Write a C program to enter any number and print all Armstrong numbers between 1 to n. How to print Armstrong numbers between given interval using loop in C program. Logic to generate Armstrong numbers in given range in C program.

**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, For loop, Nested loop, If else

Must know –

### What is Armstrong number?

An Armstrong number is an

-digit number that is equal to the sum of the *n**n*^{th} powers of its digits. Below are few examples of Armstrong numbers:

6 = 6^{1} = 6

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

## Logic to generate Armstrong number from 1 to n

Below is the step by step descriptive logic to generate Armstrong numbers:

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

Read more - Program to check Armstrong number

## Program to generate 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 the last digit
lastDigit = num % 10;
// Find sum
sum = sum + 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
sum = sum + 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, 370, 371, 407, 1634, 8208, 9474,

Happy coding 😉

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