# Armstrong number c program

Armstrong number c program: c programming code to check whether a number is Armstrong or not. Armstrong number is a number which is equal to sum of digits raise to the power total number of digits in the number. Some Armstrong numbers are: 0, 1, 2, 3, 153, 370, 407, 1634, 8208 etc. Read more about Armstrong numbers at Wikipedia. We will consider base 10 numbers in our program. Algorithm to check Armstrong is: First we calculate number of digits in our program and then compute sum of individual digits raise to the power number of digits. If this sum equals input number then number is Armstrong otherwise not an Armstrong number.

Examples:
7 = 7^1
371 = 3^3 + 7^3 + 1^3 (27 + 343 +1)
8208 = 8^4 + 2^4 +0^4 + 8^4 (4096 + 16 + 0 + 4096).
1741725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 +5^7 (1 + 823543 + 16384 + 1 + 823543 +128 + 78125)

## C programming code

```#include <stdio.h>

int power(int, int);

int main()
{
int n, sum = 0, temp, remainder, digits = 0;

printf("Input an integer\n");
scanf("%d", &n);

temp = n;
// Count number of digits
while (temp != 0) {
digits++;
temp = temp/10;
}

temp = n;

while (temp != 0) {
remainder = temp%10;
sum = sum + power(remainder, digits);
temp = temp/10;
}

if (n == sum)
printf("%d is an Armstrong number.\n", n);
else
printf("%d is not an Armstrong number.\n", n);

return 0;
}

int power(int n, int r) {
int c, p = 1;

for (c = 1; c <= r; c++)
p = p*n;

return p;
}```

Output of program:

## C program to check Armstrong number using function

We will use long long data type in our program so that we can check numbers up to 2^64-1.

```#include <stdio.h>

int check_armstrong(long long);
long long power(int, int);

int main () {
long long n;

printf("Input a number\n");
scanf("%lld", &n);

if (check_armstrong(n) == 1)
printf("%lld is an armstrong number.\n", n);
else
printf("%lld is not an armstrong number.\n", n);

return 0;
}

int check_armstrong(long long n) {
long long sum = 0, temp;
int remainder, digits = 0;

temp = n;

while (temp != 0) {
digits++;
temp = temp/10;
}

temp = n;

while (temp != 0) {
remainder = temp%10;
sum = sum + power(remainder, digits);
temp = temp/10;
}

if (n == sum)
return 1;
else
return 0;
}

long long power(int n, int r) {
int c;
long long p = 1;

for (c = 1; c <= r; c++)
p = p*n;

return p;
}```

Output of program:

```Input a number
35641594208964132
35641594208964132 is an Armstrong number.```