Bus stop method - dividing by a 1 digit number (no remainders)

963 ÷ 3 To solve this sum we use the bus stop method. The number to be divided by goes at the front and the largest number goes inside.

First we solve 9 ÷ 3. 3 goes into 9 3 times so we write 3 above the 9.

Next we solve 6 ÷ 3. 3 goes into 6 twice so we write 2 above the 6.

Finally we solve 3 ÷ 3. 3 goes into 3 once so we write 1 above the 3. This shows us that 963 ÷ 3 = 321.

Bus stop method - dividing by a 1 digit number

685 ÷ 6 To solve this sum we use the bus stop method. The number to be divided by goes at the front and the largest number goes inside.

When using this method we work left to right. First we solve 6 ÷ 6 = 1 When using this method we work left to right. First we solve 6 ÷ 6 = 1. This is written above the 6.

Next we look at the next number along which is 8 Next we look at the next number along which is 8. Because 8 is not in the 6 times table there will be a remainder. We think about how many multiples (lots of) 6 are there to get to 8. There is 1 (1x6 = 6) with 2 left over (remainders). We write the 1 above the 8 and the 2 remainders are written next to the next number along.

Now because of the remainders from the previous step, the 5 now becomes 25. This is because the 8 is actually 8 tens so the 2 remainders were actually 2 tens. 2 tens + 5 = 25. So how many 6’s go into 25? 4 x 6 = 24 4 is the closest multiple of 6 to 25. So we write 4 above and we have remainder 1 because there is 1 left over after 4 x 6 = 24. We write it as r1. Finally 685 ÷ 6 = 114 r1

Bus stop method - dividing by a 2 digit number

583 ÷ 11 To solve this sum we use the bus stop method. The number to be divided by goes at the front and the largest number goes inside.

When using this method we work left to right. First we solve 5 ÷ 11 When using this method we work left to right. First we solve 5 ÷ 11. When dividing the largest number is always first otherwise we can’t solve the sum. We can’t solve 5 ÷ 11 so we put 0 above the 5. We now have 5 remainders. The 5 remainders are written next to the 8.

The remainder 5 is now written with the 8 making it 58 The remainder 5 is now written with the 8 making it 58. We now solve 58 ÷ 11. The closest multiple (lots) of 11 to 58 is 55 this is 5 x 11. So we write the 5 above. 58 – 55 = 3 so we have remainder 3. This is written next to the 3.

The remainder 3 is written next to the 3 making it 33 The remainder 3 is written next to the 3 making it 33. We now solve 33 ÷ 11. Because 33 is a multiple of 11 it equals 3 with no remainders ( 3 x 11 = 33). This 3 is written above the 3. So 583 ÷ 11 = 53