Algorithm for Armstrong Number with in Range:

step 1:Read range

step 2: Set count=0, b= 1

step 3: Repeat through step-11 while b less than or equals to range

step 4: Set sum=0

step 5: Set c=b

step 6: Repeat through step-9 while c greater than 0

step 7: a=c%10

step 8: sum=sum+(a*a*a)

step 9: c=c/10

step 10: if sum equals to b then

print b count=count+1

step 11: b=b+1

step 12: Exit

Here is the Java Example for Armstrong Number with in Range:

import java.util.Scanner; 
 public class ArmstrongNumberWithinRange 
 { 
                public static void main(String args[]) 
             { 
                Scanner s=new Scanner(System.in); 
                System.out.print("Enter the Range :"); 
                int range=s.nextInt(); 
                int a,b,c,sum,count=0; 
                b=1; 
                while(b<=range) 
                     { 
                         sum=0; 
                         c=b; 
                         while(c>0) 
                              { 
                                 a=c%10; 
                                 sum=sum+(a*a*a); 
                                 c=c/10; 
                              } 
                          if(sum==b) 
                        { 
                          System.out.println(b+" is a Armstrong Number"); 
                          count=count+1; 
                        } 
                           b++; 
                      } 
 System.out.println("Total Armstrong Number Present With in that Range is "+count); 
                 } 
 } 
 

Algorithm for Armstrong Number:

step 1:Set s=0

step 2: Read num

step 3: Set b=num

step 4: Repeat through step-7 while num greater than 0

step 5: r=num%10

step 6: s=s+(r*r*r)

step 7: num=num/10

step 8: If b equals to s then print “it is an Armstrong number”

Else print “it is not an Armstrong number”

step 9: Exit

Here is the Java Example for Armstrong Number:

import java.util.Scanner; 
 public class ArmstrongNumber 
 { 
                  public static void main(String args[]) 
                { 
                     Scanner s1=new Scanner(System.in); 
                     int num,b,s=0,r; 
                     System.out.println("Enter A Number"); 
                     num =s1.nextInt(); 
                     b= num; 
                     while(num>0) 
                          { 
                             r= num%10; 
                             s=s+(r*r*r); 
                             num = num/10; 
                          } 
                             if (b==s) 
               System.out.println(b+" is an Armstrong Number"); 
                             else 
               System.out.println(b+" not an Armstrong Number"); 
                  } 
 } 
  

Algorithm For Prime Numbers between Range:

step 1: Set num=1

step 2: Read range

step 3: Repeat through step-9 while num less than or equals to range

step 4: Initialize b= 1, c=0

step 5: Repeat through step-7 while b less than or equals to num

step 6: If (num mod b) equals to 0 then set c=c+1

step 7: b=b+1

step 8: If c equals to 2 then print num

step 9: num =num +1

step 10: Exit

Here is the Java Example for Prime Numbers between Range:

import java.util.Scanner; 
 public class PrimeNumberBetweenRange 
 { 
                   public static void main(String args[]) 
           { 
                    int num =1,b,c,range; 
                  Scanner s=new Scanner(System.in); 
                  System.out.println("Enter The Range \n"); 
                  range=s.nextInt(); 
                  while(num <=range) 
                        { 
                            b=1; 
                            c=0; 
                            while(b<= num) 
                                 { 
                                   if((num%b)==0) 
                                     c=c+1; 
                                     b++; 
                                 } 
                                    if(c==2) 
                       System.out.println(num +" is a prime number"); 
                                     num ++; 
                         } 
              } 
 } 
 

Suppose, the value entered is say 121. It is assigned to an integer variable num. Now, num is divided by all the numbers from 2 to 8 (because a number can be divided by 1 and itself) (% rehrrns remainder). If the number is divided by any number from 2 to 8, its remainder will come to be 0 and in that case the message “Number is not prime” is displayed on the screen. 121 is not divisible by 2 but is no doubt divisible by 3 which returns remainder as 0 and hence it is not a prime number.

Algorithm For Prime Number Program:

step 1: Read num

step 2: Set b=l, c=0

step 3: Repeat through step-5 while (b <= num)

step 4: If (num mod b) equals to 0 then set c = c + 1

step 5: b = b + l

step 6: If c equals to 2 then print “num is prime”

Else print “num is not prime”

step 7: Exit

Here is the Java Example for Prime Number Program:

import java.util.Scanner; 
 public class PrimeNumber 
 { 
                 public static void main(String args[]) 
              { 
                   int num,b,c; 
                   Scanner s=new Scanner(System.in); 
                   System.out.println("Enter A Number"); 
                   num =s.nextInt(); 
                   b=1; 
                   c=0; 
                    while(b<= num) 
                       { 
                           if((num%b)==0) 
                              c=c+1; 
                              b++; 
                       } 
                        if(c==2) 
                        System.out.println(num +" is a prime number"); 
                        else 
                        System.out.println(num +" is not a prime number"); 
              } 
 } 
 

A Perfect Number is sum of its positive divisors, excluding the number itself. The First Perfect number is 6, For example 6 is a perfect number as 1,2 and3 are its divisors and the sum of divisors= ( 1 + 2 + 3)  = 6.

Algorithm for Perfect Number Program:

step 1: Read num

step 2: Set b=1, s=0

step 3: Repeat through step-5 until (b < num)

step 4: If (num mod b) equals to 0 then s=s+b

step 5: b=b+1

step 6: If s equals to num then print “num is perfect”

Else print “num is not perfect”

Step 7: Exit

Here is the Java Example for Perfect Number Program:

import java.util.Scanner; 
 class PerfectNumber 
 { 
     public static void main(String args[]) 
     { 
         Scanner sl=new Scanner(System.in); 
         System.out.println("Enter A Number"); 
         int num=sl.nextInt(); 
         
         int b=1, s=0; 
         
         while(b<num ) 
         { 
             if(num%b == 0) 
             { 
                 s += b; 
                
             } 
             b++; 
         } 
         
         if(s == num) 
         { 
             System.out.println(num+" is a perfect number"); 
         } 
        else 
         { 
            System.out.println(num+" is not a perfect number"); 
         }        
     } 
 } 
 

The Java Example to check if a Number is a palindrome or not.  A Number is palindrome if it remains unchanged when it is reversed, for example “151” is a palindrome as its reverse is “151”. [Read more…] about Palindrome Number in Java Example.

Factorial Number is the product of natural numbers from one to that particular number.  Mathematically, n! = 1 * 2 * 3 * …. * (n-1) * n

Algorithm for Factorial of the Given Number:

step 1: Set fact= 1

step 2: Read b

step 3: Set c=b

step 4: Repeat through step-6 until (c > 0)

step 5:  fact=fact*c

step 6:  c=c-1

step 7: Print fact

step 8: Exit

Here is the Java Example for Factorial of the Given Number:

import java.util.Scanner;
public class FactorialNumber
{
         public static void main(String ars[])
       {
               int fact=1,b,c;
            Scanner s=new Scanner(System.in);
            System.out.print("Enter A Number for get Factorial : ");
            b=s.nextInt();
            c=b;
            while(c>0)
                  {
                    fact=fact*c;
                            c-- ;
                   }
            System.out.println();
            System.out.println("Factorial of "+b+" is : "+fact);
         }
}
 

/* Factorial Program in Java using BufferedReader */ 
 import java.io.*;
class FactorialBufferedReader
{
    public static void main(String args[] )
    throws IOException
    {
        BufferedReader fact = new BufferedReader(new InputStreamReader (System.in));
        String m;
        int n,s=1,i;
        System.out.print("Enter a Number : ");
        m=fact.readLine();
        n=Integer.parseInt(m);
        for(i=1;i<=n;i++)
        {
            s=s*i;
        }
        System.out.println("Factorial is : "+s);
    }
} 
  

Fibonacci Series is series of Natural Number in the Sequence of: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55…., The first two number in Fibonacci series are 0 and 1, where next number is equivalent to sum of previous two number. in this example we use Recursion method , Recursion means calling the same function again and again to reduce the complexity of the problem solved. [Read more…] about Fibonacci Series Using Recursion in Java Example

step 1: Read n, x

step 2: Set r= 1

step 3: If value of n is 1 then print n and goto step – 8

step 4: Repeat through step-6 until j > 0

step 5: r=r*i

step 6: Update j as j=j-1

step 7: Print r

step 8: Exit

Here is the Java Example for the program RecursionPower

import java.util.Scanner; 
 public class RecursionPower 
 { 
                  public static void main(String args[]) 
         { 
                         int n, x, power; 
                  Scanner s=new Scanner(System.in); 
                  System.out.println("Enter values for n and x"); 
                  n=s.nextInt(); 
                  x=s.nextInt(); 
                  power=rec(n,x); 
                  System.out.println("\npower IS :"+power); 
           } 
                     static int rec(int i,int j) 
                  { 
                          int r=1;  
                           if(i==1) 
                             return i; 
                           else 
                                                   { 
                              while(j>0) 
                                   { 
                                        r=r*i; 
                                           j--; 
                                   } 
                                      return r; 
                            } 
                    } 
 }