might not know.

Calculate 903\div7

**Times Tables**

Hmmmm, we are dividing by 7 so need to use the 7 times table. We know our 7 times table, but we don’t know how many 7s make 903. Let’s try counting up instead.

Counting up

We are dividing by 7, so need to count up in 7s.

1\times7=7

2\times7=14

3\times7=21

4\times7=28

5\times7=35

…

Ok, this method will eventually give us an answer (129) but it is going to take us aaaaaages. Luckily though, there is quicker method. The Bus Stop Method. Once you are comfortable with smaller divisions and remainders this method will be super quick, but I am going to use “counting up” as well here to show how it works.

To start this method we need to draw the “bus stop” (which is where it gets its name from):

Next, we need to put the number we are dividing inside:

And the number we are dividing by on the outside:

And now, now we can start dividing the number inside the bus stop! We start by dividing the number on the left.

9\div7

We are dividing by 7, so count up in 7s.

1\times7=7

2\times7=14

So, we can see that the answer will be between 1 and 2. We take the smaller one (1) and then do the subtraction to figure out the remainder.

9-7=2

So, our answer is 1 remainder 2. The number of 7s we want (1) goes on top, and the remainder (2) moves to the right one place, beside the next number.

And now we can move on to divide the next number. We need to be careful though, because moving our remainder of 2 across has turned the 0 into a 20.

20\div7

We are dividing by 7, so count up in 7s.

1\times7=7

2\times7=14

3\times7=21

So, we can see that the answer will be between 2 and 3. We take the smaller one (2) and then do the subtraction to figure out the remainder.

20-14=6

So, our answer is 2 remainder 6. The number of 7s we want (2) goes on top, and the remainder (6) moves to the right one place, beside the next number.

And now we can move on to divide the last number. We need to be careful though, because moving our remainder of 6 across has turned the 3 into 63.

63\div7

Here is where knowing your times tables comes in handy. We know that nine 7s going into 63, so the answer must be

63\div7 =9

Because we don’t have a remainder, we can just put the 9 on top!

And finally, we just need to take the number on top of the bus stop as our answer.

903\div7 =129

But, what would have happened if our last number hadn’t divided so nicely? Let’s changedit ever so slightly by adding 2 to our starting number.

Calculate 905\div7

As a bus stop, this would look like:

Because we are dividing by 7 again and the first two numbers in the bus stop are the same as before, we can skip through to the last step:

So, now, we are doing 65\div7. This is where having a strong knowledge of times tables can be helpful. We know from the previous example that nine 7s make 63 and that if we add another seven to make ten 7s, that will give us 70. This is too much. So, we only want 9 of them. But now, we need to figure out our remainder by doing the subtraction.

65-63=2

So, our remainder is 2. Now, we know that the 9 has to go on top of the bus stop, but where does this remainder 2 go? Well, we have three methods.

Leave as a remainder

To leave it with a remainder we just write a small “r” and the remainder next to it.

And now we can take the number on the top of the bus stop as our answer and the 2 as the remainder.

905\div7 =129\text{ remainder }2

Turn the remainder into a fraction

Just like before we can turn the remainder into a fraction by putting it over the number we are dividing by.

905\div7 =129\frac{2}{7}

Continue dividing

To continue dividing we just need to continue the step of moving the remainder to the right. But I know what you’re thinking “There are no more numbers we can move the remainder next to”. Well, we can add some! What comes after units? Decimals! We can add a decimal point and 0s without it affecting our starting number, and then just move our remainder along like before.

So, we can see that we have added a decimal point, a 0, and moved our remainder across. Now, we are dividing 20 by 7.

We are dividing by 7, so count up in 7s.

1\times7=7

2\times7=14

3\times7=21

So, we can see that the answer will be between 2 and 3. We take the smaller one (2) and then do the subtraction to figure out the remainder.

20-14=6

So, our answer is 2 remainder 6. The number of 7s we want (2) goes on top, and the remainder (6) moves to the right one place, beside the next number.

Each time we have a remainder, we add another 0.

This time we are doing 60\div7.

We are dividing by 7, so count up in 7s.

1\times7=7

2\times7=14

3\times7=21

4\times7=28

5\times7=35

6\times7=42

7\times7=49

8\times7=56

9\times7=63

So, we can see that the answer will be between 8 and 9. We take the smaller one (8) and then do the subtraction to figure out the remainder.

60-56=4

So, our answer is 8 remainder 4. The number of 7s we want (8) goes on top, and the remainder (4) moves to the right one place, beside the next new 0.

**Note:** This division, will actually go on forever and gives a recurring decimal.

## Long Division

**Things to Remember:**

• Long division is useful when we are dealing with big numbers or times tables we don’t know.

• The number we are dividing by goes on the outside (left).

• The number we are dividing goes on the inside (right).

• Having a strong knowledge of multiplication and times tables will help here.

Long division, as the name would suggest, is longer version of short division. Although it looks different, we are actually doing the exact same thing. However, it is good to know/see both methods so you can choose the method which works best for you.

Calculate 2585\div11

To start with, we set up Long Division just like the bus stop method, with the number we are dividing by on the outside and the number we are dividing on the inside.

Again, just like division we need to start by dividing the first number inside the bus stop by the number on the outside. However, we know that 11 can’t go into 2, so we need to put a 0 top.

Unlike the bus stop method, we don’t usually cross out any numbers or move them across in long division, we just group them in our heads. So, the 2 and 5 make 25.

Our next step is to see how many 11s we can fit into 25, which we can do by counting up again.

1\times11=11

2\times11=22

3\times11=33

So, we can see that we can fit two 11s into 25 before we go over. We put the number of 11s we can fit into 25 (2) on top above the 5, and write the 22 underneath.

Our next step is to turn the 25 and 22 into a column subtraction to find the remainder.

Which we know how to!

And now we need to move the next number along (the 8) down to our remainder.

Our next step is to see how many 11s we can fit into 38, which we can do by counting up again.

1\times11=11

2\times11=22

3\times11=33

4\times11=44

So, we can see that we can fit three 11s into 38 before we go over. We put the number of 11s we can fit into 38 (3) on top above the 8, and write the 33 underneath the 38.

We once again turn this into a column subtraction, which gives us our remainder.

And now we need move the next number along down to make 55.

Now, we need to see how many 11s we can fit into 55.

1\times11=11

2\times11=22

3\times11=33

4\times11=44

5\times11=55

Luckily 11 goes into 55 perfectly, 5 times, so we just need to put this 5 on top and we’re done!

And our answer is the number on top. Remember though, we can ignore the 0 because it is at the front!

2585\div11=235

Just like the bus stop method though, things can get a little trickier if this last number doesn’t divide nicely. Let’s change our starting number slightly to see what happens.

Calculate 2584\div11

Because all but the last number are the same, this long division method will look the same up until the last stage where we will bring a 4 down instead of the 5.

This time we need to see how many 11s will go into 54.

1\times11=11

2\times11=22

3\times11=33

4\times11=44

5\times11=55

So, we can see that we can fit four 11s into 54 before we go over. We put the number of 11s we can fit into 54 (3) on top above the 8, and write the 44 underneath the 54.

Which we turn into a column subtraction again.

With this last remainder we have three options.

**Leave as a remainder**

If we leave this 10 as the remainder we get the following answer:

2584\div11=234\text{ remainder }10

**Turn into a fraction**

To turn the remainder into a fraction, we just need to put it over the number we were dividing by (11).

2584\div11=234\frac{10}{11}

**Keep on dividing**

Usually after a column subtraction we would bring the next number down, but we don’t have one… we can however add one. Like with the bus stop method, we can add a decimal point and a 0.

Bringing this 0 down then gives us a 100 to divide by 11.

Like before we need to see how many 11s we can fit into 100.

1\times11=11

2\times11=22

3\times11=33

4\times11=44

5\times11=55

6\times11=66

7\times11=88

8\times11=88

9\times11=99

10\times11=110

So, we can see that we can fit nine 11s into 100 before we go over. We put the number of 11s we can fit into 100 (9) on top above the 0, and write the 99 underneath the 100.

This then becomes our column subtraction.

We have run out of numbers again though, so we need to add another 0. Each time you run out of numbers when dividing, just add another 0 and bring it down.

Like before we need to see how many 11s we can fit into 10.

1\times11=11

We can’t fit any 11s into 10. We put the number of 11s we can fit into 100 (0) on top above the 0, and write the 0 underneath the 10 to make another column subtraction.

If we were to continue this division this division it will actually give a recurring decimal of 234.9090909090… Try completing it to see if you can figure out why.

## Example Questions

__What is Division?__

**Question 1:**

Calculate 35\div7

WORKED SOLUTION:

**Times Tables**

This question is asking “How many 7s go in to 35?” We know that five 7s go into 35, so the answer must be 5.

35\div7=5

**Counting Up**

Here we are dividing by 7, so we want to count up in 7s:

1\times7=7

2\times7=14

3\times7=21

4\times7=28

5\times7=35

So, we can see that we need five 7s.

35\div7=5

__What is Division?__

**Question 2:**

Johannes, Arifah, and Claire have 18 chocolates that they share equally. How many does each one get?

Looking at this question we can see that there are 3 people who are sharing 18 chocolates, so we are splitting up the 18 chocolates up into groups of 3. In maths, this would look like:

18\div3

**Times Tables**

This question is asking “How many 3s go in to 18?” We know that six 3s go into 18, so the answer must be 6.

18\div3=6

**Counting Up**

Here we are dividing by 3, so we want to count up in 3s:

1\times3=3

2\times3=6

3\times3=9

4\times3=12

5\times3=15

6\times3=18

So, we can see that we need six 3s.

18\div3=6

__What is Division?__

**Question 3:**

Calculate 29\div6, give your answer as a remainder and a fraction.

__Counting Up__

Here we are dividing by 6, so we want to count up in 6s:

1\times6=6

2\times6=12

3\times6=18

4\times6=24

5\times6=30

So, we can see that the answer will be between 4 and 5. We take the smaller one (4) and then do the subtraction to figure out the remainder.

29-24=5

So, our remainder is 5.

__Leave as a remainder__

29\div6=4\text{ remainder }5

__Turn into a fraction__

29\div6=4\frac{5}{6}

**Bust Stop Method (Short Division)**

**Question 1:**

Use the bus stop method to calculate 1599\div13.

We set this division up as before, with the number we are dividing by on the outside and the number we are dividing on the inside.

And now we just need to go through doing our divisions.

1\div13

Well, 1 is smaller than 13, so we know that zero 13s go into it and we have the 1 left over. That means we put 0 on top and move the 1 across.

The 1 that we moved across turned the 5 into 15, so we are doing 15\div13 now. We are dividing by 13 so need to count up in 13s.

1\times13=13

2\times13=26

So, we can see that the answer will be between 1 and 2. We take the smaller one (1) and then do the subtraction to figure out the remainder.

15-13=2

So, our answer is 1 remainder 2. The number of 13s we want (1) goes on top, and the remainder (2) moves to the right one place, beside the next number.

The 2 that we moved across turned the 9 into 29, so we are doing 29\div13 now. We are dividing by 13 so need to count up in 13s.

1\times13=13

2\times13=26

3\times13=39

So, we can see that the answer will be between 2 and 3. We take the smaller one (2) and then do the subtraction to figure out the remainder.

29-26=3

So, our answer is 2 remainder 3. The number of 13s we want (2) goes on top, and the remainder (3) moves to the right one place, beside the next number.

The 3 that we moved across turned the 9 into 39, so we are doing 39\div13 now. We are dividing by 13 so need to count up in 13s.

1\times13=13

2\times13=26

3\times13=39

Here we can see that we need exactly three 13s, with no remainder, so just need to put the 3 on top to finish.

The number on top of the bus stop is our answer. Because the 0 is at the front of the answer, we can just ignore it!

1599\div13=123

**Bust Stop Method (Short Division)**

**Question 2:**

Use the bus stop method to calculate 8320\div23, giving your answer as a mixed number.

We set this division up as before, with the number we are dividing by on the outside and the number we are dividing on the inside.

And now we just need to go through doing our divisions.

8\div23

Well, 8 is smaller than 23, so we know that zero 23s go into it and we have the 8 left over. That means we put 0 on top and move the 8 across.

The 8 that we moved across turned the 3 into 83, so we are doing 83\div23 now. We are dividing by 23 so need to count up in 23s.

1\times23=23

2\times23=46

3\times23=69

4\times23=92

So, we can see that the answer will be between 3 and 4. We take the smaller one (3) and then do the subtraction to figure out the remainder.

83-69=14

So, our answer is 3 remainder 14. The number of 23s we want (3) goes on top, and the remainder (14) moves to the right one place, beside the next number.

The 14 that we moved across turned the 2 into 142, so we are doing 142\div23 now. We are dividing by 23 so need to count up in 23s.

1\times23=23

2\times23=46

3\times23=69

4\times23=92

5\times23=115

6\times23=138

7\times23=161

So, we can see that the answer will be between 6 and 7. We take the smaller one (6) and then do the subtraction to figure out the remainder.

142-138=4

So, our answer is 6 remainder 4. The number of 23s we want (6) goes on top, and the remainder (4) moves to the right one place, beside the next number.

The 4 that we moved across turned the 0 into 40, so we are doing 40\div23 now. We are dividing by 23 so need to count up in 23s.

1\times23=23

2\times23=46

So, we can see that the answer will be between 1 and 2. We take the smaller one (1) and then do the subtraction to figure out the remainder.

40-23=17

So, our answer is 1 remainder 17. The number of 23s we want (1) goes on top and have a remainder of 17.

Now, to get our answer as a mixed number, we take the answer on top of the bus stop and turn the remainder into a fraction. To turn the remainder in to a fraction, we just need to put it over the number we were dividing by (23).

**Remember:** Because the 0 is at the start of the answer we can ignore it.

8320\div23=361\frac{17}{23}

**Long Division**

**Question 1:**

Use long division to calculate 3525\div15

Start by setting up your question, with the number you are dividing by on the outside and the number you are dividing on the inside.

We can’t fit any 15s in 3, so we write a 0 on top and think of it 3 and 5 together as 35.

We need to see how many 15s we can fit into 35.

1\times15=15

2\times15=30

3\times15=45

We can fit two 15s into 35, so we put the 2 above the 5 and the 30 underneath the 35 to create the first column subtraction.

We now need to bring the 2 down next to the 5 to make 52.

We need to see how many 15s we can fit into 52.

1\times15=15

2\times15=30

3\times15=45

4\times15=60

We can fit three 15s into 52, so we put the 3 above the 2 and the 45 underneath the 52 to create the second column subtraction.

We now bring the 5 down next to the 7.

We need to see how many 15s we can fit into 75.

1\times15=15

2\times15=30

3\times15=45

4\times15=60

5\times15=75

Because 15 goes into 75 five times perfectly, we just need to put a 5 on top and we’re done!

**Remember:** Because the 0 is at the front of the answer, we can ignore it!

3525\times15=235

**Long Division**

**Question 2:**

Use long division to calculate 4320\div25. Give your answer as a decimal

Start by setting up your question, with the number you are dividing by on the outside and the number you are dividing on the inside.

We can’t fit any 25s in 4, so we write a 0 on top and think of 4 and 3 together as 43.

We need to see how many 25s we can fit into 43.

1\times25=25

2\times25=50

We can fit one 25 into 43, so we put the 1 above the 3 and the 25 underneath the 43 to create the first column subtraction.

We now need to bring the 2 down next to the 18 to make 182.

We need to see how many 25s we can fit into 182.

1\times25=25

2\times25=50

3\times25=75

4\times25=100

5\times25=125

6\times25=150

7\times25=175

8\times25=200

We can fit seven 25s into 182, so we put the 7 above the 2 and the 175 underneath the 182 to create the second column subtraction.

We now bring the 0 down next to the 7.

We need to see how many 25s we can fit into 70.

1\times25=25

2\times25=50

3\times25=75

We can fit two 25s into 75, so we put the 2 above the 0 and the 50 underneath the 70 to create another column subtraction.

Usually, this is where we would bring a number down next to the 20, but we don’t have one. We need to add a decimal and a 0.

We can now bring the 0 down next to the 20.

We need to see how many 25s we can fit into 200.

1\times25=25

2\times25=50

3\times25=75

4\times25=100

5\times25=125

6\times25=150

7\times25=175

8\times25=200

Because 25 goes into 200 eight times perfectly, we just need to put an 8 above the 0 and we’re done!

**Remember:** Because the 0 is at the front of the answer, we can ignore it!

4320\times25=172.8