C Program to check a Number is Armstrong or not

June 29, 2023

Program to check a number is Armstrong or not

In this program, we will find the number is Armstrong or not where the number should be entered by the user, we will look at different ways to write Armstrong number in C.

Most websites have given incorrect definitions/codes for Armstrong of a number.

Number-is-Armstrong-or-not

Very Important

For Armstrong number in C on most websites, Armstrong number definition is given wrong. They all assume any number where the sum of the cubes of its digits is equal to the number itself is Armstrong Number.

However, Armstrong number is any number following the given rule –

  • abcd… = an + bn + cn + dn + …
  • Where n is the order(length/digits in number)

Also, Check Armstrong Number in a given Range in C

Armstrong Number in C
While loop in C

C Code:-

Run
#include <stdio.h>
#include <math.h>
 int order(int x)
 {
  int len = 0;
  while (x)
  {
    len++;
    x = x/10;
  }
  return len;
 }
  int getArmstrongSum(int num, int order){

  if(num == 0)
  return 0;

  int digit = num % 10;

  return pow(digit, order) + getArmstrongSum(num/10, order);
 }


// Driver Code
int main ()
{
  int num, len;
  num=1634;
  printf("The number is:%d\n",num); 

  // function to get order(length)
  len = order(num);

  // check if Armstrong
  if (num == getArmstrongSum(num, len))
    printf("%d is Armstrong", num);
  else
    printf("%d is Not Armstrong", num);
 
 }

 

Output:-

The number is: 1634
1634 is Armstrong

 

Method 2

This method uses recursion in C

Code

Run 
#include <stdio.h>
#include <math.h>


// Armstrong number is any number following the given rule
// abcd... = a^n + b^n + c^n + d^n + ...
// Where n is the order(length/digits in number)

// Example = 153 (order/length = 3)
// 153 = 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153

// Example = 1634 (order/length = 4)
// 1634 = 1^4 + 6^4 + 3^4 + 4^4 = 1 + 1296 + 81 + 256 = 1634

// number of digits in a number is order
int order(int x)
{
    int len = 0;
    while (x)
    {
        len++;
        x = x/10;
    }
    return len;
}

int armstrong(int num, int len){

    int sum = 0, temp, digit;
    temp = num;
    
    // loop to extract digit, find power & add to sum
    while(temp != 0)
    {
        // extract digit
        digit = temp % 10;

        // add power to sum
        sum = sum + pow(digit,len);
        temp /= 10;
    };

    return num == sum;
}

// Driver Code
int main ()
{
    //variables initialization
    int num = 1634, len;
 
    // function to get order(length)
    len = order(num);
    
    // check if Armstrong
    if (armstrong(num, len))
        printf("%d is armstrong\n",num);
    else
         printf("%d is not armstrong\n",num);


    return 0;
}

Output

The number is: 1634
1634 is Armstrong

Program to reverse a number

Program to check palindrome or not

Armstrong number in a given range

Fibonacci series up to n numbers

Nth term of fibonacci series

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Related Pages

While loop in C

6 comments on “C Program to check a Number is Armstrong or not”


  • ngali

    #include
    #include
    int main()
    {
    int n,count=0,rem,num=0;
    scanf(“%d”,&n);
    int temp=n;
    int k=n;

    while(n!=0)
    {
    n=n/10;
    count+=1;
    }

    while(temp!=0)
    {
    rem = temp%10;
    num+= pow(rem,count);
    temp = temp/10;
    }

    if(k==num)
    {
    printf(“%d is an Armstrong number”,num);
    }
    else
    {
    printf(“%d is not a Armstrong number”,num);
    }
    }


  • NAINARU SIVARAMAKRISHNA

    #include
    int main()
    {
    int num=153,sum=0,rem=0;
    int val=num;
    while(num>0){
    rem=num%10;
    rem=pow(rem,3);
    sum=sum+rem;
    num/=10;
    }
    if(val==sum){
    printf(“armstrong number”);
    }
    else{
    printf(“not armstrong number”);
    }
    }


  • SOUMYADIP

    #include
    #include

    int main()
    {
    int n,sum=0,rev=0;
    scanf(“%d”,&n);
    int temp=n;
    int x=n;
    int count=0;
    while(n!=0){
    int r=n%10;
    rev=rev*10+r;
    n=n/10;
    count++;
    }
    printf(“%d”,count);

    int y=0;
    int z=0;
    while(temp!=0){

    z=temp%10;
    y=y+(pow(z, count));
    //printf(“%d\n”,y);
    temp=temp/10;
    }
    printf(“\n%d”,y);
    if(x==y){
    printf(“\nnum is a Armstrong”);
    }
    else{
    printf(“\nnum is not a Armstrong”);
    }

    return 0;
    }


  • Arnab

    #include
    #include
    int main()
    {
    int num,i=0,num1,num2,sum=0,rem;
    printf(“Enter Number\n”);
    scanf(“%d”,&num);
    if(num>0 && num<10)
    printf("Armstrong Number");
    else
    {
    num1=num;
    while(num1!=0)
    {
    num1=num1/10;
    i++;
    }
    num2=num;
    while(num2!=0)
    {
    rem=num2%10;
    num2=num2/10;
    sum=sum+pow(rem,i);
    }
    if(sum==num)
    printf("Armstrong Number");
    else
    printf("Not Armstrong Number");
    }
    return 0;
    }


  • Sujay

    simple version
    #include
    int main() {
    int num, originalNum, remainder, result = 0;
    printf(“Enter a three-digit integer: “);
    scanf(“%d”, &num);
    originalNum = num;

    while (originalNum != 0) {
    // remainder contains the last digit
    remainder = originalNum % 10;

    result += remainder * remainder * remainder;

    // removing last digit from the orignal number
    originalNum /= 10;
    }

    if (result == num)
    printf(“%d is an Armstrong number.”, num);
    else
    printf(“%d is not an Armstrong number.”, num);

    return 0;
    }