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Generating a Random Pin of 5 Digits

Posted by: admin December 28, 2021 Leave a comment


I would like to generate Unique Pins based on a random number I have found this on Stack Overflow How to generate a random five digit number Java. That Question uses time to generate the numbers so therefore you get a lot of duplicates.

Here is my code

public int createRandomPin() {
    int k = random.nextInt(Integer.SIZE);
    k = (k + 1) * 9999;
    if (9999 > k || k > 99999) {
        //then regenerate 
    } else {
        return k;

My Question

Java Compiler then gives a warning missing “return”. As well I need to restructure the code so that if it isn’t a 5 digit pin it generates again before it “returns k”.


I would suggest to generate random number using SecureRandom class and then as it needs to be of 5 digits create a random number between 0 and 99999 using random.nextInt(100000) , here 0 is inclusive and 100000 is exclusive and then format it into 5 digit by appending zero.

SecureRandom random = new SecureRandom();
int num = random.nextInt(100000);
String formatted = String.format("%05d", num); 

I hope this solves your problem

Edit: This post incorrectly said 10,000, has been edited to say 100,000


Integer.SIZE yields the number in bits of the int data type. This nothing has to do with the range span for a random value.

Using it as the argument of Random.nextInt doesn’t make any sense (actually it generates a random value in range [0,32)).

Just generate a int value = random.nextInt(100000) so that you will obtain a value in [0,99999].

Now your definition of 5 digits pin is not precise, 40 could be interpreted as 00040 so it’s still 5 digits if you pad it. You must take this thing into account, since forcing 5 “visible” digits implies generating a number in range [10000,99999] but this is just an awkward solution.


RandomUtils.randomNumeric(int count) or RandomUtils.randomNumeric(int inLength, int maxLength) from apache common lang3 can also be used for the same purpose.