# Java Program to find Armstrong numbers with Example

Armstrong number Example in Java
How to check if a number is an Armstrong number or not? or write a Java program to find Armstrong's 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 schools, colleges, and computer courses to build programming logic among Students. An Armstrong number is a 3 digit number for which the sum of cube of its digits is equal to the number itself.

One of the popular examples of the Armstrong number is 153 as 153= 1+ 125+27 which is equal to 1^3+5^3+3^3 which is equal to the sum of cube of its digits 1, 5, and 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 a complete code example of a Java program to check if any 3 digit number is an Armstrong number or not.

If you are going for a Java interview, then be prepared for some follow-up questions like finding prime numbers, or finding Armstrong numbers of more than 3 digits.

Btw, good knowledge of essential programming concepts, operators, algorithms, data structure, and logic is important for any software engineer, particularly to do well on technical interviews.  If you are preparing for a tech interview then you should double down on these topics.

If you need recommendations to refresh your Data Structure and algorithm skills then I highly recommend checking out Data Structures and Algorithms: Deep Dive Using Java course on Udemy. It's a hands-on course and covers all essential data structures. It's also very affordable and you can get in just \$10 on Udemy flash sales which happen every now and then.

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

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

Btw, this is not the same program as print all Armstrong numbers 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 the next number.

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

These programs will help you to build the coding sense required to solve coding problems on interviews and your day-to-day life. It's also one of the important skills to map a new coding problem into an already known one and essential to pass the coding interview on companies like Google, Facebook, Amazon, and Microsoft.

If you are preparing for those companies then I also suggest you go through Grokking the Coding Interview: Patterns for Coding Questions on Educative, an interactive portal for coding interviews to learn some 16 useful coding patterns like Sliding Window, Two Pointers, Fast and Slow Pointers, Merge Intervals, Cyclic Sort, and Top K elements that can help you to solve zones of frequently asked coding problems.

### 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 a pretty simple Java program and if you look closely it just gets digit by digit by using the remainder operator and reduces the number by 1 digit after dividing it by 10. Let me know if you find any bug on this Java program to check for the Armstrong number.

Other Programming Interview problems you may like to solve
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Thanks for reading this article so far. If you find this logic-based Java coding problem from Google and my explanation useful then please share it with your friends and colleagues. If you have any questions or feedback then please drop a note.

P. S. - If you are preparing for a programming job interview, then you must prepare for an all-important topic like data structure, String, array, etc. One course which can help you with this task is the Grokking the Coding Interview: Patterns for Coding Questions course. It contains popular coding interview patterns which will help you to solve most of the problems in your coding interviews.

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. Feel free to comment, ask questions if you have any doubt.