# C program to find Armstrong numbers between 1 to n

Write a C program to print all Armstrong numbers between 1 to n. How to print Armstrong numbers between given interval using loop in C program. Logic to print Armstrong numbers in given range in C programming.

Example

Input

```Enter lower limit: 1
Enter upper limit: 1000```

Output

```Armstrong number between 1 to 1000 are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 370, 371, 407```

## Required knowledge

Must know –

### What is Armstrong number?

An Armstrong number is a n-digit number that is equal to the sum of nth power of its digits. For example,
6 = 61 = 6
371 = 33 + 73 + 13 = 371

## Logic to find all Armstrong number between 1 to n

Step by step descriptive logic to generate Armstrong numbers:

1. Input upper limit to print Armstrong number from user. Store it in some variable say end.
2. Run a loop from 1 to end, increment 1 in each iteration. The loop structure should look like `for(i=1; i<=end; i++)`.
3. Inside the loop print current number i, if it is Armstrong number.

## Program to find all Armstrong numbers from 1 to n

``````
/**
* C program to print Armstrong numbers from 1 to n
*/
#include <stdio.h>
#include <math.h>

int main()
{
int num, lastDigit, digits, sum, i, end;

/* Input upper limit from user */
printf("Enter upper limit: ");
scanf("%d", &end);

printf("Armstrong number between 1 to %d are: \n", end);

for(i=1; i<=end; i++)
{
sum = 0;

/* Copy the value of num for processing */
num = i;

/* Find total digits in num */
digits = (int) log10(num) + 1;

/* Calculate sum of power of digits */
while(num > 0)
{
/* Extract last digit */
lastDigit = num % 10;

// Find sum of power of digits
// Use ceil() function to overcome any rounding errors by pow()
sum = sum + ceil(pow(lastDigit, digits));

/* Remove the last digit */
num = num / 10;
}

/* Check for Armstrong number */
if(i == sum)
{
printf("%d, ", i);
}

}

return 0;
}``````

Once, you are done with generating Armstrong numbers from 1 to n. You can easily modify the logic to work it for given ranges. Below program generates Armstrong numbers in a given range.

## Program to find Armstrong numbers in given range

``````/**
* C program to generate Armstrong numbers in a given range
*/
#include <stdio.h>
#include <math.h>

int main()
{
int num, lastDigit, digits, sum, i;
int start, end;

/* Input lower and upper limit from user */
printf("Enter lower limit: ");
scanf("%d", &start);
printf("Enter upper limit: ");
scanf("%d", &end);

printf("Armstrong number between %d to %d are: \n", start, end);

for(i=start; i<=end; i++)
{
sum = 0;

/* Copy the value of num for processing */
num = i;

/* Find total digits in num */
digits = (int) log10(num) + 1;

/* Calculate sum of power of digits */
while(num > 0)
{
/* Extract the last digit */
lastDigit = num % 10;

// Find sum of power of digits
// Use ceil() function to overcome any rounding errors by pow()
sum = sum + ceil(pow(lastDigit, digits));

/* Remove the last digit */
num = num / 10;
}

/* Check for Armstrong number */
if(i == sum)
{
printf("%d, ", i);
}
}

return 0;
}``````

Output

```Enter lower limit: 1
Enter upper limit: 10000
Armstrong number between 1 to 10000 are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474,```

Happy coding 😉 