A-level Computing/AQA/Problem Solving, Programming, Data Representation and Practical Exercise/Fundamentals of Programming/Modulo arithmetic





Modular arithmetic is all about finding the remainder from long division (MOD), and the total number of times that a number goes into a division (DIV). Let's take a look at a quick example of 10 divided by 7 (you might want to remind yourself about long division): 1 r 3 7)10    7      3 Hopefully that wasn't too hard. We now need to introduce some terminology, MOD and DIV:
 * MOD = finds the remainder from long division i.e. 10 MOD 7 = 3
 * DIV = finds the number of divides in long division i.e. 10 DIV 7 = 1

Try these examples, working out the MOD and DIV of each:

7 / 2

MOD = 1 DIV = 3

17 / 5

MOD = 2 DIV = 3

8 / 2

MOD = 0 DIV = 4

6 / 9

MOD = 6 DIV = 0 (9 does not divide into 6 at all!)

Now try these explicit calculations:

11 MOD 8

= 3

8 MOD 4

= 0

6 DIV 5

= 1

600 DIV 20

= 30

Hopefully you are now pretty good with MOD and DIV, but what exactly is the point of all this? A very common example in past exam paper has been using the MOD and DIV to work out a binary equivalent of a denary number.

Try the code out and see if it works. Try to write a trace table and see how it works for the number 67 (again another popular question in exams):

Output: 67 in binary = 01000011

Another common use is in finding out whether a number is odd or even using the MOD function. We know that MOD returns the remainder from a division sum. So for example 4 MOD 2 = 0, 5 MOD 2 = 1, 6 MOD 2 = 0 and so on. By modding something with 2 we can work out whether it is odd or not due to the return value.