C Program to Check Armstrong Number

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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

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

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:

int main()
    int n, nu, num=0, rem;
    printf("Enter any positive number: ");
    scanf("%d", &n);
        rem = nu%10;
        num = num + (rem*rem*rem);
        nu = nu/10;
        printf("\nIt's an Armstrong Number");
        printf("\nIt's not an Armstrong Number");
    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

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

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);
    for(i=n1; i<=n2; i++)
        sum = 0;
        temp = i;
            rem = temp%10;
            sum = sum + (rem*rem*rem);
            temp = temp/10;
        if(sum == i)
            printf("%d is an Armstrong number.\n", i);
    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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