C Exercises: Check whether an n digits number is Armstrong or not
C For Loop: Exercise-59 with Solution
Write a C program to check the Armstrong number of n digits.
This C program checks whether a given n-digit number is an Armstrong number or not. It prompts the user for an integer input, calculates the number of digits in the input number, and then computes the sum of cubes of individual digits raised to the power of 'n'. Finally, it compares this result with the original number to determine whether it's an Armstrong number or not, printing the result accordingly.
Visual Presentation:
Sample Solution:
C Code:
#include <stdio.h> // Include the standard input/output header file.
#include <math.h> // Include the math header file.
int main()
{
// Variable declarations
int n1, onum, r, result = 0, n = 0 ;
// Prompting user for input
printf("\n\n Check whether an n digits number is armstrong or not :\n");
printf("-----------------------------------------------------------\n");
printf(" Input an integer : ");
scanf("%d", &n1);
// Store the original number for later comparison
onum = n1;
// Count the number of digits in the input number
while (onum != 0)
{
onum /= 10;
++n;
}
// Reset onum to the original number for further processing
onum = n1;
// Calculate the sum of cubes of individual digits raised to the power of 'n'
while (onum != 0)
{
r = onum % 10;
result += pow(r, n);
onum /= 10;
}
// Check if the result is equal to the original number
if(result == n1)
printf(" %d is an Armstrong number.\n\n", n1);
else
printf(" %d is not an Armstrong number.\n\n", n1);
return 0;
}
Output:
Check whether an n digits number is Armstrong or not : ----------------------------------------------------------- Input an integer : 1634 1634 is an Armstrong number.
Flowchart:
C Programming Code Editor:
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