Write a program to display all perfect numbers between 1 - 100

[The_Big_K]

Here's another programming question I've asked in interviews to CS/IT Toppers in interviews when they tell me they are interested in Maths & Programming :

Write a program to display all perfect numbers between 1 - 100

Computer Language Choice: Whatever, doesn't matter. I'm happy even if you can come up with an algorithm.

PS: If you are interested in mathematics and don't know what Perfect Numbers are, you don't impress me.

[c.deepak257]

2^(n-1)(2^n -1)
this means (2 raise to the power of (n-1)) into (2 raise to the power n)-1

this is the euclid equation for calculating the perfect numbers. The first value of the n is 2 because n-1 should be greater than 1 and n is a prime number i.e. 2,3,5,7 and so on.
use this formula to calculate the perfect numbers between 1 and any number.try yourself. This is the hint.

(P.S.= ASSUMED THAT U KNOW WHAT IS PERFECT NUMBER)

[The_Big_K]

I'd not ask if I didn't know.

[c.deepak257]

perfect number are those in which the whose divisors proper sum is equal to the number.for eg: 6 has 1,2,3 as the proper divisors.sum of 1,2, and 3 is 6.

[sriramchandrk]

Hi,

This program finds all perfect numbers within 1 to 100

I have hardoded n = 100!
The inner for loop finds all factors for a number

This is simple way any one would do, but if you think of speed, then you should use euclids formula.



Thanks & Regards
Sriram

#include <iostream>
using namespace std;
main()
{
    int n = 100,sum = 0;
    cout << "Searching for perfect number less than 100..." << endl;
    for(int num = 1; num <= n; num++)
    {
    sum = 0;
    for(int i = 1; i < num; i++)
    {
       if(!(num%i))
       {
          sum+=i;
       }
    }
    if(sum == num)
        cout <<  num << " Is perfect number" << endl;
    }
}

Test run
>./a.out
Searching for perfect number less than 100...
6 Is perfect number
28 Is perfect number

[OrAnGeWorX]

it's funny i was asked this same question a week ago in C and it's been kinda haunting me... the truth of the matter is that i don't understand the maths behind it... i was never math savvy so when it gets to things like these i get confused really easy and my code ... let's not go there.. for now at least
though the question had a different twist, user would enter a number, that creates a table with n amount of rows to be filled by the user with his choice numbers to be tested as perfect numbers, y/n, and the non-perfect numbers need to be multiplied together.

i believe there's another way to do it with modulo (the perfect number calculation itself) but i'm sorta stuck...

here it is

// loop to fill table with numbers to be checked.
      for (i = 0; i < nbrMax; i++)
          {
          scanf ("%d", &tableNbr[i]);
          }

      // calculations to find perfect numbers
      sum = 0;
      for (j = 0; j < nbrMax; j++)
          {
            for (k = 1; k < tableau[j]; k++)
                {
                    if(tableau[j]%k==0)
                        {
                        sum += k;
                        }
                }
          }

[Ashutosh_shukla]

Hi OrAnGeWorX what I could make out of your question is that user enters N numbers and you want the perfect numbers. You also want the product of non perfect numbers.I have tried something I hope you will go through it and reply me.
The code is :

#include<iostream.h>
#include<conio.h>
#define MAXSIZE 100
void main()
{
 int a[MAXSIZE],n,i,j,sum;
 long int prod=1;
 clrscr();
 cout<<"Enter the number of elements to be checked : ";
 cin>>n;
 cout<<"Enter the elements : ";
 for(i=0;i<n;i++)
  cin>>a[i];
 cout<<"The perfect nos are : \n";
 for(i=0;i<n;i++)
 {
  sum=0;
  for(j=1;j<a[i];j++)
   if(a[i]%j==0)
    sum+=j;
  if(sum==a[i])
   cout<<a[i]<<endl;
  else
   prod*=a[i];
 }
 cout<<"The product of non perfect numbers is : "<<prod;
 getch();
}

[Raviteja.g]

what about powerful number?
can you design the program which generates powerful number less than 100

[OrAnGeWorX]

Originally Posted by Ashutosh_shukla

Hi OrAnGeWorX what I could make out of your question is that user enters N numbers and you want the perfect numbers. You also want the product of non perfect numbers.I have tried something I hope you will go through it and reply me.
The code is :

#include<iostream.h>
#include<conio.h>
#define MAXSIZE 100
void main()
{
 int a[MAXSIZE],n,i,j,sum;
 long int prod=1;
 clrscr();
 cout<<"Enter the number of elements to be checked : ";
 cin>>n;
 cout<<"Enter the elements : ";
 for(i=0;i<n;i++)
  cin>>a[i];
 cout<<"The perfect nos are : \n";
 for(i=0;i<n;i++)
 {
  sum=0;
  for(j=1;j<a[i];j++)
   if(a[i]%j==0)
    sum+=j;
  if(sum==a[i])
   cout<<a[i]<<endl;
  else
   prod*=a[i];
 }
 cout<<"The product of non perfect numbers is : "<<prod;
 getch();
}

Ashutosh_shukla, thanks for replying.. i should rephrase myself as far as the problem and coding this in C shouldn't be problematic (i guess)

The user enters N for number of integers he'd like to check if or not they are perfect.
stage 1: how many numbers do u want to check?
user enters up to 10 and hits enter
stage 2: program now waits for user to input his n numbers that will be saved in a table.
stage 3: program goes through table[0] to table[n-1] and calculates if integer in each position is perfect or not, display a message accordingly, table[n] = integer is / is not a perfect number and in that same loop calculate the product (multiplication) of the non perfect numbers
finally to display that last number.

Thanks in advance.
Marc
i'll try convertin this to C and see if that works.... having some problems with dev-c++, compiling is fine but when i execute, program is crashing with windows send report window..

[Ashutosh_shukla]

Hi I have got the definition of powerful no from wikipedia as follows:
A powerful number is a positive integer m that for every prime number p dividing m, p^2 also divides m. Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a^2 * b^3, where a and b are positive integers.

The code that gives powerful nos less than 1000 is :

#include<iostream.h>
#include<conio.h>
int check_prime(int n)
{
 for(int j=2;j<n;j++)
 {
  if(n%j==0)
  {
   return 0;
  }
 }
 return 1;
}
void main()
{
 int i,j;
 cout<<"the powerful nos upto 100 are : \n";
 for(i=1;i<=1000;i++)
 {
  for(j=2;j<=i;j++)
   if(check_prime(j))
   {
    if((i%j==0 && i%(j*j)!=0)||(i%j!=0 && i%(j*j)==0))
     break;
   }
  if(j>i)
   cout<<i<<"\t";
 }
 getch();
}