C++ Examples

C++ Program - Check Armstrong Number

A positive natural number is known as Armstrong number of order n if it can be expressed as the sum of each digits of the number raised to the power of n. Mathematically, it can be expressed as:

Examples:

153 = 1³+5³+3³ = 1+125+27 = 153

371 = 3³+7³+1³ = 27+343+1 = 371

1634 = 1⁴+6⁴+3⁴+4⁴ = 1+1296+81+256 = 1634

Method 1: Check Armstrong Number

In the example below, the MyNum is checked for Armstrong number using function called ArmStrongNum(). The function requires two parameters, first the number and second the number of digits in it. It calculates power of the digit using the Pow() function. See the example below for syntax:

#include <iostream>
using namespace std;

static int Pow(int, int);
static void ArmStrongNum(int);

//Calculate power of a digit
static int Pow(int MyNum, int n) {
  int x = 1;
  while(n > 0) {
    x = x*MyNum;
    n--;
  }
  return x;
}

static void ArmStrongNum(int MyNum, int Order) {
  int y = MyNum;
  int sum = 0;

  while (y > 0) {
    int x = y % 10;
    sum = sum + Pow(x, Order);
    y = y/10;
  }

  if (MyNum == sum){
    cout<<MyNum<<" is a Armstrong Number.\n";
  } else {
    cout<<MyNum<<" is not a Armstrong Number.\n";
  }
}

int main() {
  ArmStrongNum(371, 3);
  ArmStrongNum(1634, 4);
  ArmStrongNum(1000, 4);
  return 0;
}

The above code will give the following output:

371 is a Armstrong Number.
1634 is a Armstrong Number.
1000 is not a Armstrong Number.

Method 2: Armstrong Number of order n

In this example, ArmStrongNum function requires only one parameter, the number itself. The number of digits in the passed parameter is estimated inside the function.

#include <iostream>
using namespace std;

static int Pow(int, int);
static void ArmStrongNum(int);

//Calculate power of a digit
static int Pow(int MyNum, int n) {
  int x = 1;
  while(n > 0) {
    x = x*MyNum;
    n--;
  }
  return x;
}

static void ArmStrongNum(int MyNum) {
  int y = MyNum;
  int sum = 0;
  int Order = 0;
  //Find number of digit in the Number
  while(y > 0) {
    Order++;
    y = y / 10;
  }

  y = MyNum;
  while (y > 0) {
    int x = y % 10;
    sum = sum + Pow(x, Order);
    y = y/10;
  }
  
  if (MyNum == sum) {
    cout<<MyNum<<" is a Armstrong Number.\n";
  } else {
    cout<<MyNum<<" is not a Armstrong Number.\n";
  }
}

int main() {
  ArmStrongNum(153);
  ArmStrongNum(9474);
  ArmStrongNum(5000);
  return 0;
}

The above code will give the following output:

153 is a Armstrong Number.
9474 is a Armstrong Number.
5000 is not a Armstrong Number.

C++ Program - Check Armstrong Number

Examples:

Method 1: Check Armstrong Number

Method 2: Armstrong Number of order n

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