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 = 153Input: 120
Output: No
120 is not a Armstrong number.
1*1*1 + 2*2*2 + 0*0*0 = 9Input: 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 = 723Input: 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:
- Let the number of digits be n.
- For every digit r in input number x, compute rn.
- 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 notbool 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 Codeint 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 notint 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 Codeint 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 notpublic 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 manipulationdef 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 numberprint(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 ydef 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 numberdef 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 notdef 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 Programx = 153print(isArmstrong(x))x = 1253print(isArmstrong(x)) |
C#
// C# program to determine whether the// number is Armstrong number or notusing 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 Codeint 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 Codeint 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 notpublic 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 = 153temp = np = 0# function to calculate# the sum of individual digitswhile (n > 0): rem = n % 10 p = (p) + (rem * rem * rem) n = n // 10if 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 notusing 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 approachlet n = 153;let temp = n;let p = 0;/*function to calculatethe 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 nth Armstrong Number
Input: 9
Output: 9Input: 10
Output: 153
Below is the program to find the nth 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 Numberint 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 Codeint 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 Numberint 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 Codeint main(){ int n = 12; printf("%d", NthArmstrong(n)); return 0;}// This Code is Contributed by 'sathiyamoorthics19' |
Java
// Java Program to find// Nth Armstrong Numberimport 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 Numberimport mathimport sys# Function to find Nth Armstrong Numberdef 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 Coden = 12print(NthArmstrong(n))# This code is contributed by chandan_jnu |
C#
// C# Program to find// Nth Armstrong Numberusing 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 Numberfunction 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 codelet 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 Numberfunction 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 Codeint 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 notpublic 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 codeint 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 librariesimport 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: passprint(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 approachfunction 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 codelet 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
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