C Program to Check Armstrong Number

In this tutorial, you will learn and get code about checking whether the given number by user (at run-time) is an Armstrong number or not. But before going through the program. Let's first understand about an Armstrong number.

How to Check for Armstrong Number ?

For a number to be an Armstrong number, the sum of cubes of all of its digits must be equal to the number itself. For example, 153 is an Armstrong number. Because

  • 153 = (1*1*1) + (5*5*5) + (3*3*3)

As you can clearly see that, the sum of cubes of all of its digit (1, 5, and 3) is equal to the number itself. So it is an Armstrong number. But the number 26 is not an Armstrong number. Because

  • 26 != (2*2*2) + (6*6*6)

Now let's move on and implement it in a C program.

Check Armstrong or Not in C

The question is, Write a program in C to check whether the given number is an Armstrong number or not. The answer to this question is:

#include<stdio.h>
#include<conio.h>
int main()
{
    int n, nu, num=0, rem;
    printf("Enter any positive number: ");
    scanf("%d", &n);
    nu=n;
    while(nu!=0)
    {
        rem = nu%10;
        num = num + (rem*rem*rem);
        nu = nu/10;
    }
    if(num==n)
        printf("\nIt's an Armstrong Number");
    else
        printf("\nIt's not an Armstrong Number");
    getch();
    return 0;
}

The above program was written under Code::Blocks IDE, therefore after successful build and run, here is the output you will get on your output screen:

c program to check number is armstrong or not

Now supply any positive number say 153 and press ENTER key to see the output that will say whether it is an Armstrong number or not as shown in the following snapshot:

c program check armstrong or not

Program Explained

  • Get any positive number as input from user say 153
  • Now initialize the number to another variable say nu for the operation
  • Include a while loop to work with each and every digit of the given number say 153
  • We have to find the remainder of the number one by one, the first remainder we will get is 3
  • Initialize 3 to rem variable and calculate cube of the rem and initialize it to num variable after summing it with the variable itself
  • We have initialize num variable with 0 at starting of the program
  • Now divide the number nu with 10
  • And check whether the value inside nu is not equal to 0
  • If condition gets true, then again come inside the while loop
  • And do the same operation until nu holds the value 0
  • Therefore at first run of the while loop, rem holds 3, num holds 0 + 3*3*3 or 27 and nu holds 15. And at second run of while loop, rem holds 5, num holds 27 + 5*5*5 or 152, and nu holds 1. And at third run, rem holds 1, num holds 152 + 1*1*1 or 153, and nu holds 0
  • At last check whether the value of num is equal to the given number (the value present inside the variable n) or not
  • If it is equal, then the number is an Armstrong number, otherwise the number is not an Armstrong number

Here is another program in C, that prints all Armstrong number between any two given three-digit number. To generate or print Armstrong numbers without caring about digit, then refer to Generate Armstrong Numbers

#include<stdio.h>
#include<conio.h>
int main()
{
    int n1, n2, i, temp, rem, sum, prod;
    printf("Enter the value of n1 (starting three-digit number): ");
    scanf("%d", &n1);
    printf("Enter the value of n2 (ending three-digit number): ");
    scanf("%d", &n2);
    printf("\n");
    for(i=n1; i<=n2; i++)
    {
        sum = 0;
        temp = i;
        while(temp>0)
        {
            rem = temp%10;
            sum = sum + (rem*rem*rem);
            temp = temp/10;
        }
        if(sum == i)
            printf("%d is an Armstrong number.\n", i);
    }
    getch();
    return 0;
}

Here is the sample run:

c program print all armstrong number

Now provide any three-digit number as starting point, say 100 and another three-digit number as ending point say 999 to print all the Armstrong numbers present in between these two numbers. Here is the second snapshot of the sample run:

find all armstrong number c

Logic used in this program is almost same as used in the above one, except that we have used for loop here to start it with n1 (starting number) and ends with n2 (ending number). In between these two numbers, we have used the same concept inside the for loop, that is if the number is found as an Armstrong number, then print it out, otherwise get to the next one to check for Armstrong.

Before check, initialize 0 to sum, and i to temp variable for the operation. Here we have used sum variable in place of num variable. You can choose something different, it is up to you.

Same Program in Other Languages

C Online Test


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