How can you use subtraction to solve a division problem?

Answer 1

Like multiplication is a quick way to do multiple additions of the same number, subtraction is a quick way to do multiple subtractions of the same number - subtraction gives how many times the divisor can be subtracted from the dividend until zero is reached (and not passed). If a number is reached (when subtracting) which is less than the divisor but greater than zero, then a fraction of the divisor needs to be subtracted - this fraction is this number (as the numerator) over the divisor (as the denominator).

The amount subtracted each time does not have to be the divisor, but can be any multiple of the divisor (sticking to multiples of powers of 10 (10, 100, 1000 etc) as the multiple is best) speeds up the multiple subtractions; if using this method, add up the multiples - this is known as "Chunking" (as 'chunks' of the divisor are removed from the dividend at each subtraction).

Example: 4966 ÷ 13

Subtracting 13 each time gives:

1: 4966 - 13 = 4953

2: 4966 - 13 = 4940

3: 4940 - 13 = 4927

...

380: 39 - 13 = 26

381: 26 - 13 = 13

382: 13 - 13 = 0

→ 4966 ÷ 13 = 382

Using chunking with powers of 10 as the multiples:

4966 - 13 × 100 = 4966 - 1300 = 3666

3666 - 13 × 100 = 3666 - 1300 = 2366

2366 - 13 × 100 = 2366 - 1300 = 1066

1066 - 13 × 10 = 1066 - 130 = 936

936 - 13 × 10 = 936 - 130 = 806

...[8 times the subtraction of 13 × 10 can be done]

156 - 13 × 10 = 156 - 130 = 26

26 - 13 × 1 = 26 - 13 = 13

13 - 13 × 1 = 13 - 13 = 0

→ 4966 ÷ 13 = 100 + 100 + 100 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 1 + 1 = 382

When chunking with powers of 10 as the multiples, but using a multiple (possibly greater than 1) of the powers of 10, leading to:

4966 - 13 × (100 × 3) = 4966 - 13 × 300 = 4966 - 3900 = 1066

1066 - 13 × (10 × 8) = 1066 - 13 × 80 = 4966 - 1040 = 26

26 - 13 × (1 × 2) = 26 - 13 × 2 = 26 - 26 = 0

→ 4966 ÷ 13 = 300 + 80 + 2

(If you look carefully, you may spot that this multiple chunking is how long division is done.)

Answer 2

You can do repeated subtraction, and see how often you need to subtract until you get a number that is less than the divisor. Basically this only works for divisions between whole numbers.