Armstrong Numbers upto 'n' numbers in JAVA
Armstrong Number in JAVA
Armstrong
Number: A number is said to be Armstrong
Number, If it is a equal to the sum of its own digits each raised to the power
of the number of digits.
Example:
Input : 153
No. of Digits : 3
Sum of digits : 13 + 53
+ 33 = 1 + 125 + 27 = 153
Output : It is an
Armstrong Number
Input : 15
No. of Digits : 2
Sum of digits : 12 + 52
= 1 + 25 = 26
Output : It is not an
Armstrong Number
Input : 1634
No. of Digits : 4
Sum of digits : 14 + 64
+ 34 + 44 = 1 + 1296 + 81 + 256 = 1634
Output : It is an
Armstrong Number
Algorithm and Program given below :-
Algorithm :-
- Input a number
- Use for loop from 0 to the input number
- Copy the value of loop variable in another variable
- Using a string variable convert the integer value into string
- Store the length of this string in another variable.
- Using while loop calculate the sum of digits raised with the power of number of digits in a variable say ‘s’.
- Compare the calculated number with the original number
- If the numbers are equal print the number.
- Initialize value of s=0 within the for loop again.
- Compile and run the program
Program :-
import java.io.*;
import java.util.*;
class Armstrong1
{
public static void main(String args[])
{
Scanner in= new Scanner(System.in);
int n,p,rev,i,l;
double s=0;
String str;
System.out.println("Enter the number");
n= in.nextInt(); // Enter Number from user
System.out.println("Armstrong Number from 0 to "+n+" are:-");
for(i=0;i<=n;i++)
{
p=i;
str= Integer.toString(i);
l= str.length();
while(p>0)
{
rev= p%10; // extracting last digit of the number
s=s+Math.pow(rev,l); // Storing digit with respective power
p=p/10; // extracting quotient of the number
}
if(s==i) // checking entered number with cubes of its digits
{
System.out.println(i);
}
s=0;
}
}// end of main method
}// end of class
Video Link of this Program : https://youtu.be/Ldsyc7hfUr4
Follow me on Instagram
All the Best :)
Keep Learning :)
Comments
Post a Comment