Java program to find whether the given number is an Armstrong number

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[@pl_0077]

Great series ever for java program

[@QAFox]

Thank you :) Note: Join our telegram group here - t.me/qafoxoriginal

[@karthikeyanravi9840]

Sir, thank you so much this is exactly I was looking for. Loved your content! 😍

[@QAFox]

Most welcome!

[@Rakesh-le6mi]

Thank you sir

[@TejendraPatel-wz2yk]

If i Enter number with 6 digit?

[@DeveshKumarprajapati-we4oi]

This program is not perfect because it will check only that number which is in between 100 to 999. To make this program perfect we have to count the digit and find the power of each digit with the help of count digit value. let's assume the number is 1234 then we have to calculate it like 1^4 + 2^4 + 3^4 + 4^4

[@QAFox]

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[@DeveshKumarprajapati-we4oi]

@@QAFox Armstrong numbers between 1 and 10000 are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474 package Checker; import java.util.Scanner; public class Checker { public static void main(String[] args) { Scanner sc=new Scanner(System.in); //Taking user input...... System.out.println("Enter the number: "); int n=sc.nextInt(); sc.close(); //calling the method for checking armstrong number int sum=armStrong(n); //checking the condition if(sum==n) { //final output System.out.println(n+ " is ARMSTRONG number."); } else { //final output System.out.println(n+ " is not ARMSTRONG number."); } } //this method is going to count the number of digit in the given number public static int countDigit(int num) { int count=0; while(num>0) { num=num/10; count++; } //returning the number of digits return count; } public static int armStrong(int num) { int sum=0; //calling the method for counting the number of digit in given number and storing it.. int pow=countDigit(num); while(num>0) { int power=1; int rem=num%10; //loop for finding power for(int i=1;i