How Do I Find Perfect Number Using Linq

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Hi...I just done a code to find a perfect number in C#..But its lengthy...So i want to develop this Using Linq...Here i added my code also...Please Help me..

C#

static void Main(string[] args)
        {
            int number, sumofnumbers, i;
            Console.WriteLine("hello");
            for (number = 2; number <= 100000; number++)
            {
                i = 1;
                sumofnumbers = 0;
                while (i < number)
                {

                    if (number % i == 0)
                        sumofnumbers = sumofnumbers + i;
                    i++;
                }
                if (sumofnumbers == number)
                    Console.WriteLine("perfect number is " + number);
            }
        }

Thanks in advance

Posted 23-Jan-15 1:07am

Shivaram_i

Updated 23-Jan-15 1:15am

jaket-cp

v2

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Richard Deeming · Accepted Answer · 2015-01-23T01:23:00

Solution 1

How about:

C#

IEnumerable<int> perfectNumbers = Enumerable.Range(2, 100000)
    .Where(number => Enumerable.Range(1, number - 1).Where(i => (number % i) == 0).Sum() == number);

foreach (int number in perfectNumbers)
{
    Console.WriteLine("perfect number is " + number);
}

Posted 23-Jan-15 1:23am

Richard Deeming

Comments

BillWoodruff 23-Jan-15 8:29am

+5 This is a Zen Koan ! Congratulations (in case I didn't say it already) on your MVP status for this year.

Shivaram_i 23-Jan-15 9:18am

What??? #BillWoodruff

BillWoodruff 23-Jan-15 11:48am

I am expressing my appreciation for Richard's very advanced example here which demonstrates the power of Linq.

Shivaram_i 23-Jan-15 9:18am

Hey thank you Richard...But its little bit confusion...Will you please explain This...??
I mean In this Line ".Where(number => Enumerable.Range(1, number - 1).Where(i => (number % i) == 0).Sum() == number);"
In first case number is 2 and i is 1 right???But how It will taken like this only...If am wrong then tell me what value variable i has been taken and how it will increment???Sorry because i am not aware of this LINQ...So

Thanks

Richard Deeming 23-Jan-15 9:25am

The LINQ version is the equivalent to your original loop. It's shorter than the original, but not necessarily easier to understand. :)

* Loop through 100000 numbers starting at 2: Enumerable.Range(2, 100000)
* For each number in the outer loop:
*- Loop through the integers from 1 to number - 1: Enumerable.Range(1, number - 1)
*- Filter it to only include numbers by which the outer number is divisible: .Where(i => (number % i) == 0)
*- Take the sum of the matching numbers in the inner loop: .Sum()
* Filter the outer loop to numbers where the number is equal to the inner sum: .Where(number => ... == number)

Matt T Heffron 23-Jan-15 19:52pm

+5!!!!