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

Basic C programming, If statement, For loop, While loop, Nested loop

Must know –

### What is Armstrong number?

An Armstrong number is a `n`-digit number that is equal to the sum of `n`^{th} power of its digits. For example,

6 = 6^{1} = 6

371 = 3^{3} + 7^{3} + 1^{3} = 371

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

Step by step descriptive logic to generate Armstrong numbers:

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

. - 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 😉

### Recommended post

- Loop programming exercises index.
- C program to print all Strong numbers between 1 to n.
- C program to print all Perfect numbers between 1 to n.
- C program to print all Prime numbers between 1 to n.
- C program to find first and last digit of a number.
- C program to find sum of first and last digit of a number.

doubt, write here. I willhelp my best.<pre><code> ----Your Source Code---- </code></pre>