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

Discussion in 'Computer Science | IT | Networking' started by Kaustubh Katdare, Oct 9, 2008.

  1. Kaustubh Katdare

    Kaustubh Katdare Administrator

    Engineering Discipline:
    Electrical
    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 :


    PS: If you are interested in mathematics and don't know what Perfect Numbers are, you don't impress me.
  2. c.deepak257

    c.deepak257 Apprentice

    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)
  3. Kaustubh Katdare

    Kaustubh Katdare Administrator

    Engineering Discipline:
    Electrical
    I'd not ask if I didn't know.
  4. c.deepak257

    c.deepak257 Apprentice

    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.
  5. sriramchandrk

    sriramchandrk Apprentice

    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

    Code:
    #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;
        }
    }
    
  6. OrAnGeWorX

    OrAnGeWorX Apprentice

    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
    Code:
    // 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;
                            }
                    }
              }
    
    
  7. Ashutosh_shukla

    Ashutosh_shukla Apprentice

    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.:cean:
    The code is :

    Code:
    #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();
    }
    
    
  8. Raviteja.g

    Raviteja.g Apprentice

    Engineering Discipline:
    Computer Science
    what about powerful number?
    can you design the program which generates powerful number less than 100
  9. OrAnGeWorX

    OrAnGeWorX Apprentice

    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..
  10. Ashutosh_shukla

    Ashutosh_shukla Apprentice

    Re: Write a program to display all perfect numbers upto 1000

    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 :
    Code:
    #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();
    }
    :cean:
  11. Kaustubh Katdare

    Kaustubh Katdare Administrator

    Engineering Discipline:
    Electrical
    Everyone - please enclose the code in
    Code:
     ... [ /code ] tags
  12. OrAnGeWorX

    OrAnGeWorX Apprentice

    ok here's what i got so far but i have an error problem... line 41

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    main()
    {
       // initialisation des variables
       int nbrMax=0;                        // Nbr d'entiers à saisir
       int produit=0;
       int tableauNbr[nbrMax];            // Tableau pour contenir les nombres entrés par l'usager
       int i, j, k;                       // Compteur boucle
       int somme;                   // variables pour calcul nbr parfait
       char condition;                    // Oui ou Non
    
       // Boucle do ... while, pour ne pas devoir repartir le programme pour chaque calcul
       do
          {
          printf ("Entrez le nombre d'entier (Max 10):\n");
          scanf ("%d", &nbrMax);
    
          // boucle pour la saisie des nombres et remplissage le tableau.
          for (i = 0; i < nbrMax; i++)
              {
              scanf ("%d", &tableauNbr[i]);
              }
    
          // Calculs pour trouver les nombres parfaits
          somme = 0;
    // after edit
          produit = tableauNbr[0];   // for multiplication by 0 later.
    
          for (j = 0; j < nbrMax; j++)            {
    
                for (k = 1; k < tableauNbr[j]; k++)
                    {
                        if(!tableauNbr[j]%k)
                            somme += k;
                    }
                    // Affichage des resultats
                        if (somme == tableauNbr[j])
                            printf("%d est un nombre parfait\n", tableauNbr[j]);
                            else {
                            printf("%d n'est pas un nombre parfait\n", tableauNbr[j]);
                            produit *= tableauNbr[j];        // tableauNbr undeclared, first use in this function error...
                            }
            }    
          
    
          printf("le produit des nombres non-parfaits est: %d", produit);
    
          // Condition requise pour repartir du début
          printf ("\nVoulez vous faire un autre calcul, (o/n)?");
          fflush (stdin);                        // Fonction pour vider stdin
          condition = toupper(getchar());        // Capitaliser la letter et soumettre à condition
          }
       while ( condition == 'O');                // Repartir du début si O
    }
    
    edit: error on line 41 corrected... typo (silly me)
    but as the code ran, it's still not doing it's job
    i also added a line to initialize the product to the first element of the table so that the multiplication in the for loop works (0*9 = 0)
  13. OrAnGeWorX

    OrAnGeWorX Apprentice

    ok i've cleaned it up a little and fixed the calculations, the results are still wrong...
    there's gotta be something with the calculations section....

    here are some results

    Entrez le nombre d'entier (Max 10):
    5
    12
    6
    28
    1
    6
    12 n'est pas un nombre parfait
    6 n'est pas un nombre parfait
    28 n'est pas un nombre parfait -- it is
    1 n'est pas un nombre parfait
    6 n'est pas un nombre parfait --- is as well
    le produit des nombres non-parfaits est: 0
    Voulez vous faire un autre calcul, (o/n)?n
    Press any key to continue . . .
  14. shack

    shack Certified CEan

    Engineering Discipline:
    Computer Science
    To generate perfect numbers:
    main()
    {
    int n,sum;
    sum=0;
    printf("enter the range");
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
    for(int j=1;j<i;j++)
    {
    if(i%j==0)
    sum=sum+j;
    }
    if(sum==i)
    printf("%d\n",i);
    sum=0;
    }
    getch();
    }
  15. Minar

    Minar Certified CEan

    Engineering Discipline:
    Electrical & Electronics
    hi, can any one tell me whats wrong in my codes here..its not showing any output.:(

    #include<stdio.h>
    #include<conio.h>
    int main(void)
    {
    int num,i,sum;
    sum=0;
    for(num=1;num<=100;num++)
    {
    for(i=1;i<=num-1;i++)
    {
    if(!(num%i))
    {
    sum +=i;
    }
    }
    if(sum==num)printf("%d",sum);

    }

    getch();
    return 0;
    }

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