Generating numbers to the multiple of 5?

Question

2.50/5 (4 votes)

See more:

Hello.

I have a text box that can accept 5 chars, representing a numeric value from 00000 to 99999.

When the user enters a value, say for example, 12345, the following function is called:

if ((NumToValidate % 5) == 0)
        return VALIDATION_SUCCESS;

Where NumToValidate is the numerical form of the 5 char input.

The function checks that the inputed value, in this case 12345 is a multiple of 5, and if it is it returns as a success.

It works well, however, I have no idea how to create a function that can loop through generateing values that are a multiple of 5, from 00000 to 99999 (All possible values)

AND, if possible, to generate a random value and find its next multiple of 5.

e.g.

Rand value generated = 1834 now find what 5th digit, if any, produces a multiple of 5.

Thank you,
Stephen

Posted 2-Aug-11 8:27am

stephen.darling

Add a Solution

5 solutions

Solution 4

First we can generate a list with all mutiples of 5 between 0 and 100000 (not included).

C#

List<int> numbers = new List<int>;
for (int i = 0; i < 100000; i += 5)
{
    numbers.Add(i);
}

Then you can take a number at a random position and remove it from the list so that you will get each number only once.

C#

Random rand = new Random();

while (numbers.Count > 0)
{
    int randomIndex = rand.Next(numbers.Count);

    int randomNumber = numbers[randomIndex];

    // Could validate that the code works...
    System.Diagnostics.Debug.Assert((randomNumer % 5) == 0);

    numbers.RemoveAt(randomIndex);

    DoSomethingWithRandomMultipleOf5(randomNumbr);
}

If you don't want number to be returned only once, it is even easier.

C#

const int maxReturnedValue = 99995;
const int multipleOf = 5;
const int randLimit = (maxReturnedValue - 1) / multipleOf;

Random rand = new Random();

while (WantAnotherRandomNumber())
{
    int randomNumber = rand.Next(randLimit) * multipleOf;

    // Confirms that the code works. Could be useful when you
    // are not too sure that you code is correct.
    System.Diagnostics.Debug.Assert((randomNumer % 5) == 0);
    System.Diagnostics.Debug.Assert(randomNumer >= 0);
    System.Diagnostics.Debug.Assert(randomNumer <= 99995);

    DoSomethingWithRandomMultipleOf5(randomNumbr);
}

Posted 2-Aug-11 13:33pm

Philippe Mori

Updated 2-Aug-11 13:37pm

v2

Solution 5

Just for completeness, see the answer here[^] that I find most elegant, but David1987 was faster (a lot).

Edit: The link changed to another post[^].

Posted 22-Aug-11 21:12pm

lukeer

Updated 23-Aug-11 22:42pm

v2

Solution 2

Your problem can be easily solved.

C#

protected void GenerateMultiples(int from, int to)
{
for (int i = from; i < to; i++)
{
     if(i % 5 == 0)
     {
        Console.WriteLine(i);
        i = i + 4; //Do this to make it faster.
     }
}
}

That is the best way I can think of.

For your second problem, remember that if a number is divisible by 5, it can only end in a 5 or a 0, so just add either to the end of your randomly generated number.

Posted 2-Aug-11 9:00am

DominicZA

Updated 2-Aug-11 20:25pm

v4

Comments

CPallini 2-Aug-11 15:40pm

This code doesn't generate all possible multiples.

DominicZA 2-Aug-11 15:53pm

why doesn't this generate all possible multiples?

CPallini 2-Aug-11 16:11pm

suppose from=2. When i=5 the program writes it and then sets i=25 (the 'Do this to make it faster'), missing the multiples 10,15,20.

DominicZA 2-Aug-11 16:32pm

Oh wow...typo from my side...answered this from my phone so I didnt even realize :P Thanks for pointing this out.

Philippe Mori 2-Aug-11 19:09pm

By doing i = i + 5 and i++, i will be incremented by 6. Then since you will align to a multiple of 5, it will incremented by 10 between each outputed number. Thus you will have either a series like 5, 15, 25, 35... or a series like 0, 10, 20, 30... depending on the initial number.

Luc Pattyn 22-Aug-11 23:00pm

i = i + 4; //Do this to make it faster.

that comment made me laugh aloud, which doesn't happen very often.
If you want multiples of 5, then step by 5, not by 1, nor by 1 then by 4.
Just get the initial value corrected, like so:
for (int i=(from+4)/5*5; i<to; i+=5) ...

There is no need for a modulo operation here!

:)

CPallini 23-Aug-11 3:27am

OK, we accept your 'further optimization' :-D

Solution 1

C#

Random random = new Random();
int num = random.Next(0, 1000);
for (int i = 0; i < 5; i++)
{
    if (num % 5 == 0)
    {
        MessageBox.Show(num.ToString());
    }
    num++;
}

Posted 2-Aug-11 8:59am

luisnike19

Updated 2-Aug-11 9:46am

v3

Comments

DominicZA 2-Aug-11 15:02pm

Although not likely, this could end up becoming an infinite loop. Not a good solution.

luisnike19 2-Aug-11 15:45pm

Sorry, answer updated.

luisnike19 2-Aug-11 15:46pm

The next random num multiple of 5, between 0 and 1000

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

N_tro_P · Accepted Answer · 2011-08-02T09:10:00

Loop for generating vals divisible by 5

C#

for(int val = 0; val < MAX; val+=5)
{
   //Because we are adding 5 and starting at 0 val must be divisible by 5... unless 
   // you modify it inside the loop.
}

To generate a random value between 0 and 99999 use the random generator combined with modulus operator.

Random rand = new Random();
int val = random.Next(MIN, MAX);

//Now get to the 'next' divisible by 5
int leftOver = val % 5;
int newVal = val + leftOver; //Adding the leftover will result in divisible by 5