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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // CPP program to find Armstrong numbers in a range #include 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; }

Java

 // 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.*/

Python

 # 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.

C#

 // 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.*/

PHP



Output:

153 370 371

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