Android Development

package com.company ; import java.util.Scanner ; public class Main { public static void main ( String [] args ) { Scanner scan = new Scanner( System . in ) ; System . out .println( "Enter a Number" ) ; int n = scan .nextInt() ; int r ; String str = "" ; while ( n > 0 ) { r = n % 10 ; n = n / 10 ; str = str + r ; } System . out .println( str ) ; char c ; for ( int i = str .length()- 1 ; i >= 0 ; i --) { c = str .charAt( i ) ; switch ( c ) { case '0' : System . out .print( "Zero " ) ; break ; case '1' : System . out .print( "One " ) ; break ; case '2' : System . out .print( "Two " ) ; b

Pallindrome Number : - If the reverse of the number is eqal to original number then , the number is pallindrome. ex:- original number = 121 , reverse number = = 121 therefore 121 is pallindrome. int r , n ; int rev = 0 ; Scanner sc = new Scanner( System . in ) ; n = sc .nextInt() ; int number = n ; // to not lost original number while ( n > 0 ) { r = n % 10 ; rev = rev * 10 + r ; n = n / 10 ; } System . out .println( rev ) ; if ( rev == number ) { System . out .println( "Number is pallindrome" ) ; } else System . out .println( "Number is not pallindrome" ) ;   output :-                        121                                121                                Number is pallindrome

       Armstorng no : If the sum of the digits of the cube of number is equal to original number then it is armstorng number .       ex : take number n = 153;           if  3*3*3 + 5*5*5 + 1*1*1 = 153 (i.e. same  number ,n )           then, the number is Armstorng number. int n , r ; int sum = 0 ; Scanner sc = new Scanner( System . in ) ; n = sc .nextInt() ; int store = n // to store the number ; while ( n > 0 ) { r = n % 10 ; sum = sum + r * r * r ; //sum of the cuber of the number n = n / 10 ; } System . out .println( sum ) ; if ( sum == store ) { System . out .println( store + " is Armstrong no" ) ; } else System . out .println( store + " is not Armstrong no" ) ; } 23564 440 23564 is not Armstrong no Process finished with exit code 0