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

Fundamentals of Structured Programming Relational operators 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 Exercise: MOD and DIV 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. Note that Visual Basic does not have a DIV function and uses A\B instead Example: Converting Denary to Binary using DIV sub convertDtoB(byVal base10 as integer) dim base2(7) as integer 'create an array to store the binary dim temp as integer = base10 for i = 7 to 0 step -1 'loop through each binary bit starting from the biggest value base2(i) = temp \ (2^i) 'temp DIV 2^i temp = temp MOD 2^i next console.write(base10 & " in binary = ") for i = 7 to 0 step -1 'loop through each binary bit starting from the biggest value console.write(base2(i)) next end sub 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): base10 temp 2^i i base2 0 1 2 3 4 5 6 7 67 67 128 7 0 3 64 6 1 32 5 0 16 4 0 8 3 0 4 2 0 1 2 1 1 0 1 0 1 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. Example: Finding out if a number is Odd using MOD Dim testNum as Integer Dim temp as Integer console.writeline("please insert a number to test:") testNum = console.readline() If testNum MOD 2 = 0 Then Console.Writeline("The number is even") Else Console.Writeline("The number is odd") End If