Program for Armstrong Numbers

Given a number x, determine whether the given number is Armstrong’s number or not.

A positive integer of n digits is called an Armstrong number of order n (order is the 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

Naive Approach

The idea is to first count the number of digits (or find the order). 

Algorithm:

1. Let the number of digits be n. 
2. For every digit r in input number x, compute rⁿ. 
3. If the sum of all such values is equal to x, then return true, else false. 

Below is the program to check whether the number is an Armstrong number or not:

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 Code
int main()
{
    int x = 153;
    cout << boolalpha << isArmstrong(x) << endl;
    x = 1253;
    cout << boolalpha << 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 Code
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 Code
    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));
    }
}

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(153))

print(isArmstrong(1253))

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))

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.

Javascript

<script>

    // JavaScript program to determine whether the
    // number is Armstrong number or not

    // Function to calculate x raised
    // to the power y
    function power(x, y)
    {
        if( y == 0)
            return 1;
        if (y % 2 == 0)
            return power(x, parseInt(y / 2, 10)) *
                   power(x, parseInt(y / 2, 10));

        return x * power(x, parseInt(y / 2, 10)) *
                   power(x, parseInt(y / 2, 10));
    }

    // Function to calculate
    // order of the number
    function order(x)
    {
        let n = 0;
        while (x != 0)
        {
            n++;
            x = parseInt(x / 10, 10);
        }
        return n;
    }

    // Function to check whether the
    // given number is Armstrong number
    // or not
    function isArmstrong(x)
    {

        // Calling order function
        let n = order(x);
        let temp = x, sum = 0;
        while (temp != 0)
        {
            let r = temp % 10;
            sum = sum + power(r, n);
            temp = parseInt(temp / 10, 10);
        }

        // If satisfies Armstrong condition
        return (sum == x);
    }

    let x = 153;
    if(isArmstrong(x))
    {
        document.write("True" + "</br>");
    }
    else{
        document.write("False" + "</br>");
    }
    x = 1253;
    if(isArmstrong(x))
    {
        document.write("True");
    }
    else{
        document.write("False");
    }

</script>

Output

true
false

Optimized Approach

Below is the shorter way to implement the above approach:

C++

// C++ program to check whether the
// number is an Armstrong number or not
#include <iostream>
using namespace std;

// Driver Code
int main()
{
    int n = 153;
    int temp = n;
    int p = 0;

    // Function to calculate the sum of
    // individual digits
    while (n > 0) {
        int rem = n % 10;
        p = (p) + (rem * rem * rem);
        n = n / 10;
    }

    // Condition to check whether the value
    // of P equals to user input or not.
    if (temp == p) {
        cout << ("Yes. It is Armstrong No.");
    }
    else {
        cout << ("No. It is not an Armstrong No.");
    }
    return 0;
}
// This code is contributed by sathiyamoorthics19

C

// C program to check whether the given
// number is an Armstrong number or not
#include <stdio.h>

// Driver Code
int main()
{
    int n = 153;
    int temp = n;
    int p = 0;

    // Function to calculate the sum of
    // individual digits
    while (n > 0) {
        int rem = n % 10;
        p = (p) + (rem * rem * rem);
        n = n / 10;
    }

    // Condition to check whether the value
    // of P equals to user input or not.
    if (temp == p) {
        printf("Yes. It is Armstrong No.");
    }
    else {
        printf("No. It is not an Armstrong No.");
    }
    return 0;
}

// This code is contributed by sathiyamoorthics19

Java

// Java program to determine whether the
// number is Armstrong number or not
public class ArmstrongNumber {
    public static void main(String[] args)
    {
        int n = 153;
        int temp = n;
        int p = 0;

        // Function to calculate the sum of
        // individual digits
        while (n > 0) {
            int rem = n % 10;
            p = (p) + (rem * rem * rem);
            n = n / 10;
        }

        // Condition to check whether the value
        // of P equals to user input or not.
        if (temp == p) {
            System.out.println("Yes. It is Armstrong No.");
        }
        else {
            System.out.println(
                "No. It is not an Armstrong No.");
        }
    }
}

Python3

n = 153
temp = n
p = 0

# function to calculate
# the sum of individual digits
while (n > 0):
    rem = n % 10
    p = (p) + (rem * rem * rem)
    n = n // 10
if temp == p:
    print("armstrong")
else:
    print("not a armstrong number")

    # This code is contributed by ksrikanth0498.

C#

// C# program to determine whether the number is
// Armstrong number or not
using System;

public class ArmstrongNumber {
    public static void Main(string[] args)
    {

        int n = 153;
        int temp = n;
        int p = 0;

        /*function to calculate
          the sum of individual digits
         */
        while (n > 0) {

            int rem = n % 10;
            p = (p) + (rem * rem * rem);
            n = n / 10;
        }

        /* condition to check whether
           the value of P equals
           to user input or not. */
        if (temp == p) {
            Console.WriteLine("Yes. It is Armstrong No.");
        }
        else {
            Console.WriteLine(
                "No. It is not an Armstrong No.");
        }
    }
}

// This code is contributed by phasing17

Javascript

// JavaScript implementation of the approach

let n = 153;
let temp = n;
let p = 0;

/*function to calculate
the sum of individual digits */
while (n > 0) {

    let rem = n % 10;
    p = (p) + (rem * rem * rem);
    n = Math.floor(n / 10);
}

/* condition to check whether
   the value of P equals
   to user input or not. */
if (temp == p) {
    console.log("Yes. It is Armstrong No.");
}
else {
    console.log("No. It is not an Armstrong No.");
}

// This code is contributed by phasing17

Output

Yes. It is Armstrong No.

Time complexity: O(log n)

Auxiliary Space:O(1)

Find n^th Armstrong Number 

Input: 9
Output: 9

Input: 10
Output: 153

Below is the program to find the n^th Armstrong number:

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 Code
int main()
{
    int n = 12;
    cout << NthArmstrong(n);
    return 0;
}

// This Code is Contributed by 'jaingyayak'

C

// C Program to find
// Nth Armstrong Number
#include <limits.h>
#include <math.h>
#include <stdio.h>

// 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 Code
int main()
{
    int n = 12;
    printf("%d", NthArmstrong(n));
    return 0;
}

// This Code is Contributed by 'sathiyamoorthics19'

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.

Javascript

<script>

// Javascript program to find Nth Armstrong Number

// Function to find Nth Armstrong Number
function NthArmstrong(n)
{
    let count = 0;

    // Upper limit from integer
    for(let i = 1; i <= Number.MAX_VALUE; i++)
    {
        let num = i, rem, digit = 0, sum = 0;

        // Copy the value for num in num
        num = i;

        // Find total digits in num
        digit = parseInt(Math.log10(num), 10) + 1;

        // Calculate sum of power of digits
        while(num > 0)
        {
            rem = num % 10;
            sum = sum + Math.pow(rem, digit);
            num = parseInt(num / 10, 10);
        }

        // Check for Armstrong number
        if (i == sum)
            count++;
        if (count == n)
            return i;
    }
    return n;
}

// Driver code
let n = 12;

document.write(NthArmstrong(n));

// This code is contributed by rameshtravel07

</script>

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

371

Time complexity: O(log n)

Auxiliary Space: O(1)

Using Numeric Strings

When considering the number as a string we can access any digit we want and perform operations. Below is the program to implement the above approach:

C++

// C++ program to check whether the
// number is Armstrong number or not
#include <bits/stdc++.h>

using namespace std;

string armstrong(int n)
{
    string number = to_string(n);

    n = number.length();
    long long output = 0;
    for (char i : number)
        output = output + (long)pow((i - '0'), n);

    if (output == stoll(number))
        return ("True");
    else
        return ("False");
}

// Driver Code
int main()
{
    cout << armstrong(153) << endl;
    cout << armstrong(1253) << endl;
}

// This code is contributed by phasing17

Java

// Java program to check whether the
// number is Armstrong number or not
public class armstrongNumber {
    public void isArmstrong(String n)
    {
        char[] s = n.toCharArray();
        int size = s.length;
        int sum = 0;

        for (char num : s) {
            int temp = 1;
            int i
                = Integer.parseInt(Character.toString(num));

            for (int j = 0; j <= size - 1; j++) {
                temp *= i;
            }
            sum += temp;
        }

        if (sum == Integer.parseInt(n)) {
            System.out.println("True");
        }
        else {
            System.out.println("False");
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        armstrongNumber am = new armstrongNumber();
        am.isArmstrong("153");
        am.isArmstrong("1253");
    }
}

Python3

def armstrong(n):
    number = str(n)

    n = len(number)
    output = 0
    for i in number:
        output = output+int(i)**n

    if output == int(number):
        return(True)
    else:
        return(False)


print(armstrong(153))
print(armstrong(1253))

C#

using System;
public class armstrongNumber {
    public void isArmstrong(String n)
    {
        char[] s = n.ToCharArray();
        int size = s.Length;
        int sum = 0;
        foreach(char num in s)
        {
            int temp = 1;
            int i = Int32.Parse(char.ToString(num));
            for (int j = 0; j <= size - 1; j++) {
                temp *= i;
            }
            sum += temp;
        }
        if (sum == Int32.Parse(n)) {
            Console.WriteLine("True");
        }
        else {
            Console.WriteLine("False");
        }
    }

    public static void Main(String[] args)
    {
        armstrongNumber am = new armstrongNumber();
        am.isArmstrong("153");
        am.isArmstrong("1253");
    }
}

// This code is contributed by umadevi9616

Javascript

<script>
function armstrong(n){
    let number = new String(n)

    n = number.length
    let output = 0
    for(let i of number)
      output = output + parseInt(i)**n

    if (output == parseInt(number))
        return("True" + "<br>")
    else
        return("False" + "<br>")
}

document.write(armstrong(153))
document.write(armstrong(1253))

// This code is contributed by _saurabh_jaiswal.
</script>

Output

True
False

Time Complexity: O(n).
Auxiliary Space: O(1).

Find all Armstrong Numbers in a Range

Below is the program to find all Armstrong numbers in a given range:

C++

// C++ program to find all Armstrong numbers
// in a given range
#include <bits/stdc++.h>
using namespace std;

void isArmstrong(int left, int right)
{
    for (int i = left; i <= right; i++) {
        int sum = 0;
        int temp = i;
        while (temp > 0) {
            // Finding the lastdigit
            int lastdigit = temp % 10;

            // Finding the sum
            sum += pow(lastdigit, 3);
            temp /= 10;
        }

        // Condition to print the number if
        // it is armstrong number
        if (sum == i) {
            cout << i << " ";
        }
    }
    cout << endl;
}

// Driver code
int main()
{
    int left = 5, right = 1000;
    isArmstrong(left, right);
    return 0;
}

// This code is contributed by 525tamannacse11

Java

// Java program to find all Armstrong numbers
// in a given range

// Importing necessary libraries
import java.util.*;

public class Main {
    public static void isArmstrong(int left, int right)
    {
        for (int i = left; i <= right; i++) {
            int sum = 0;
            int temp = i;
            while (temp > 0) {
                // Finding the last digit
                int lastdigit = temp % 10;

                // Finding the sum
                sum += Math.pow(lastdigit, 3);
                temp /= 10;
            }

            // Condition to print the number if it
            // is an Armstrong number
            if (sum == i) {
                System.out.print(i + " ");
            }
        }
        System.out.println();
    }

    // Driver code
    public static void main(String[] args)
    {
        int left = 5, right = 1000;
        isArmstrong(left, right);
    }
}

Python3

def armstrong(n):
    number = str(n)

    n = len(number)
    output = 0
    for i in number:
        output = output+int(i)**n

    if output == int(number):
        return(True)
    else:
        return(False)


arm_list = []

nums = range(10, 1000)

for i in nums:
    if armstrong(i):
        arm_list.append(i)
    else:
        pass

print(arm_list)

C#

using System;

class MainClass {
    static void isArmstrong(int left, int right)
    {
        for (int i = left; i <= right; i++) {
            int sum = 0;
            int temp = i;
            while (temp > 0) {
                // finding the lastdigit
                int lastdigit = temp % 10;
                // finding the sum
                sum += (int)Math.Pow(lastdigit, 3);
                temp /= 10;
            }
            /* Condition to print the number if it is
             * armstrong number */
            if (sum == i) {
                Console.Write(i + " ");
            }
        }
        Console.WriteLine();
    }

    public static void Main(string[] args)
    {
        int left = 5, right = 1000;
        isArmstrong(left, right);
    }
}

Javascript

// JavaScript code to implement the approach
function isArmstrong(left, right) {
    for (let i = left; i <= right; i++) {
        let sum = 0;
        let temp = i;
        while (temp > 0) {
            // finding the lastdigit
            let lastdigit = temp % 10;
            // finding the sum
            sum += Math.pow(lastdigit, 3);
            temp = Math.floor(temp / 10);
        }
        /* Condition to print the number if it is armstrong number */
        if (sum === i) {
            process.stdout.write(i + " ");
        }
    }
    console.log("\n");
}

// Driver code

let left = 5,
    right = 1000;
isArmstrong(left, right);

// This code is contributed by phasing17

Output

153 370 371 407 

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

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