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A number is called **perfect number**, positive integer number that is equal to the sum of the its possible factor.

For example:

6 is a perfect number or not.

The possible factors of 6 is 1, 2, 3. So these are the possible factors of 6 So now add(sum) these factors 1 + 2 + 3 = 6

6 is a **perfect** number.

```
// check perfect number in c
#include<stdio.h>
int main(){
int n;
printf("Enter the Number : ");
scanf("%d", &n);
int sum = 0;
for(int i=1; i<n; i++){
if(n%i==0){
sum = sum + i;
}
}
if(sum == n){
printf("%d is the Perfect Number",n);
}else{
printf("%d is Not Perfect Number",n);
}
return 0;
}
```

- Initialized a variable n, which stores the value of given by user.
- Initialized a another variable sum. Which stores the sum of factors of given number. Here we initial assigned the value is zero (0).
- Start the for loop, in for loop we initialized a variable i which is initial assigned the value is one (1), this loop is work until the given condition is will be false. Condition is (i < n). When this condition is true then repeat the step 4.
- If (n % i == 0) is true the go to and execute this piece of code (sum = sum + i).
- After complete this process we compare the given value n is and sum if the both are equal then we print given number is
**perfect number**otherwise print given number is**not perfect number**.

1. Start

2. Declared n as integer

3. Read n from the user

4. Set sum = 0

5. Iterate from 1 to n and go to step 6

6. Check n % i == 0 then step 7

7. Set sum = sum + i

[End loop]

8. if sum == n

PRINT "perfect number"

9. else

PRINT "not a perfect number"

10. End

1. INCLUDE stdio.h

2. FUNCTION main()

1. DECLARED n AS INTEGER

2. PRINT "Enter the Number: "

3. READ n FROM USER

4. SET sum = 0

5. FOR i to n - 1

1. IF n % i == 0

SET sum = sum + i

6. IF sum == n

PRINT "perfect number"

7. ELSE

PRINT "not perfect number"

8. RETUTRN 0

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