Java program to check number is armstrong or not | Learn Coding 

I agree 💯

Yes you are right

same bro

Aap ka har explanation easy to understand h...Mere liye samjh na bohat asan ho raha h...Keep it up sir, Mera prob solve kar dijiye samaj me nahi a raha kya kare...

@@futureworld1807 kya problem hai bolo?

it is working

Yes u r right

yes bhai

No not working

Please Share kr do 🙏🙏🙏

@@LearnCodingOfficial why don't you teach for n digits 😞 I recommend your channel to all my friend and now they said me that he didn't teach for n digits 😞😞😞 Please make a video for n digits

import java.util.Scanner; public class Main{ public static void main(String[] args){ Scanner in=new Scanner(System.in); int n=in.nextInt(); int p=n, q=n, count=0, check=0, value=n; while(p!=0){ int rem=p%10; count++; p=p/10; } while(n!=0){ int rev=n%10; check=((int)(Math.pow(rev, count))+check); n=n/10; } if(check==value){ System.out.println(value+" is a Armstrong number"); }else System.out.println(value + " is not a Armstrong number"); } }

while(n>0) { rev=(rev*10)+ n%10; n=n/10; } System.out.println(rev);

public static void main(String[] args) { System.out.println("enter a number"); int n ,z, rev=0; Scanner sc = new Scanner(System.in); n = sc.nextInt(); // z=n; while(n>0) { rev=(rev*10)+ n%10; n=n/10; } s.o.p(rev);

import java.util.*; public class Main { public static void main(String[] args) { int n,arm=0,rem,d=0; Scanner sc=new Scanner(System.in); System.out.println("Enter the Number"); n=sc.nextInt(); while(n>0){ n=n/10; d++; } int c=n; while(n>0 ) { rem=n%10; arm=(rem*rem*rem)+arm; n=n/10; } if(c==arm){ System.out.println("Armstrong number"); } else System.out.println("Not Armstrong number"); } }