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C Exercises: Check whether a given number is a perfect number or not

C For Loop: Exercise-27 with Solution

Write a C program to check whether a given number is a 'Perfect' number or not.

Pictorial Presentation:

Check whether a given number is a perfect number or not

Sample Solution:

C Code:

/*Perfect number is a positive number which sum of all positive divisors excluding that number is equal to that number. For example 6 is perfect number since divisor of 6 are 1, 2 and 3.  Sum of its divisor is 1 + 2+ 3 = 6*/
#include <stdio.h>

void  main()
{
  int n,i,sum;
  int mn,mx;

  printf("Input the  number : ");
  scanf("%d",&n);
    sum = 0;
 printf("The positive divisor  : ");
    for (i=1;i<n;i++)
      {
      if(n%i==0)
         {
         sum=sum+i;
         printf("%d  ",i);
         }
       }
printf("\nThe sum of the divisor is : %d",sum);
    if(sum==n)
      printf("\nSo, the number is perfect.");
    else
      printf("\nSo, the number is not perfect.");
printf("\n");
}

Sample Output:

Input the  number : 56                                                                                        
The positive divisor  : 1  2  4  7  8  14  28                                                                 
The sum of the divisor is : 64                                                                                
So, the number is not perfect.  

Flowchart:

Flowchart : Check whether a given number is perfect number or not

C Programming Code Editor:

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Previous: Write a program in C to find the sum of the series 1 +11 + 111 + 1111 + .. n terms.
Next: Write a c program to find the perfect numbers within a given number of range.

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C Programming: Tips of the Day

What does (x ^ 0x1) != 0 mean?

The XOR operation (x ^ 0x1) inverts bit 0. So the expression effectively means: if bit 0 of x is 0, or any other bit of x is 1, then the expression is true.

Conversely the expression is false if x == 1.

So the test is the same as:

if (x != 1)

and is therefore (arguably) unnecessarily obfuscated.

Ref :https://bit.ly/2NIisQM





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