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/*Program to find whether given no. is Armstrong or not. Example : Input - 153 Output - 1^3 + 5^3 + 3^3 = 153, so it is Armstrong no. */ class Armstrong{ public static void main(String args[]){ int num = Integer.parseInt(args[0]); int n = num; //use to check at last time int check=0,remainder; while(num > 0){ remainder = num % 10; check = check + (int)Math.pow(remainder,3); num = num / 10; } if(check == n) System.out.println(n+" is an Armstrong Number"); else System.out.println(n+" is not a Armstrong Number"); } }

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Write a java program to check a number is Armstrong number?

import java.io.*; public class Jerry { public static void main(String as[]) throws IOException { int k=Integer.parseInt(as[0]); int n=k; int d=0,s=0; while(n>0) { d=n%10; s=s+(d*d*d); n=n/10; } if(k==s) System.out.println("Armstrong number"); else System.out.println("not Armstrong number"); } }


To find strong number in java?

sir , what is perfect no?


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How do you find out whether Java is installed on Linux?

Enter "java -version" into a terminal. If Java is installed, it will tell you the version number. If it is not installed, it will say "command not found."


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List all Armstrong numbers between 1 to 5000?

Armstrong numbers are those numbers which are equal to the sum of the digits of the number each raised to the power of the number of digits in the number itself. There are 14 Armstrong numbers in the range 1-5000, which are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634.


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If by system number you mean the version number of the Java language on the client computer, you can useSystem.getProperty("java.version"); .Or, if you want the Java virtual machine version number you can use System.getProperty("java.vm.version"); .Finally, if you want the version number of the operating system, you can use System.getProperty("os.version"); .


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Flow chart for Armstrong number?

armstrong number


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How do you write a Java program to see whether a number is Armstrong or not... using Buffered Reader and stuff ....?

import java.io.*; class Armstrong { public static void main()throws IOException { BufferedReader in=new BufferedReader(new InputStreamReader(System.in)); System.out.println("Enter a number"); int a=Integer.parseInt(in.readLine()); int n1=a,rev=0,d=0; while(a!=0) { d=a%10; rev=rev+d*d*d; a=a/10; } if(rev==n1) System.out.println("It is a armstrong number "); else System.out.println("It is not a armstrong number "); } }