JavaScript Question

Question

6.24 (Perfect Numbers) An integer number is said to be a perfect number if its factors, including 1 (but not the number itself), sum to the number. For example, 6 is a perfect number, because 6 = 1 + 2 + 3. Write a method isPerfect that determines whether parameter number is a perfect number. Use this method in an application that displays all the perfect numbers between 1 and and integer entered by the user. Display the factors of each perfect number to confirm that the number is indeed perfect. Challenge the computing power of your computer by testing numbers much larger than 1000. Display the results.

Output(must have exact structure)

Enter·the·number·up·to·which·you·would·like·to·look·for·perfect·numbers:1000↵ Looking·for·perfect·numbers·from·1·to·1000↵ 6·is·a·perfect·number·it's·factors·are:·1·2·3·↵ 28·is·a·perfect·number·it's·factors·are:·1·2·4·7·14·↵ 496·is·a·perfect·number·it's·factors·are:·1·2·4·8·16·31·62·124·248·↵

Answer 1

import java.util.ArrayList;

import java.util.Scanner;

class PerfectNumber

{

private Scanner scanner;

PerfectNumber() { scanner = new Scanner(System.in); }

//returns the arrayList of factors of N, including 1 but not N;

// ex. GetFactors(32,factors); ---> factors = {1,2,4,8,16}

public void GetFactors(long N, ArrayList<Long> factors)

{

factors.add(1L);

for (long iLoop=2; iLoop<N/2+1; iLoop++)

{

if (N % iLoop==0)

{

factors.add(iLoop);

}

}

}

boolean IsPerfectNumber( long nLongIntNum, ArrayList<Long> factors)

{

GetFactors(nLongIntNum,factors);

int N = factors.size();

boolean booleanReturn=false;

long sumFactors = 0;

for (int iLoop=0; iLoop<N; iLoop++)

{

sumFactors += (long) factors.get(iLoop);

}

return (booleanReturn = (sumFactors==nLongIntNum));

}

void DisplayPerfectNumber( long nLongIntNum, ArrayList<Long> factors)

{

System.out.print( nLongIntNum + " is a perfect number. It's factors are :");

int N = factors.size();

for (int iLoop=0; iLoop<N; iLoop++)

{

System.out.print(factors.get(iLoop)+ " ");

}

System.out.println("");

}

public void Go()

{

long N=-1;

ArrayList<Long> factors = new ArrayList<Long>();

// 2 and 3 are not perfect numbers

while (N<3)

{

System.out.print(" Enter the number up to which you would like to look for perfect numbers :>");

N = scanner.nextLong();

}

System.out.println("Scanning for perfect numbers between 2 and " + N);

for (long iLoop=3; iLoop<=N; iLoop++)

{

if (iLoop%10000==0) { System.out.println(" checking " + iLoop + "...."); }

if (IsPerfectNumber(iLoop,factors))

{

DisplayPerfectNumber(iLoop,factors);

}

factors.clear();

}

}

public static void main (String args[])

{

new PerfectNumber().Go();

}

}