Armstrong Numbers between two integers

A positive integer with digits a, b, c, d… is called an Armstrong number of order n if following condition is satisfied.

abcd... = an + bn + cn + dn +...
153 = 1*1*1 + 5*5*5 + 3*3*3  
    =  1 + 125 + 27
    =  153        
Therefore, 153 is an Armstrong number.

Examples:

Input : 100 400
Output :153 370 371
Explanation : 100 and 400 are given 
two integers.(interval)
  153 = 1*1*1 + 5*5*5 + 3*3*3 
      = 1 + 125 + 27
      =  153  
  370 = 3*3*3 + 7*7*7 + 0
      = 27 + 343 
      = 370
  371 = 3*3*3 + 7*7*7 + 1*1*1
      = 27 + 343 +1
      = 371



The approach implemented below is simple. We traverse through all numbers in given range. For every number, we first count number of digits in it. Let the number of digits in current number be n. Them we find sum of n-th power of all digits. If sum is equal to i, we print the number.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to find Armstrong numbers in a range
#include <bits/stdc++.h>
using namespace std;
  
// Prints Armstrong Numbers in given range
void findArmstrong(int low, int high)
{
    for (int i = low+1; i < high; ++i) {
  
        // number of digits calculation
        int x = i;
        int n = 0;
        while (x != 0) {
            x /= 10;
            ++n;
        }
  
        // compute sum of nth power of 
        // its digits
        int pow_sum = 0; 
        x = i;
        while (x != 0) {
            int digit = x % 10;
            pow_sum += pow(digit, n);
            x /= 10;
        }
  
        // checks if number i is equal to the
        // sum of nth power of its digits
        if (pow_sum == i) 
            cout << i << " ";     
    }
}
  
// Driver code
int main()
{
    int num1 = 100;
    int num2 = 400;
    findArmstrong(num1, num2);
    cout << '\n';
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// JAVA program to find Armstrong 
// numbers in a range
import java.io.*;
import java.math.*;
  
class GFG {
      
    // Prints Armstrong Numbers in given range
    static void findArmstrong(int low, int high)
    {
        for (int i = low + 1; i < high; ++i) {
       
            // number of digits calculation
            int x = i;
            int n = 0;
            while (x != 0) {
                x /= 10;
                ++n;
            }
       
            // compute sum of nth power of 
            // its digits
            int pow_sum = 0
            x = i;
            while (x != 0) {
                int digit = x % 10;
                pow_sum += Math.pow(digit, n);
                x /= 10;
            }
       
            // checks if number i is equal 
            // to the sum of nth power of
            // its digits
            if (pow_sum == i) 
                System.out.print(i + " ");     
        }
    }
       
    // Driver code
    public static void main(String args[])
    {
        int num1 = 100;
        int num2 = 400;
        findArmstrong(num1, num2);
        System.out.println();
    }
}
  
/*This code is contributed by Nikita Tiwari.*/

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# PYTHON program to find Armstrong 
# numbers in a range
import math
  
# Prints Armstrong Numbers in given range
def findArmstrong(low, high) :
      
    for i in range(low + 1, high) :
          
        # number of digits calculation
        x = i
        n = 0
        while (x != 0) :
            x = x / 10
            n = n + 1
              
        # compute sum of nth power of 
        pow_sum = 0 
        x = i
        while (x != 0) :
            digit = x % 10
            pow_sum = pow_sum + math.pow(digit, n)
            x = x / 10
              
        # checks if number i is equal to
        # the sum of nth power of its digits
        if (pow_sum == i) :
            print(str(i) + " "),
  
# Driver code
num1 = 100
num2 = 400
findArmstrong(num1, num2)
print("")
  
# This code is contributed by Nikita Tiwari.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find Armstrong
// numbers in a range
using System;
  
class GFG {
  
    // Prints Armstrong Numbers in given range
    static void findArmstrong(int low, int high)
    {
        for (int i = low + 1; i < high; ++i) {
  
            // number of digits calculation
            int x = i;
            int n = 0;
            while (x != 0) {
                x /= 10;
                ++n;
            }
  
            // compute sum of nth power of
            // its digits
            int pow_sum = 0;
            x = i;
            while (x != 0) {
                int digit = x % 10;
                pow_sum += (int)Math.Pow(digit, n);
                x /= 10;
            }
  
            // checks if number i is equal
            // to the sum of nth power of
            // its digits
            if (pow_sum == i)
                Console.Write(i + " ");
        }
    }
  
    // Driver code
    public static void Main()
    {
        int num1 = 100;
        int num2 = 400;
        findArmstrong(num1, num2);
        Console.WriteLine();
    }
}
  
/*This code is contributed by vt_m.*/

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find 
// Armstrong numbers 
// in a range
  
// Prints Armstrong
// Numbers in given range
function findArmstrong($low, $high)
{
    for ($i = $low + 1;
         $i < $high; ++$i
    {
  
        // number of digits
        // calculation
        $x = $i;
        $n = 0;
        while ($x != 0) 
        {
            $x = (int)($x / 10);
            ++$n;
        }
  
        // compute sum of nth 
        // power of its digits
        $pow_sum = 0; 
        $x = $i;
        while ($x != 0) 
        {
            $digit = $x % 10;
            $pow_sum += (int)(pow($digit, $n));
            $x = (int)($x / 10);
        }
  
        // checks if number i is
        // equal to the sum of 
        // nth power of its digits
        if ($pow_sum == $i
            echo $i . " ";     
    }
}
  
// Driver code
$num1 = 100;
$num2 = 400;
findArmstrong($num1, $num2);
  
// This code is contributed by mits
?>

chevron_right



Output:

153 370 371

This article is contributed by Aditya Ranjan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : Mithun Kumar



Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.