Armstrong Number in Java

Armstrong Number Java Program

The following Java program prints all the Armstrong numbers up to the specified limit.

ArmstrongNumberExample1.java

import java.util.Scanner;  
import java.lang.Math;  
public class ArmstsrongNumberExample  
{  
//function to check if the number is Armstrong or not  
static boolean isArmstrong(int n)   
{   
int temp, digits=0, last=0, sum=0;   
//assigning n into a temp variable  
temp=n;   
//loop execute until the condition becomes false  
while(temp>0)    
{   
temp = temp/10;   
digits++;   
}   
temp = n;   
while(temp>0)   
{   
//determines the last digit from the number      
last = temp % 10;   
//calculates the power of a number up to digit times and add the resultant to the sum variable  
sum +=  (Math.pow(last, digits));   
//removes the last digit   
temp = temp/10;   
}  
//compares the sum with n  
if(n==sum)   
//returns if sum and n are equal  
return true;      
//returns false if sum and n are not equal  
else return false;   
}   
//driver code  
public static void main(String args[])     
{     
int num;   
Scanner sc= new Scanner(System.in);  
System.out.print("Enter the limit: ");  
//reads the limit from the user  
num=sc.nextInt();  
System.out.println("Armstrong Number up to "+ num + " are: ");  
for(int i=0; i<=num; i++)  
//function calling  
if(isArmstrong(i))  
//prints the armstrong numbers  
System.out.print(i+ ", ");  
}   
}  

Output:

Enter the limit: 999
Armstrong Number up to 999 are: 
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407

Let’s create another Java program that checks if the given number is an Armstrong number or not.

ArmstrongNumberExample2.java

import java.util.Scanner;  
import java.lang.Math;  
public class ArmstsrongNumberExample2  
{  
//function to check if the number is Armstrong or not  
static boolean isArmstrong(int n)   
{   
int temp, digits=0, last=0, sum=0;   
//assigning n into a temp variable  
temp=n;   
//loop execute until the condition becomes false  
while(temp>0)    
{   
temp = temp/10;   
digits++;   
}   
temp = n;   
while(temp>0)   
{   
//determines the last digit from the number      
last = temp % 10;   
//calculates the power of a number up to digit times and add the resultant to the sum variable  
sum +=  (Math.pow(last, digits));   
//removes the last digit   
temp = temp/10;   
}  
//compares the sum with n  
if(n==sum)   
//returns if sum and n are equal  
return true;      
//returns false if sum and n are not equal  
else return false;   
}   
//driver code  
public static void  main(String args[])     
{     
int num;   
Scanner sc= new Scanner(System.in);  
System.out.print("Enter the number: ");  
//reads the limit from the user  
num=sc.nextInt();  
if(isArmstrong(num))  
{  
System.out.print("Armstrong ");  
}  
else   
{  
System.out.print("Not Armstrong ");  
}  
}   
}  

Output 1:

Enter the number: 2
Armstrong

Output 2:

Enter the number: 1675
Not Armstrong