Program for Armstrong Numbers

Difficulty Level : Easy
Last Updated : 09 Nov, 2020

Given a number x, determine whether the given number is Armstrong number or not. A positive integer of n digits is called an Armstrong number of order n (order is number of digits) if.

abcd... = pow(a,n) + pow(b,n) + pow(c,n) + pow(d,n) + ....

Example:

Input : 153
Output : Yes
153 is an Armstrong number.
1*1*1 + 5*5*5 + 3*3*3 = 153

Input : 120
Output : No
120 is not a Armstrong number.
1*1*1 + 2*2*2 + 0*0*0 = 9

Input : 1253
Output : No
1253 is not a Armstrong Number
1*1*1*1 + 2*2*2*2 + 5*5*5*5 + 3*3*3*3 = 723

Input : 1634
Output : Yes
1*1*1*1 + 6*6*6*6 + 3*3*3*3 + 4*4*4*4 = 1634

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is to first count number digits (or find order). Let the number of digits be n. For every digit r in input number x, compute rⁿ. If sum of all such values is equal to n, then return true, else false.

C++

// C++ program to determine whether the number is 
// Armstrong number or not 
#include<bits/stdc++.h> 
using namespace std; 
  
/* Function to calculate x raised to the power y */
int power(int x, unsigned int y) 
{ 
    if( y == 0) 
        return 1; 
    if (y%2 == 0) 
        return power(x, y/2)*power(x, y/2); 
    return x*power(x, y/2)*power(x, y/2); 
} 
  
/* Function to calculate order of the number */
int order(int x) 
{ 
    int n = 0; 
    while (x) 
    { 
        n++; 
        x = x/10; 
    } 
    return n; 
} 
  
// Function to check whether the given number is 
// Armstrong number or not 
bool isArmstrong(int x) 
{ 
    // Calling order function 
    int n = order(x); 
    int temp = x, sum = 0; 
    while (temp) 
    { 
        int r = temp%10; 
        sum += power(r, n); 
        temp = temp/10; 
    } 
  
    // If satisfies Armstrong condition 
    return (sum == x); 
} 
  
// Driver Program 
int main() 
{ 
    int x = 153; 
    cout << isArmstrong(x) << endl; 
    x = 1253; 
    cout << isArmstrong(x) << endl; 
    return 0; 
} 

C

// C program to find Armstrong number 
  
#include <stdio.h> 
  
/* Function to calculate x raised to the power y */
int power(int x, unsigned int y) 
{ 
    if (y == 0) 
        return 1; 
    if (y % 2 == 0) 
        return power(x, y / 2) * power(x, y / 2); 
    return x * power(x, y / 2) * power(x, y / 2); 
} 
  
/* Function to calculate order of the number */
int order(int x) 
{ 
    int n = 0; 
    while (x) { 
        n++; 
        x = x / 10; 
    } 
    return n; 
} 
  
// Function to check whether the given number is 
// Armstrong number or not 
int isArmstrong(int x) 
{ 
    // Calling order function 
    int n = order(x); 
    int temp = x, sum = 0; 
    while (temp) { 
        int r = temp % 10; 
        sum += power(r, n); 
        temp = temp / 10; 
    } 
  
    // If satisfies Armstrong condition 
    if (sum == x) 
        return 1; 
    else
        return 0; 
} 
  
// Driver Program 
int main() 
{ 
    int x = 153; 
    if (isArmstrong(x) == 1) 
        printf("True\n"); 
    else
        printf("False\n"); 
  
    x = 1253; 
    if (isArmstrong(x) == 1) 
        printf("True\n"); 
    else
        printf("False\n"); 
  
    return 0; 
} 

Java

// Java program to determine whether the number is 
// Armstrong number or not 
public class Armstrong 
{ 
    /* Function to calculate x raised to the 
       power y */
    int power(int x, long y) 
    { 
        if( y == 0) 
            return 1; 
        if (y%2 == 0) 
            return power(x, y/2)*power(x, y/2); 
        return x*power(x, y/2)*power(x, y/2); 
    } 
  
    /* Function to calculate order of the number */
    int order(int x) 
    { 
        int n = 0; 
        while (x != 0) 
        { 
            n++; 
            x = x/10; 
        } 
        return n; 
    } 
  
    // Function to check whether the given number is 
    // Armstrong number or not 
    boolean isArmstrong (int x) 
    { 
        // Calling order function 
        int n = order(x); 
        int temp=x, sum=0; 
        while (temp!=0) 
        { 
            int r = temp%10; 
            sum = sum + power(r,n); 
            temp = temp/10; 
        } 
  
        // If satisfies Armstrong condition 
        return (sum == x); 
    } 
  
    // Driver Program 
    public static void main(String[] args) 
    { 
        Armstrong ob = new Armstrong(); 
        int x = 153; 
        System.out.println(ob.isArmstrong(x)); 
        x = 1253; 
        System.out.println(ob.isArmstrong(x)); 
    } 
} 

Python

# Python program to determine whether the number is 
# Armstrong number or not 
  
# Function to calculate x raised to the power y 
def power(x, y): 
    if y==0: 
        return 1
    if y%2==0: 
        return power(x, y/2)*power(x, y/2) 
    return x*power(x, y/2)*power(x, y/2) 
  
# Function to calculate order of the number 
def order(x): 
  
    # variable to store of the number 
    n = 0
    while (x!=0): 
        n = n+1
        x = x/10
    return n 
  
# Function to check whether the given number is 
# Armstrong number or not 
def isArmstrong (x): 
    n = order(x) 
    temp = x 
    sum1 = 0
    while (temp!=0): 
        r = temp%10
        sum1 = sum1 + power(r, n) 
        temp = temp/10
  
    # If condition satisfies 
    return (sum1 == x) 
  
  
# Driver Program 
x = 153
print(isArmstrong(x)) 
x = 1253
print(isArmstrong(x)) 

Output:

1
0

Python3

# python3 >= 3.6 for typehint support  
# This example avoids the complexity of ordering 
# through type conversions & string manipulation  
  
def isArmstrong(val:int) -> bool: 
      
    """val will be tested to see if its an Armstrong number.  
    Arguments: 
        val {int} -- positive integer only.  
    Returns: 
        bool -- true is /false isn't 
    """
      
    # break the int into its respective digits 
    parts = [int(_) for _ in str(val)]  
      
    # begin test. 
    counter = 0
    for _ in parts: 
        counter += _**3
    return ( counter == val )  
  
# Check Armstrong number 
print(isArmstrong(100)) 
  
print(isArmstrong(153)) 
  
# Get all the Armstrong number in range(1000) 
print([ _ for _ in range(1000) if isArmstrong(_)]) 

C#

// C# program to determine whether the  
// number is Armstrong number or not 
using System; 
  
public class GFG 
{ 
      
    // Function to calculate x raised 
    // to the power y 
    int power(int x, long y) 
    { 
        if( y == 0) 
            return 1; 
        if (y % 2 == 0) 
            return power(x, y / 2) *  
                   power(x, y / 2); 
                     
        return x * power(x, y / 2) * 
                   power(x, y / 2); 
    } 
  
    // Function to calculate  
    // order of the number 
    int order(int x) 
    { 
        int n = 0; 
        while (x != 0) 
        { 
            n++; 
            x = x / 10; 
        } 
        return n; 
    } 
  
    // Function to check whether the  
    // given number is Armstrong number 
    // or not 
    bool isArmstrong (int x) 
    { 
          
        // Calling order function 
        int n = order(x); 
        int temp = x, sum = 0; 
        while (temp != 0) 
        { 
            int r = temp % 10; 
            sum = sum + power(r, n); 
            temp = temp / 10; 
        } 
  
        // If satisfies Armstrong condition 
        return (sum == x); 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        GFG ob = new GFG(); 
        int x = 153; 
        Console.WriteLine(ob.isArmstrong(x)); 
        x = 1253; 
        Console.WriteLine(ob.isArmstrong(x)); 
    } 
} 
  
// This code is contributed by Nitin Mittal. 

Output:

True
False

Find n^th Armstrong number

Input  : 9
Output : 9

Input  : 10
Output : 153

C++

// C++ Program to find  
// Nth Armstrong Number 
#include<bits/stdc++.h> 
#include<math.h> 
using namespace std; 
  
// Function to find Nth Armstrong Number 
int NthArmstrong(int n) 
{ 
    int count=0; 
      
    // upper limit from integer  
    for(int i = 1; i <= INT_MAX; i++) 
    { 
        int num=i, rem, digit=0, sum=0; 
          
        //Copy the value for num in num  
        num = i; 
          
        // Find total digits in num  
        digit = (int) log10(num) + 1; 
          
        // Calculate sum of power of digits  
        while(num > 0) 
        {  
            rem = num % 10; 
            sum = sum + pow(rem,digit); 
            num = num / 10; 
        } 
        // Check for Armstrong number  
        if(i == sum) 
            count++; 
        if(count==n) 
            return i; 
    } 
} 
  
// Driver Function 
int main() 
{ 
    int n = 12; 
    cout<<NthArmstrong(n); 
    return 0; 
} 
  
  
// This Code is Contributed by 'jaingyayak' 

Java

// Java Program to find  
// Nth Armstrong Number 
import java.lang.Math; 
  
class GFG 
{ 
      
// Function to find Nth Armstrong Number 
static int NthArmstrong(int n) 
{ 
    int count = 0; 
      
    // upper limit from integer  
    for(int i = 1; i <= Integer.MAX_VALUE; i++) 
    { 
        int num = i, rem, digit = 0, sum = 0; 
          
        //Copy the value for num in num  
        num = i; 
          
        // Find total digits in num  
        digit = (int) Math.log10(num) + 1; 
          
        // Calculate sum of power of digits  
        while(num > 0) 
        {  
            rem = num % 10; 
            sum = sum + (int)Math.pow(rem, digit); 
            num = num / 10; 
        } 
          
        // Check for Armstrong number  
        if(i == sum) 
            count++; 
        if(count == n) 
            return i; 
    } 
    return n; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int n = 12; 
    System.out.println(NthArmstrong(n)); 
} 
} 
  
  
// This code is contributed by Code_Mech. 

Python3

# Python3 Program to find Nth Armstrong Number 
import math; 
import sys; 
  
# Function to find Nth Armstrong Number 
def NthArmstrong(n): 
  
    count = 0; 
      
    # upper limit from integer  
    for i in range(1, sys.maxsize): 
  
        num = i;  
        rem = 0; 
        digit = 0; 
        sum = 0; 
          
        # Copy the value for num in num  
        num = i; 
          
        # Find total digits in num  
        digit = int(math.log10(num) + 1); 
          
        # Calculate sum of power of digits  
        while(num > 0): 
            rem = num % 10; 
            sum = sum + pow(rem, digit); 
            num = num // 10; 
          
        # Check for Armstrong number  
        if(i == sum): 
            count += 1; 
        if(count == n): 
            return i; 
  
# Driver Code 
n = 12; 
print(NthArmstrong(n)); 
  
# This code is contributed by chandan_jnu 

C#

// C# Program to find  
// Nth Armstrong Number 
using System; 
  
class GFG 
{ 
      
// Function to find Nth Armstrong Number 
static int NthArmstrong(int n) 
{ 
    int count = 0; 
      
    // upper limit from integer  
    for(int i = 1; i <= int.MaxValue; i++) 
    { 
        int num = i, rem, digit = 0, sum = 0; 
          
        // Copy the value for num in num  
        num = i; 
          
        // Find total digits in num  
        digit = (int) Math.Log10(num) + 1; 
          
        // Calculate sum of power of digits  
        while(num > 0) 
        {  
            rem = num % 10; 
            sum = sum + (int)Math.Pow(rem, digit); 
            num = num / 10; 
        } 
          
        // Check for Armstrong number  
        if(i == sum) 
            count++; 
        if(count == n) 
            return i; 
    } 
    return n; 
} 
  
// Driver Code 
public static void Main() 
{ 
    int n = 12; 
    Console.WriteLine(NthArmstrong(n)); 
} 
} 
  
// This code is contributed by Code_Mech. 

PHP

<?php 
// PHP Program to find  
// Nth Armstrong Number 
  
// Function to find 
// Nth Armstrong Number 
function NthArmstrong($n) 
{ 
    $count = 0; 
      
    // upper limit 
    // from integer  
    for($i = 1;  
        $i <= PHP_INT_MAX; $i++) 
    { 
        $num = $i;  
        $rem; 
        $digit = 0; 
        $sum = 0; 
          
        // Copy the value  
        // for num in num  
        $num = $i; 
          
        // Find total  
        // digits in num  
        $digit = (int) log10($num) + 1; 
          
        // Calculate sum of 
        // power of digits  
        while($num > 0) 
        {  
            $rem = $num % 10; 
            $sum = $sum + pow($rem,  
                              $digit); 
            $num = (int)$num / 10; 
        } 
          
        // Check for  
        // Armstrong number  
        if($i == $sum) 
            $count++; 
        if($count == $n) 
            return $i; 
    } 
} 
  
// Driver Code 
$n = 12; 
echo NthArmstrong($n); 
  
// This Code is Contributed  
// by akt_mit 
?> 

Output:

References:
http://www.cs.mtu.edu/~shene/COURSES/cs201/NOTES/chap04/arms.html
http://www.programiz.com/c-programming/examples/check-armstrong-number

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Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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PHP