# Java Program to find Armstrong numbers with Example

Armstrong number Example in Java
How to check if a number is Armstrong number or not? or write a Java program to find Armstrong number? This is a common Java interview question asked on campus interviews and fresher level interviews. This is also a popular Java programming exercise on various school, colleges and computer courses to build programming logic among Students. An Armstrong number is a 3 digit number for which sum of cube of its digits is equal to the number itself. An example of Armstrong number is 153 as 153= 1+ 125+27 which is equal to 1^3+5^3+3^3. One more example of the Armstrong number is 371 because it is the sum of 27 + 343 + 1 which is equal to 3^3 + 7^3 + 1^3 . In this Java program example, we will see complete code example of Java program to check if any 3 digit number is Armstrong number or not. If you are going for Java interview, then be prepare for some follow-up questions e.g. finding prime numbers, or finding Armstrong number of more than 3 digits.

By the way, this Java program is in continuation of our earlier programming exercise like

## How to check if number is Armstrong number in Java

Here is complete code for checking if a number is Armstrong number or not. It uses a method called isArmstrong(int number) to implement logic for checking if a number is Armstrong nor not.

Btw, this is not the same program as print all Armstrong number between 0 and 999 but you can use this logic to solve that question as well. All you need to do is loop till 1000 and check if the number is Armstrong or not. If yes then print otherwise move to next number.

Java Program to Find Armstrong Number
```import java.util.Scanner;

/**
* Simple Java Program to check or find if a number is Armstrong number or not.
* An Armstrong number of three digit is a number whose sum of cubes of its digit is equal
* to its number. For example 153 is an Armstrong number of 3 digit because 1^3+5^3+3^3 or   1+125+27=153
* @author Javin
*/
public class ArmstrongTest{

public static void main(String args[]) {

//input number to check if its Armstrong number
System.out.println("Please enter a 3 digit number to find if
its an Armstrong number:");
int number = new Scanner(System.in).nextInt();

//printing result
if(isArmStrong(number)){
System.out.println("Number : " + number + " is an Armstrong number");
}else{
System.out.println("Number : " + number + " is not an Armstrong number");
}

}

/*
* @return true if number is Armstrong number or return false
*/
private static boolean isArmStrong(int number) {
int result = 0;
int orig = number;
while(number != 0){
int remainder = number%10;
result = result + remainder*remainder*remainder;
number = number/10;
}
//number is Armstrong return true
if(orig == result){
return true;
}

return false;
}

}

Output:
Please enter a 3 digit number to find if its an Armstrong number:
153
Number : 153 is an Armstrong number
Please enter a 3 digit number to find if its an Armstrong number:
153
Number : 153 is an Armstrong number
Please enter a 3 digit number to find if its an Armstrong number:
371
Number : 371 is an Armstrong number```

That's all on How to check if a number is Armstrong in Java. It’s pretty simple Java program and if you look closely it just gets digit by digit by using remainder operator and reduce number by 1 digit after dividing it by 10. Let me know if you find any bug on this Java program for checking Armstrong number.

Further Learning
The Coding Interview Bootcamp: Algorithms + Data Structures
Data Structures and Algorithms: Deep Dive Using Java
Algorithms and Data Structures - Part 1 and 2

Other Java programming problems you may like :

1. Good tutorial, How to find Armstrong number with more than 3 digits ?

1. /*
Simple Java Program to check or find if a number is Armstrong number or not.
[ An Armstrong number of any number of digit is a number whose sum of each digits raised to the power the total no. of digits is equal to it (the original number). For example, 9474 is an Armstrong number of 4 digits because 9^4 + 4^4 + 7^4 + 4^4 = 6561 + 256 + 2401 = 9474 ]
*/

import java.io.*;
import java.util.*;

public class ArmstrongNumber
{
public static void main(String args[])
{

// Input number to check if its Armstrong number
System.out.println("Please enter a number to find if it is an Armstrong number: ");
int number = new Scanner(System.in).nextInt();

// Printing result
if(isArmStrong(number))
{
System.out.println(number + " is an Armstrong number");
}
else
{
System.out.println(number + " is not an Armstrong number");
}

} // End of main()

// Function returning true if number is Armstrong number or returning false

private static boolean isArmStrong(int number)
{
int result = 0;
int dc=0;
int orig = number;
while(orig != 0)
{
dc++;
orig = orig/10;
}
orig = number;
while (orig!=0)
{
result = result + (int)Math.pow(orig%10, dc);
orig = orig/10;
}
if(number == result)
return true;
else
return false;
} // End of isArmStrong()
}

OUTPUT:
Please enter a number to find if it is an Armstrong number:
9474
9474 is an Armstrong number

Please enter a number to find if it is an Armstrong number:
8208
8208 is an Armstrong number

2. package com.example.test.nover;
import java.util.List;
import java.util.ArrayList;
public class ArmstrongNumber {

public void findNumber() {
List numList = new ArrayList();
int newNum = 0;
String numVal= "";
int length = 0;
for(int i = 1000; i<=9999; i++) {
numVal = (numVal+i).trim();
length = numVal.length();
for(int j=0;j<length;j++) {
char charVal = numVal.charAt(j);
int charNum = charVal-'0';
charNum = (int) Math.pow(charNum,4);
//int charNo = (int) Math.pow(charNum,4);
newNum = newNum + charNum;
}
if(i==newNum)
newNum = 0;
numVal="";
}
for(int i: numList) {
System.out.println("NUMBERS FOUND ARE " +i);
}
}

public static void main(String args[]) {
ArmstrongNumber numVer = new ArmstrongNumber();
numVer.findNumber();
}
}

3. Using String will generalize to find ArmStrong Number for 2 digits, 3 digits or 4 digits since we can capture the length and use Math.pow(number,length) which will make the code generic.

2. According to Abhirupa Mitra's code numbers from 1 to 9 would be displayed as Armstrong numbers which is not correct. What is the correct definition of Armstrong numbers? Within 1 to 10 only 1 is an armstrong number.

1. Yes, you are right. The correct definition of Armstrong number is:-
ex. 153. if 1^3+5^3+3^3 = 153 then it is an Armstrong number. In words, a number whose sum is equal to the sum of the cubes of its digits then it is an Armstrong number

3. The definition of an Armstrong Number was not that clear to me. I looked up here and figured out what it is: http://www.cs.mtu.edu/~shene/COURSES/cs201/NOTES/chap04/arms.html

4. I'd probably do it like this:

public boolean isArmstrong(int i) {
return (i >= 100) && (i <= 999) && (sumCubeDigits(i) == i);
}

private int sumCubeDigits(int digits) {
if (digits == 0)
return 0;
else
return cube(digits % 10) + sumCubeDigits(digits / 10);
}

private int cube(int i) {
return i * i * i;
}

and if it turned out to be too slow, like this:

public boolean isArmstrong(int i) {
if ((i < 100) || (i > 999)) {
return false;
} else {
int a = i % 10;
int b = (i / 10) % 10;
int c = i / 100;
int sumOfCubes = (a*a*a) + (b*b*b) + (c*c*c);
return (i == sumOfCubes);
}
}

5. I'd do it like this....

public static void armstrong(int number){
String numToString = String.valueOf(number);
int number1;
int multiplication = 0;
for (int i=0; i<=numToString.length()-1; i++){
number1 = Character.getNumericValue(numToString.charAt(i));
multiplication = multiplication + number1 * number1 * number1;
}
if (number == multiplication) {
System.out.println("Number : " + number + " is an Armstrong number");
} else {
System.out.println("Number : " + number + " is not an Armstrong number");
}
}

6. I found this program interesting and quite simple to understand! They have explained few elements on their page which helped me to grasp the topic.'
Its better to use a single function t understand the Java program better.
import java.io.*;

public class armstrong
{
public static void main(String args[]) throws Exception
{
int num,temp,sum=0,rem=0;
System.out.println("\nEnter an Integer:");
temp=num;
while(num>0)
{
rem=num%10;
sum=sum+(rem*rem*rem);
num=num/10;
}
if(temp==sum)
{
System.out.println("\n" + temp + " is an Armstrong Number\n");
}
else
{
System.out.println("\n" + temp + " is Not an Armstrong Number\n");
}
}
}
Source: http://www.codingalpha.com/armstrong-number-java-program/

7. Hi All,

Just generalised the very first answer while trying to solve Armstrong example. Hope it is useful :)

import java.util.Scanner;

public class ArmstrongNumbers {

public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
Long n=sc.nextLong();
Long orignal=n;

long rem;
long result=0;
String forLength=n.toString();
int length = String.valueOf(forLength).length();
System.out.println("Length "+length);

while(n!=0){
rem=n%10;
System.out.println("Rem "+rem);
result=(long) (result+Math.pow(rem, length));
System.out.println("Result " +result);
n=n/10;
}
if(result == orignal){
System.out.println("Armstrong");
}
else{
System.out.println("Not Armstrong");
}
}

}

8. 