You can use the Random class:
int randomNumber = new Random().nextInt(9000) + 1000;
Top answer 1 of 3
8
You can use the Random class:
int randomNumber = new Random().nextInt(9000) + 1000;
2 of 3
1
Working solution - different number each time
Random r = new Random();
int fourDigit = 1000 + r.nextInt(10000);
System.out.println(fourDigit);
Broken - same number every time
Random r = new Random(123); // <---- uses same seed every time !
int fourDigit = 1000 + r.nextInt(9000);
System.out.println(fourDigit);
How to do in Java
howtodoinjava.com › home › java 8 – generate random number in range
Java 8 - Generate Random Number in Range
September 6, 2023 - The following Java program generates a few random long values between 1000 (origin) and 9999 (bound).
Write a program that generates five random numbers. The number
should be between 1000 and 9999. Then write these number into a
file named “codes.txt”
in java (bouns points if you can give it to me in a pdf file for
java ). :D
Answer to Write a program that generates five random numbers. More on chegg.com
1
February 28, 2021How to generate a random four digit number?
Does the random package also guarantee that each time the method is invoked a different number will be generated than the one previous. No. In fact, very explicitly, no. If you can't get the same thing twice in a row, it won't be what's commonly considered random. I would suggest a few different things. As others have said, leading zeros should be done in your presentation layer. Sampling-with-rejection is likely your best bet for the problem exactly as described. ie, Integer newRandom; do { newRandom = gen.nextInt(9999); } while(newRandom == oldRandom); I suspect what you really want is a permutation of the numbers 0 through 9999. I'd do it thus: List arr = new ArrayList(10000); for(int i = 0; i < 10000; i++) { arr.add(String.format("%04d", i)); } Collections.shuffle(arr); // Now keep popping items off the end of the list with: String nextStr = arr.remove(0); // And when the array gets empty, populate a new one. It's worth adding that you still have a nontrivial probability of two numbers being produced back to back when you generate a fresh array More on reddit.com
15
0
February 9, 2014how can I generate a random number of any particular number of digits (say 5) in java? - Stack Overflow
I want to generate a random number of any particular digit (say 5, i.e., 10000 to 99999) in java. How can I do that? More on stackoverflow.com
write a program that generates five random numbers the number should be between 1000 and 9999 th
Numerade offers video solutions for the most popular textbooks More on numerade.com
1
February 6, 2023Videos
JanBask Training
janbasktraining.com › community › java › what-are-some-methods-for-generating-numbers-between-the-range-of-1000-and-9999
What are some methods for generating numbers between the range of 1000 and 9999? | JanBask Training Community
November 8, 2023 - For example, generating a random between 0 and 8999 and then adding 1000 to the result will produce a required four-digit number between 1000 and 8999. Always ensure that the number should fall between 1000 and 9999 if you're also trying this ...
Java67
java67.com › 2015 › 01 › how-to-get-random-number-between-0-and-1-java.html
How to Generate Random Number between 1 to 10 - Java Example | Java67
*/ public static int getRandom(int ... 1 and 100 : 29 Random number between 1 and 100 : 37 Random value between 1000 and 9999 : 3517 Random value between 1000 and 9999 : 3581...
CopyProgramming
copyprogramming.com › howto › java-random-number-between-1000-and-9999
Random: Generating a Random Java Number within the Range of 1000 and 9999
May 4, 2023 - This approach works effectively ... to generate 4 digit random numbers in java from 0000 to 9999, Random r = new Random(); String randomNumber = String.format("d", (Object) Integer.valueOf(r.nextInt(1001))); System.out...
GeeksforGeeks
geeksforgeeks.org › java › generating-random-numbers-in-java
Generating Random Numbers in Java - GeeksforGeeks
April 24, 2025 - // Demonstrates how to generate random integers, // doubles, and booleans using ThreadLocalRandom import java.util.concurrent.ThreadLocalRandom; public class Geeks { public static void main(String[] args) { // Generate random integers in range 0 to 999 // Random number between 0 and 999 int r1 = ThreadLocalRandom.current().nextInt(1000); // Random number between 0 and 999 int r2 = ThreadLocalRandom.current().nextInt(1000); // Print random integers System.out.println("Random Integers: " + r1); System.out.println("Random Integers: " + r2); // Generate Random doubles between 0.0 (inclusive) // an
JustAnswer
justanswer.com › computer-programming › ghhyy-java-code-write-method-generates-digit-random.html
How to Generate a 4-Digit Random Number in Java | Expert Q&A
To generate a 4-digit random number in Java, use the Random class or Math.random(). For example, use (int)(Math.random() * 9000) + 1000 to ensure the number ranges from 1000 to 9999, avoiding leading zeros.
Techdive
techdive.in › java › java-random-number-generator
Java - Random Number Generator | TechDive.in
November 17, 2010 - /** * This method returns a random number between smallest n digit number and largest n digit number. For example if * n=4, then a random number would be generated between 1000 and 9999.
Chegg
chegg.com › engineering › computer science › computer science questions and answers › write a program that generates five random numbers. the number
should be between 1000 and 9999. then write these number into a
file named “codes.txt”
in java (bouns points if you can give it to me in a pdf file for
java ). :d
Solved Write a program that generates five random numbers. | Chegg.com
February 28, 2021 - To generate any random number between 1000 to 9999 , we can simple generate any random number between 0 and 8999 and add 1000 to it. Using this tehcnique , here is a Java code.
Reddit
reddit.com › r/java › how to generate a random four digit number?
r/java on Reddit: How to generate a random four digit number?
February 9, 2014 -
I am trying to generate a random four digit number between 0 and 9999 and so far have the code Random generator = new Random(); regNo = generator.nextInt(9999)
this works great but is there a way to ensure that if the number is less than four digits than it is still in four digit format i.e. 18 should be 0018.
Does the random package also guarantee that each time the method is invoked a different number will be generated than the one previous.
Any help would be much appreciated. Cheers
Top answer 1 of 4
7
Does the random package also guarantee that each time the method is invoked a different number will be generated than the one previous. No. In fact, very explicitly, no. If you can't get the same thing twice in a row, it won't be what's commonly considered random. I would suggest a few different things. As others have said, leading zeros should be done in your presentation layer. Sampling-with-rejection is likely your best bet for the problem exactly as described. ie, Integer newRandom; do { newRandom = gen.nextInt(9999); } while(newRandom == oldRandom); I suspect what you really want is a permutation of the numbers 0 through 9999. I'd do it thus: List arr = new ArrayList(10000); for(int i = 0; i < 10000; i++) { arr.add(String.format("%04d", i)); } Collections.shuffle(arr); // Now keep popping items off the end of the list with: String nextStr = arr.remove(0); // And when the array gets empty, populate a new one. It's worth adding that you still have a nontrivial probability of two numbers being produced back to back when you generate a fresh array
2 of 4
1
Look at apache commons RandonStringUtils. There is a static method, I belive, randomNumeric that demonstrates how to do what you've described.
Mkyong
mkyong.com › home › java › java – generate random integers in a range
Java - Generate random integers in a range - Mkyong.com
August 19, 2015 - In this article, we will show you three ways to generate random integers in a range · This Random().nextInt(int bound) generates a random integer from 0 (inclusive) to bound (exclusive)
CopyProgramming
copyprogramming.com › howto › java-random-number-between-1000-and-99
Java: Generating a random Java number within the range of 1000 to 99
August 4, 2023 - The goal is to determine whether the integer represented by i is divisible by either 100 or 1000. Determine whether i modulo 10 is set to 0 . ... Produce a random number within a specific range, and repeat the process until a non-"banned" number is selected.
Number Generator
numbergenerator.org › randomnumbergenerator › 1000-9999
Random Number Between 1000 And 9999 - Number Generator
Lets you pick a number between 1000 and 9999. Use the start/stop to achieve true randomness and add the luck factor.
Top answer 1 of 5
8
Create a random integer in range 0,10^k - 10^(k-1), and add 10^(k-1) - in your example. range 0,90000, and add 10000 to it.
Use the Random.nextInt(int) method for it.
Something like:
new Random().nextInt(90000) + 10000
2 of 5
0
Simple.
- generate a random number upto 89999
- Add 10000 to each generated number.
Baeldung
baeldung.com › home › java › java numbers › generating random numbers in a range in java
Generating Random Numbers in a Range in Java | Baeldung
May 11, 2024 - Learn about alternative ways of generating random numbers within a range in Java.
Study.com
study.com › courses › business courses › business 104: information systems and computer applications
Java: Generate Random Number Between 1 & 100 - Lesson | Study.com
January 6, 2018 - The Random class has a method to generate a pseudo-random number, nextInt(int n) , between 0 and the specified value (n). We changed this to a range between 1 and 100. Java's Random generates a pseudorandom number, meaning that the actual bits ...
Top answer 1 of 5
27
There's a better way to get random numbers, and that's with java.util.Random. Math.random() returns a double (floating-point) value, but based on your request of a 3-digit number, I'm going to assume that what you really want is an integer. So here's what you do.
// initialize a Random object somewhere; you should only need one
Random random = new Random();
// generate a random integer from 0 to 899, then add 100
int x = random.nextInt(900) + 100;
2 of 5
6
You are not correct. That can produce numbers under 100. The most familiar way is random.nextInt(900)+100. Here random is an instance of Random.
Before Java 7 there was no reason to create more than one instance of Random in your application for this purpose. Now there's no reason to have any, as the best way is
int value = ThreadLocalRandom.current().nextInt(100, 1000);
Top answer 1 of 4
15
private long generateRandomNumber(int n) {
long min = (long) Math.pow(10, n - 1);
return ThreadLocalRandom.current().nextLong(min, min * 10);
}
nextLong produces random numbers between lower bound inclusive and upper bound exclusive so calling it with parameters (1_000, 10_000) for example results in numbers 1000 to 9999.
Old Random did not get those nice new features unfortunately. But there is basically no reason to continue to use it anyways.
2 of 4
5
public static int randomInt(int digits) {
int minimum = (int) Math.pow(10, digits - 1); // minimum value with 2 digits is 10 (10^1)
int maximum = (int) Math.pow(10, digits) - 1; // maximum value with 2 digits is 99 (10^2 - 1)
Random random = new Random();
return minimum + random.nextInt((maximum - minimum) + 1);
}
Setmeapp
setmeapp.com › rr1ur16 › java-random-number-between-1000-and-9999.html
Setmeapp
We cannot provide a description for this page right now