Hi i with this perfect number program Please help

[killerkhangg]

Perfect Number Program Java Methods

A perfect number is a positive integer equal to the sum of all its divisors (including 1 but excluding the number itself). For example:
28 = 1 + 2 + 4 + 7 + 14
According to Euclid, any solution to (2n-1) that results in a prime number will indicate that (2n-1)(2n-1) will be a perfect number. Thus, since 23-1 = 7, which is prime, 23-1(23-1) = 28, which is a perfect number. All even perfect numbers can be found in this fashion.

Write a program that will find the first n even perfect numbers, by testing the first formula to find primes, then using the second formula to find perfect numbers when the first reveals a prime answer. Use a long type rather than an int type to hold your responses. Find the first six perfect numbers.

This is what i got so far. please tell me what I am doing wrong

/**
* @(#)perfectnumber.java
*
*
* @author
* @version 1.00 2011/11/27
*/

import java.util.Scanner;
public class perfectnumber {

public static void main (String args[])
{
Scanner input = new Scanner(System.in);
int Number;
int Sum=0;
System.out.println("Place an interger in the range of 2 to 100");
Number=input.nextInt();
if(Number<2 || Number>100){
System.out.println("Number does not match the range!");
}
else
for(int i=2; i<Number; i++)
{
for(int j=1; j<i; j++){
if(i%j==0) //i mod j=0
Sum+=j; //part of the sum
}
if(Sum==i){
System.out.println("Perfect Number:" + i);
for(int factor=1; factor<1; factor++){
if(i%factor==0)
System.out.print(factor + "+");
}
System.out.println("="+ i);
}
Sum=0;
}
}

}

 

[DrGreg]

According to Euclid, any solution to (2n-1) that results in a prime number will indicate that (2n-1)(2n-1) will be a perfect number. Thus, since 23-1 = 7, which is prime, 23-1(23-1) = 28, which is a perfect number.

That makes no sense at all. What you meant to say was

According to Euclid, any solution to (2ⁿ−1) that results in a prime number will indicate that (2ⁿ⁻¹)(2ⁿ−1) will be a perfect number. Thus, since 2³−1 = 7, which is prime, 2³⁻¹(2³−1) = 28, which is a perfect number.​

There's a big difference between what you wrote and what I wrote. If you want help, you need to get the question right first.

 

[killerkhangg]

sorry that's what my teacher gave me. i was really confuse on how to write a program putting that equation on

 

[Mark44]

Homework Statement

A perfect number is a positive integer equal to the sum of all its divisors (including 1 but excluding the number itself). For example:
28 = 1 + 2 + 4 + 7 + 14
According to Euclid, any solution to (2n-1) that results in a prime number will indicate that (2n-1)(2n-1) will be a perfect number. Thus, since 23-1 = 7, which is prime, 23-1(23-1) = 28, which is a perfect number. All even perfect numbers can be found in this fashion.
Write a program that will find the first n even perfect numbers, by testing the first formula to find primes, then using the second formula to find perfect numbers when the first reveals a prime answer. Use a long type rather than an int type to hold your responses. Find the first six perfect numbers.

Homework Equations

The Attempt at a Solution

this is what i have so far, please tell me what i am doing wrong

When you post code here at Physics Forums, please put your code between [noparse]

 ...

[/noparse] tags. I have done that for you.

The problem description says to use the long type. Your variables are of type int.

Also, indenting your code makes it easier to read. The code you posted was not indented at all, and this makes it more difficult to see what is included in loops and so on.

Instead of asking us to tell you what you are doing wrong, tell us what the program is doing that is incorrect.

/**
* @(#)perfectnumber.java
*
*
* @author
* @version 1.00 2011/11/27
*/

import java.util.Scanner;
public class perfectnumber {

  public static void main (String args[])
  {
    Scanner input = new Scanner(System.in);
    int Number;
    int Sum=0;
    System.out.println("Place an interger in the range of 2 to 100");
    Number=input.nextInt();
    if(Number<2 || Number>100){
      System.out.println("Number does not match the range!");
    }
    else
      for(int i=2; i<Number; i++)
      {
        for(int j=1; j<i; j++){
          if(i%j==0) //i mod j=0
           Sum+=j; //part of the sum
        }
        if(Sum==i){
          System.out.println("Perfect Number:" + i);
          for(int factor=1; factor<1; factor++){
            if(i%factor==0)
              System.out.print(factor + "+");
          }
          System.out.println("="+ i);
        }
        Sum=0;
      }
    }
}

 

[killerkhangg]

This program that i wrote only find the first 4 perfect number and i am suppose to find the first 6 perfect number. And my teacher expect me to put those two equations above on the program which i really don't know how that works. Thank you very much. It would really help if you guys can tell me

 

[Mark44]

Did you see what DrGreg wrote? Your program should calculate 2ⁿ - 1 for various values of n, check each to see if it's a prime. If so, then 2^{n - 1}(2ⁿ - 1) will be a perfect number.

As I noted in my earlier post, you should be using longs, not ints.

 

[killerkhangg]

I know i see what he wrote. I switch the variables to long but how would you put it into code. I'm guessing Math.squrt.2n-1? i missed class for 1 week due to intestine problems so i didn't get these notes. Sorry

 

[Mark44]

I know i see what he wrote. I switch the variables to long but how would you put it into code. I'm guessing Math.squrt.2n-1?

?
The name of the function is sqrt, but you wouldn't use it to calculate powers of 2. You can write a function that calculates powers of 2, as in the following:

long powerOfTwo(int exponent)
{
  long result = 1;
  for (int i = 0; i < exponent; i++)
  {
     result *= 2;
  }
  return result;
}

i missed class for 1 week due to intestine problems so i didn't get these notes. Sorry