Perfect number

Question

This is in the exercise Question 15. I have no understanding about perfect numbers. Please help me with this program.

Vulpes · Answer

Perfect numbers are positive integers which are the sum of all the number's positive divisors except the number itself. It follows that 1 will always be a divisor of a perfect number and so this is why it's being added to the factors list.

Maha · Answer

I have withdrawn a thread because I have solved the problem.

Maha · Answer

Thanks

Vulpes · Answer

Yes that works fine :)

Maha · Answer

Your program is slightly modified to show the difference between perfect number and non-perfect number. Please see whether it is correct. using System; class Factors { static void Main() { while (true) { Console.Write(Input a positive number or 0 to quit: ); int i = int.Parse(Console.ReadLine()); if (i Console.WriteLine(The factors of {0} are as follows i); Console.Write( 1); int sum = 1; for (int j = 2; j { if (i % j == 0) { Console.Write( + {0} j); sum = sum + j; } } Console.WriteLine( = {0} sum); if (i == sum) Console.WriteLine(i +  is a perfect number); else Console.WriteLine(i +  is not a perfect number); Console.WriteLine(
); } } }

Vulpes · Answer

That's absolutely right Maha.

Maha · Answer

From your example I come to a conclusion that factors of a particular number is within the range of half the value of that number that is why we have to divide by two. I don't know whether my conclusion is correct. For example factors of number 16(=i) is within the range of 1 to 8 that is why i/2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Vulpes · Answer

Well what you could do is to input a number and then use the same logic to output all its factors (other than itself) which you can then check manually to ensure it's giving the right answers. The following is off the top of my head as I don't have .NET on this machine: using System; class Factors { static void Main() { while(true) { Console.Write(Input a positive number or 0 to quit: ); int i = int.Parse(Console.ReadLine()); if (i Console.WriteLine(The factors of {0} are as follows i); Console.Write( 1 ); for(int j = 2; j { if (i % j == 0) Console.Write({0}  j); } Console.WriteLine(); } } }

Maha · Answer

I wish to know whether it is possible to simplify like following program to demonstrate the above concept. using System; class Program { static void Main(string[] args) { for (int i = 2; i { for (int j = 2; j { Console.Write(j +  );//2 2 2 3 2 3 2 3 4 2 3 4 2 3 4 5 } } Console.Read(); } }

Vulpes · Answer

Suppose on the contrary that a number i could have a factor > i/2 other than itself. Then the remaining factor of the number would have to be somewhere between 1 and 2 which is impossible as we're only dealing with integers. So we can therefore conclude that all factors must be

Maha · Answer

How to know that we have to consider numbers = 2) of a number i.

Vulpes · Answer

To find all the factors (>= 2) of a number i we only need to consider numbers If the number is even the highest possible factor will be i / 2. If the number is odd the highest possible factor will be i / 2 - 1. So if we iterate from 2 up to i / 2 and test all numbers in that range we will capture all the remaining factors (apart from 1) whether the number is odd or even.

Maha · Answer

A perfect number is a whole number an integer greater than zero; and when you add up all of the factors less than that number you get that number. These are the conditions. All perfect numbers can be divided by 1 without any remainder. Therefore adding factors.Add(1); can be understood. Please explain the function of the highlighted one. using System; using System.Collections.Generic; class Perfect { static void Main() { List factors = new List (); Console.WriteLine(The perfect numbers less than or equal to 1000 are :
); for (int i = 2; i { factors.Clear(); factors.Add(1); for (int j = 2; j { if (i % j == 0) factors.Add(j); } int sum = 0; for (int k = 0; k { sum += factors[k]; } if (sum == i) { for (int l = 0; l { Console.Write({0} +  factors[l]); } Console.Write({0} = {1}
 factors[factors.Count - 1] i); } } Console.Read(); } }

Maha · Answer

Thank you very much for your help.

Vulpes · Answer

Here's a program I wrote a while back in response to a similar question: using System; using System.Collections.Generic; class Perfect { static void Main() { List factors = new List (); Console.WriteLine(The perfect numbers less than or equal to 1000 are :
); for (int i = 2; i { factors.Clear(); factors.Add(1); for(int j = 2; j { if (i % j == 0) factors.Add(j); } int sum = 0; for (int k = 0; k { sum += factors[k]; } if (sum == i) { for (int l = 0; l { Console.Write({0} +  factors[l]); } Console.Write({0} = {1}
 factors[factors.Count - 1] i); } } } }