 If you appreciate these tips, please consider a small donation: General Tip How to Determine the Factors of a Specified Number in Visual Basic Code Tested with VB6 Courtesy of Marshall Ellis A factor of a number is a number that when multiplied with another factor of the same number will result in the number. It sounds more confusing than it is. For instance, factors of 9 are 1, 3 and 9. Factors of 5 are 1 and 5. Factors of 20 are 1, 2, 4, 5, 10 and 20. Determining factors is related to detecting Prime Numbers, as a prime is any number with factors of only 1 and itself. So how do we tell VB to list the factors of a number? Sounds tough! It's really quite easy, using the MOD function – modular division – and a bit of logic. Where regular division concerns itself with getting the absolute result down to decimal places, modular division gives you the remainder if you stop dividing before getting to decimals. For instance: Regular division 5/2 = 2.5 Modular division 5/2 = 2 with a remainder of 1, or 5 Mod 2 = 1 Any number that is a factor of a specified number will give a Mod result of 0, because it will divide evenly. Taking the number 6 as an example: 6 mod 1 = 0 6 mod 2 = 0 6 mod 3 = 0 6 mod 4 = 2 6 mod 5 = 1 6 mod 6 = 0 Thus factors of 6 are 1, 2, 3, and 6. By now you will probably see where I am going with this, and what the code might look like. So without further dull preliminary explanation, on to the sample code... 1 – Create a new standard VB project. 2 – On the form place a textbox, a command button and a listbox. 3 – Set the text property of the textbox to "1" simply to avoid it ever evaluating to zero without trying. We will concern ourselves here only with positive numbers 1 and greater. I'll leave generating factors of negative numbers as an exercise for the reader. Obviously you should always code or error-trap to account for inappropriate input, which in a minor way has been done below. 4 – In the Click event of the command button place the following code: 'Get the number we want to test from the textbox Dim lngNumber As Long lngNumber = Text1.Text 'Clear the listbox of anything that might remain from prior results List1.Clear 'If the number isn't a positive integer, go no further If lngNumber < 0 Then Exit Sub 'Now loop through 1 to the specified number For i = 1 To lngNumber    If lngNumber Mod i = 0 Then     'Mod is 0 so the number is evenly divisible _      by i, making the current i a factor      lstFactor.AddItem i   End If Next 5 – Run the program, enter a number in the textbox and test. Note that the code could also be placed in a routine callable from anywhere it might be needed in a program.

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