5.16 Armstrong Number in Java

An Armstrong number of three digits is an integer such that the sum of the cubes of its digits is equal to the number itself. For example, 153 is an Armstrong number since 1^3 + 5^3 + 3^3 = 1+ 125+ 27 = 153.
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[@deepapn5049]

Thank you very much sir tomorrow is my java exam i didnt understood the concept which my tchr taught me but you made me understtand it thank u very much god bless u

[@Free_Bird78]

This is not a right method because it only implies when a number is of 3 digit only like 153, 370, 371 but if the number is of 4 digit like 1634(which is an armstrong number), it will show ,number is not an armstrong number' because it cubes the digit rather than multiplying it 4 times. Basically, there is another logic or program which implies in every digit number.

[@ASHOKkumar-pj3oj]

Bro give r*r*r*r instead of r*r*r simple

[@naveentandon993]

Thnku so much Sir for clearing the concept..... may God bless u always ❤❤🙏🙏

[@Anilkumar-ec8sj]

/* code */ public class Armstrong { public static void main(String args[]){ int n =370 , r,sum=0; int temp = n ; while(n>0){ r=n%10; n=n/10; sum = sum+r*r*r; } if(temp==sum){ System.out.println("Armstrong number"); } else { System.out.println("Not Armstrong Number"); } } }

[@maksimvolkov8892]

Nice, thank you for this video

[@SA-ie6lr]

What if we need to check a 4 digit number is armstrong or not?

[@vinaygoraigorai8553]

import java .util .*; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); System.out.print("input no."); int n = sc.nextInt(); int value = n; int r = 0, sum = 0, c = 0; for (int i = n; i > 0; i /= 10) { c++; } while (n > 0) { r = n % 10; sum += (Math.pow(r, c)); n = n / 10; } if (sum == value) System.out.print("armstrong no."); else System.out.print("not"); } }

[@mallikarjunareddy6122]

public class Armstrong { public static void main(String args[]) { int n= 153; int temp=n; int sum=0; while(n>0) { int r=n%10; n=n/10; sum = sum+r*r*r;//Note :-if n=a number of digits then that maney r's has to multipliy. } //Note:for ex:n=153,n has a 3 digits 1,5 and 3...so r*r*r.,n=1634,n has a 4 digits 1,6,3 and 4...so r*r*r. if(temp==sum) System.out.println(temp+ " is a armstrong"); else { System.out.println(temp+ " is not an armstrong"); } } }

[@ashutoshrai2717]

Yes this is right program for only 3 digit number not for 4 digits

[@mr.adityasingh9012]

Where -> sum = sum + r*r*r*r this will work on four digit....if you want check 5digit then....sum = sum+r*r*r*r*r

[@NaveenKumar-be8xl]

Simply super sir🙏

[@sivapradeep977]

int n=153; str N1=n; int length =len(N1); int ANS,FIN_ANS=0; for(i=0;i

[@abdulrahim7763]

tried without while loop..this is working fine for me (give any number within the limit of integer data type) Scanner sc=new Scanner(System.in); System.out.println("Enter any number to find out if it is a Armstrong number or not"); int input=sc.nextInt(); String s=""; String count=s.valueOf(input); int input_size = count.length(); int digit; double Armstrong=0; double result=0; int actual=input; if(input>0) { for(int i=1;i

[@tanveerjafri3052]

int num; Scanner scanner = new Scanner(System.in); System.out.println("Enter the number for ArmStrong Number"); num = scanner.nextInt(); int temp = num; int save = 0; int sum = 0; save = num%10; sum = (num/10)%10; num = (num/10)/10; num = num*num*num + sum*sum*sum + save*save*save; if (temp == num) System.out.println("This is ArmStrong Number: " + num); else System.out.println("This is not a ArmStrong Number");

[@drake71000]

Armstrong number can be of more than 3 digits.

[@Aulin001]

Yes

[@rejaulmallik9397]

Yaah of course lol🤣🤣🤣😃😂😂you don't know this

[@mohammadadil456]

I am new to java I was tasked to write a program to find Armstrong number my code: int a=125; do {System.out.println((a%10)*(a%10)*(a%10)); a=a/10; } while(a!=0); Output: 125 8 1 can someone explain my mistake.

[@digjeetkaur9519]

your explaination is superb , can u please also explain loops more how they work again and again

[@simplecoding5435]

this method is wrong armstrong number is not like that suppose if we have 4 digit no then what to do

[@Naidu-wv1dy]

explaination is good but this code will works three digit number only

[@akashgokul6818]

This prg is not applicable for 4digits values

[@pasupuletikoteswararao9184]

It is for only 3 digits What if we have n number

[@aviroxi]

this code is take input from the user and find whether given number is armstrong or not import java.util.*; //armstrong number or not? public class armstrongnumber{ public static void main(String[] args){ Scanner sc= new Scanner(System.in); int x=sc.nextInt(); int t=x,s=x; int n=0,sum=0,ind=0; while(x>0){ ++n; x=x/10; } while(t>0){ ind=t%10; sum=sum+power(ind,n); t=t/10; } if(s==sum) System.out.print("true"); else System.out.print("false"); } static int power(int N, int P) { if (P == 0) return 1; else return N * power(N, P - 1); } }

[@skasif4476]

This program is valid for 3 digit numbers only...can you generalize for all the numbers we wish to check??

[@aviroxi]

this code is take input from the user and find whether given number is armstrong or not import java.util.*; //armstrong number or not? public class armstrongnumber{ public static void main(String[] args){ Scanner sc= new Scanner(System.in); int x=sc.nextInt(); int t=x,s=x; int n=0,sum=0,ind=0; while(x>0){ ++n; x=x/10; } while(t>0){ ind=t%10; sum=sum+power(ind,n); t=t/10; } if(s==sum) System.out.print("true"); else System.out.print("false"); } static int power(int N, int P) { if (P == 0) return 1; else return N * power(N, P - 1); } }

[@murliengineering6130]

Is we take a 9474 it's not working because that power is 9^4+4^4+7^4+4^4=9474

[@dhanalaskhmidhanu3468]

I really not understand what you did ?????

[@dharunsivakumar7584]

Wow.. I thought I am the only one watching this in 2020.

[@dhanalaskhmidhanu3468]

@@dharunsivakumar7584 nice

[@ozlemcetin5868]

I wish the calculation (r*r*r) could be dynamic because the number of multiplications is determined by the digit count. For instance, when determining if 1345 is an Armstrong number, as it has four digits, it should be r*r*r*r.

[@sekharraj525]

If you given numbers 4digit

[@ujjwalgoyal7215]

Hello sir , This program is not working for 4 digit numbers.

[@aviroxi]

this code is take input from the user and find whether given number is armstrong or not import java.util.*; //armstrong number or not? public class armstrongnumber{ public static void main(String[] args){ Scanner sc= new Scanner(System.in); int x=sc.nextInt(); int t=x,s=x; int n=0,sum=0,ind=0; while(x>0){ ++n; x=x/10; } while(t>0){ ind=t%10; sum=sum+power(ind,n); t=t/10; } if(s==sum) System.out.print("true"); else System.out.print("false"); } static int power(int N, int P) { if (P == 0) return 1; else return N * power(N, P - 1); } }

[@aadilhussain4556]

Sir it is applicable only for 3 digit numbers if we have no like 8208, then we have write the condition like this sum=sum+ rev*rev*rev*rev; but sir is it not good practies

[@shubhammhatre5089]

but what if the number is 4 digit then we will have to use num raise to 4, so how will we change the r*r*r*r that time @telusko

[@elonmusk.0]

/* improvised status: below code works for a number with any number of digits user wishes(i.e. user input - any number of digits) */ import java.lang.Math; import java.util.*; public class armstrong_num { public boolean check_armstrong(int n){ int sum = 0,i=0,r,t,dgts,m; // variable 't' and 'dgts' are used to store original value of 'n' because we need to compare changed value with original value since later in loops 'n' value is changed. t = n; dgts = n; while (dgts!=0){ // This loop is to count the total digits in 'n'. e.g. n=123, so 'i' will be 3. We will use this 'i' to as the power of r. So with this we can find any armstrong number not just 3 digit number. dgts = dgts / 10; i++; } System.out.println("Total digits in numbers: "+i); while (n>0){ r = n % 10; //seperate last digit and stored in 'r' n = n / 10; //n will no longer have last digit (it will have digits other than last one) m = (int)Math.pow(r, i); // (r) ^ (i) // if n = 2345 then 'i' will be 4(from upper while loop) sum += m; //System.out.println(sum); // this shows 'r' with power of 'i' one by one. } if(t == sum){ return true; } return false; } public static void main(String[] args) { armstrong_num an = new armstrong_num(); Scanner sc = new Scanner(System.in); System.out.println("Enter a number to check whether it is armstrong number or not"); int n = sc.nextInt(); //int n = 9474; //e.g. 9474 , 1634 are armstrong numbers boolean b = an.check_armstrong(n); //if function(check_armstrong) is static, we can call function directly i.e. no need for an object in main i.e. 'an'. if(b){//true System.out.println(n+" is a armstrong number."); } else{ System.out.println(n+" is not a armstrong number"); } } }

[@NirajPRoxx]

Thank you :)

[@balapramothc.m.]

why only cube

[@mangeshshelke1846]

If user enter number 4 digits then?.. It's just hard coded value

[@shuvshaw9594]

int x=153,j=x,r,sum=0; while(x>0) { r=x%10; sum=sum+(r*r*r); x=x/10; } if(sum==j) System.out.println("y"); else System.out.println("n");

[@murliengineering6130]

Sir this program is not work for 4 digit

[@fatemehrahmani6740]

Thank u very much for your tutorial.

[@अभयसिंहयादव]

Its is applicablr only for 3 digit number It is not useful How to calculate 4 /2/5 digit number are armstrong

[@aviroxi]

this code is take input from the user and find whether given number is armstrong or not import java.util.*; //armstrong number or not? public class armstrongnumber{ public static void main(String[] args){ Scanner sc= new Scanner(System.in); int x=sc.nextInt(); int t=x,s=x; int n=0,sum=0,ind=0; while(x>0){ ++n; x=x/10; } while(t>0){ ind=t%10; sum=sum+power(ind,n); t=t/10; } if(s==sum) System.out.print("true"); else System.out.print("false"); } static int power(int N, int P) { if (P == 0) return 1; else return N * power(N, P - 1); } }

[@anushreebachhav9299]

why didnt we simply do, (n==sum) ?

[@aviroxi]

to check if they are same, if yes then it is an armstrong number

[@RohitRaj-bt7qx]

If we check (n == sum) it won't work because.. After completion of the while loop, As we wrote that n = n/10; The value will be 0.153 (in case of number 153) So.. We stored the given input in temp to check it at the end of the program. As if we wrote (n == sum) 0.153 never matches to 153.

[@atulgupta3407]

why you have so many browsers , i think you do sussy things

[@yasseralkindi7350]

brilliant, if you can add some explanation of the math logic. code wise, obviously its perfect!

[@DAME-RAJATHS]

I wish you had done it n^n not n^3

[@sekharraj525]

How to work on scanner class

[@perpetual23]

it's nt working fr the input 1 to 9.. shows nt armstrng..

[@SmartProgramming]

import java.util.Scanner; class ArmstrongNumber { public static void main(String args[]) { Scanner s=new Scanner(System.in); System.out.println("Enter The Number : "); int no=s.nextInt(); int t1=no; int leng=0; while(t1 != 0) { leng=leng+1; t1=t1/10; } int t2=no; int arm=0; int rem; while(t2 != 0) { int mul=1; rem=t2%10; for(int i=1;i

[@Dinesh_kuku]

Superbly explained!

[@sumeetsajjan]

Thank u sir now easy to understand

[@shampachakraborty2515]

Thanks

[@mr_mouli_10]

chala thanks andi meku

[@karnataka8863]

Used r x r x r for 3digit no if user use any no then how we come to know the power ?

[@ts9dream]

We use another loop to get the size of number and then we use math.pow(digit,size)

[@InInferno]

@@ts9dream thank you

[@PraveenKumar-uj2ih]

Clear Explanation ......Thanks Bro

[@sanketkeluskar5583]

really good :)

[@dragonbooster5618]

Nice explanation sir.Thank you so much

[@mayankmittal9269]

Thank you

[@karan_yt_family]

Ossam

[@jaishukla5054]

8208 is Also an Armstrong number but it's doesn't work

[@aviroxi]

this code is take input from the user and find whether given number is armstrong or not import java.util.*; //armstrong number or not? public class armstrongnumber{ public static void main(String[] args){ Scanner sc= new Scanner(System.in); int x=sc.nextInt(); int t=x,s=x; int n=0,sum=0,ind=0; while(x>0){ ++n; x=x/10; } while(t>0){ ind=t%10; sum=sum+power(ind,n); t=t/10; } if(s==sum) System.out.print("true"); else System.out.print("false"); } static int power(int N, int P) { if (P == 0) return 1; else return N * power(N, P - 1); } }

[@argonxgaming2045]

Thankyou sir:)

[@algorithmgaming7416]

Why did we use while loop

[@shivaprakashreddyyeredla1705]

I didn't understand this vedio 😭😭

[@thegrimare2917]

sorry dude it does not work

[@SmartProgramming]

import java.util.Scanner; class ArmstrongNumber { public static void main(String args[]) { Scanner s=new Scanner(System.in); System.out.println("Enter The Number : "); int no=s.nextInt(); int t1=no; int leng=0; while(t1 != 0) { leng=leng+1; t1=t1/10; } int t2=no; int arm=0; int rem; while(t2 != 0) { int mul=1; rem=t2%10; for(int i=1;i

[@shuvshaw9594]

very helpful

[@abdulravoof6444]

thanks bro

[@sudhakar3991]

This logic is not working......dont refer this

[@anusmitasen2009]

It's completely correct, check your program

[@Shiko99FF]

​​@@anusmitasen2009No he is right what if we have 4 digit number then this logic will not going to work on that 4 digit number

[@koushikaisrat3670]

what about if the number n has four or five digits ?

[@karnveerayush]

+koushika israt It will make no difference.

[@AbhiShek-nv3vq]

It'll definitely make a difference. We'll have to multiply the remainder the number of times there are digits in the number. e.g. 9474 is an armstrong number which is equal to: 9*9*9*9+ 4*4*4*4+ 7*7*7*7+ 4*4*4*4= 9474.

[@drake71000]

Or even more than that. If we are not sure how many digits are there in a number than what to do?

[@NirajPRoxx]

we have to multiply each digit only three times in order to get cube.

[@SmartProgramming]

import java.util.Scanner; class ArmstrongNumber { public static void main(String args[]) { Scanner s=new Scanner(System.in); System.out.println("Enter The Number : "); int no=s.nextInt(); int t1=no; int leng=0; while(t1 != 0) { leng=leng+1; t1=t1/10; } int t2=no; int arm=0; int rem; while(t2 != 0) { int mul=1; rem=t2%10; for(int i=1;i

[@venkateswarluguntaka7370]

❤

[@sumitchaturvedi5933]

thank u sir

[@saikumarn3547]

//To check whether given number is armstrong number or not import java.util.Scanner; class ArmstrongNumber { public static void main(String[] args) { Scanner scanner=null; int n=0,r=0,sum=0,temp=0; scanner=new Scanner(System.in); System.out.println("Enter number to check armstrong or not : "); n=scanner.nextInt(); temp=n; int length=(int)Math.log10(temp)+1; while(n>0){ r=n%10; n=n/10; sum=sum+((int)Math.pow(r,length)); } if(temp==sum){ System.out.println(temp+" is armstrong number"); } else{ System.out.println(temp+" is not a armstrong number"); } } }

[@ravinsr69]

It does not works for 4 digits or more

[@PramodKumar1207]

import java.util.Scanner; public class arm { public static void main(String[] args) { // TODO Auto-generated method stub System.out.println("-- Armstrong number --"); System.out.println("Enter the no"); Scanner sc=new Scanner(System.in); int n = sc.nextInt(); int original=n; int count=0; int arm=0; while(n>0) { n =n/10; count++; } System.out.println("Length of the number: "+count); n = original; while(n>0) { int rem = n%10; n=n/10; int temp=1; for(int i=count;i>0;i--) { temp=temp*rem; } arm=arm+temp; } if(arm == original) { System.out.println(original+ " is an Armstrong number"); } else { System.out.println(original+ " is not an Armstrong number"); } } }

[@Fbielijah]

Video is not good

[@pranavgupta9132]

CAN YOU GIVE PROGRAM TO PRINT MAGIC NUMBERS FROM 1 TO 1000 ?

[@AbhiShek-nv3vq]

What Magic numbers??

[@tharunchinna3375]

for any digits use this code class Main{ public static void main(String[] args) { int d = 54748; int k =d; int t =0; String g = Integer.toString(d); int m =g.length(); while(d>0){ int f = d%10; t+=Math.pow(f,m); d=d/10;} if(t==k){ System.out.println("it an armstrong number"); }else{ System.out.println("it ! an armstrong number"); } } }

[@Sumit.Kr.singh93]

import java.util.*; public class Main { public static void main(String[] args) { // int n = 371; Scanner scn =new Scanner(System.in); int n=scn.nextInt(); int nod = numOfDigits(n); int ans = 0; int backup = n; while (n > 0) { int digit = n % 10; ans = ans + (int)Math.pow(digit, nod); n = n / 10; } if (backup == ans) { System.out.println("true"); } else { System.out.println("false"); } } public static int numOfDigits(int n) { int ans = 0; while (n != 0) { n =n/ 10; ans++; } return ans; } } this will work ffor your every input.