# C program to print all Perfect numbers from 1 to n

Learn how to write a C program to print all perfect numbers from 1 to n. This guide includes an explanation of perfect numbers, a step-by-step algorithm, and complete code with examples.

A perfect number is a positive integer that is equal to the sum of its proper divisors, excluding itself. Proper divisors of a number are the positive divisors other than the number itself. For example, 28 is a perfect number because its proper divisors are 1, 2, 4, 7, and 14, and their sum is 1 + 2 + 4 + 7 + 14 = 28.

In this article, we will write a C program to find and print all perfect numbers in the range from 1 to a user-specified number `n`

. We'll start by understanding the concept of perfect numbers in more detail, then outline the algorithm, and finally provide the complete C program with an explanation.

## Understanding Perfect Numbers

A perfect number is a number that satisfies the following condition:

Sum of proper divisors=Number

For example:

- 6 is a perfect number because its proper divisors are 1, 2, and 3, and their sum is 1 + 2 + 3 = 6.
- 28 is a perfect number because its proper divisors are 1, 2, 4, 7, and 14, and their sum is 1 + 2 + 4 + 7 + 14 = 28.

## Algorithm to Find Perfect Numbers from 1 to n

**Input the number n**: Get the upper limit from the user.**Iterate through each number**from 1 to n.**Find proper divisors**: Iterate through all numbers from 1 to the current number - 1 and find the divisors.**Sum the divisors**: Calculate the sum of all proper divisors.**Check if the sum**is equal to the original number.**Print the number**if it is a perfect number.

## Write a C program to print all Perfect numbers from 1 to n

Here's the complete C program to print all perfect numbers from 1 to n:

```
#include <stdio.h>
int main() {
int n, num, sum;
// Input the upper limit from the user
printf("Enter an upper limit: ");
scanf("%d", &n);
printf("Perfect numbers between 1 and %d are:\n", n);
// Iterate through each number from 1 to n
for (num = 1; num <= n; num++) {
sum = 0;
// Find and sum all proper divisors of the number
for (int i = 1; i <= num / 2; i++) {
if (num % i == 0) {
sum += i;
}
}
// Check if the sum of divisors is equal to the original number
if (sum == num && num != 0) {
printf("%d\n", num);
}
}
return 0;
}
```

**Output**

```
Enter an upper limit: 100
Perfect numbers between 1 and 100 are:
6
28
```

## Explanation of the Code

**Input the Upper Limit**:`scanf("%d", &n);`

reads the upper limit ( n ) from the user.

**Iterate through Each Number**:- The
`for`

loop iterates through each number from 1 to ( n ).

- The
**Find and Sum Proper Divisors**:- The nested
`for`

loop iterates through all numbers from 1 to`num / 2`

to find the divisors. `if (num % i == 0)`

checks if`i`

is a divisor of`num`

.`sum += i;`

adds the divisor to the sum.

- The nested
**Check for the Perfect Number**:- The condition
`if (sum == num && num != 0)`

checks if the sum of the proper divisors equals the original number. If true, it prints that the number is perfect.

- The condition

## Example Runs

Let's look at some example runs to see how the program works:

### Example 1

```
Enter an upper limit: 1000
Perfect numbers between 1 and 1000 are:
6
28
496
```

- Calculation for 6: Proper divisors are 1, 2, 3. Sum = 1 + 2 + 3 = 6.
- Calculation for 28: Proper divisors are 1, 2, 4, 7, 14. Sum = 1 + 2 + 4 + 7 + 14 = 28.
- Calculation for 496: Proper divisors are 1, 2, 4, 8, 16, 31, 62, 124, 248. Sum = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496.

### Example 2

```
Enter an upper limit: 500
Perfect numbers between 1 and 500 are:
6
28
```