Skip to main content

KAPREKAR NUMBER

PROGRAM : KAPREKAR NUMBER

Kaprekar number  is a non-negative integer, the representation of whose square  can be split into two parts that add up to the original number again. For instance, 45 is a Kaprekar number, because 452 = 2025 and 20+25 = 45. And if second splitted part have only 0s then it is not kaprekar number for example 100*100=10000 splitted part is 10+000 thus second part have 0s only

Some Examples are: 999
999*999=998001=>998+001=999
1, 9, 45, 55, 99, 297, 703, 999, 2223, 2728, 4879, 4950, 5050, 5292, 7272, 7777, 9999, 17344, 22222, 38962, 77778, 82656, 95121, 99999, 142857, 148149, 181819, 187110 
are some examples of Kaprekar Number

SYNTAX

import java.io.*;
class kaprekar
{
    public static void main(String args[])throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n;int d=0,c=0;int n1;
        System.out.println("ENTER A NUMBER");
        n=Integer.parseInt(br.readLine());
        int square=n*n;int d1=0,d2=0;n1=n;
        int sq=square;
        String s=Integer.toString(square);
        while(sq!=0)
        {
            d=sq%10;
            c++;
            sq=sq/10;
        }
    
    if(c%2==0)
    {
        int x=(int)Math.pow(10,c/2);
        d2=(square%(int)x);
        d1=square/(int)x;
        System.out.print(d1+"+"+d2);
        if(d2==0)System.out.println("No The Number is not kaprekar Number\n As the second part of its square have only 0s");
        else if((d1+d2)==(n))
        System.out.print("="+(n)+"\nYES The Number is kaprekar Number");
        else
        System.out.println("!="+(n)+"\nNo The Number is not kaprekar Number");
    }
    else
    {   int y=(int)Math.pow(10,(c+1)/2);
        d2=(square%(int)y);
        d1=square/(int)y;
        System.out.print(d1+"+"+d2);
        if(d2==0)System.out.println(" \nNo The Number is not kaprekar Number\n As the second part of its square have only 0s");
        else if((d1+d2)==(n))
        System.out.print("="+(n)+"\nYES The Number is kaprekar Number");
        else
        System.out.print("!="+(n)+"\nNo The Number is not kaprekar Number");
}
}
}


SCREENSHOTS




Comments

Popular posts from this blog

Arranging according to potential.

PROGRAM: Requested by Sir  Ankur Acharya   ARRANGING WORDS IN A STRING ACCORDING TO THEIR POTENTIAL The encryption of alphabets are to be done as follows: A = 1 B = 2 C = 3 . . . Z = 26 The potential of a word is found by adding the encrypted value of the alphabets. Example: KITE Potential = 11 + 9 + 20 + 5 = 45 Accept a sentence which is terminated by either “ . ” , “ ? ” or “ ! ”. Each word of sentence is separated by single space. Decode the words according to their potential and arrange them in ascending order. Output the result in format given below: Example 1 INPUT : THE SKY IS THE LIMIT. POTENTIAL : THE = 33 SKY = 55 IS = 28 THE = 33 LIMIT = 63 OUTPUT : IS THE THE SKY LIMIT Example 2 INPUT : LOOK BEFORE YOU LEAP. POTENTIAL : LOOK = 53 BEFORE = 51 YOU = 61 LEAP = 34 OUTPUT : LEAP BEFORE LOOK YOU SYNTAX import java.io.*; class Potential {     public static void main(String args[])throws IOException     {         BufferedReader br=new BufferedReader(new Inp

SMITH NUMBER

PROGRAM: SMITH NUMBER A   Smith number  is a composite number, the sum of whose digits is the sum of the digits of its prime factors obtained as a result of prime factorization (excluding 1). The first few such numbers are 4, 22, 58, 85, 94, 121 ……………….. For Example 58=> 5+8=13 2+(2+9) of 29 =13 Since sum of digits=sum of digits of prime factors 58 is a SMITH NUMBER SYNTAX: import java.io.*; class smith_no { static int sum,sum1;     public static void main(String args[])throws IOException     {         BufferedReader br=new BufferedReader(new InputStreamReader(System.in));         int n;int c=0;         System.out.println("Enter a Number");         n=Integer.parseInt(br.readLine());         int n1=n;         while(n1!=0){         int d=n1%10;         sum1+=d;         n1=n1/10;}         for(int i=2;i<n;i++)         {             if(n%i==0)             {                 for(int j=2;j<i;j++)                 {                     if(i

MAGICAL/ MAGIC NUMBER USING RECURSION

PROGRAM :  MAGICAL NUMBER BY RECURSION This  program  checks if a  number  is a  Magic number  in  JAVA . A  number  is said to be a  Magic number  if the sum of its digits are calculated till a single digit is obtained by recursively adding the sum of its digits. If the single digit comes to be 1 then the  number  is a  magic number . Example: 10999 => 1+0+9+9+9=28=>2+8=>10=>1+0= 1                  199    =>1+9+9 =>19=> 1+9=>10=>1+0= 1 SYNTAX import java.io.*; class magic  {    static int c;     public static void main(String arsg[])throws IOException     {         BufferedReader br=new BufferedReader(new InputStreamReader(System.in));         int n;int d=0;int m;         System.out.println("ENter a NUmber");         n=Integer.parseInt(br.readLine());         m=n;         while(n!=0)         {d=n%10;             n=n/10;             c++;         }         magical(m);              }     public static void magical(int x)