If you look up the definition of a perfect number in Wikipedia it states - "A Perfect number is defined as a positive integer which is the sum of its proper positive divisors (not including the number itself)". For example, 6 is a perfect number because its divisors are 1,2, and 3. 1+2+3 = 6. In this C# Corner BLOG, we show you how to use LINQ to search for all perfect numbers up to 10000.
Euclid proved that
2n-1(2n - 1)
gives an even perfect number whenever 2n-1 is prime.
Here are some other cool facts about perfect numbers:
1) if you sum the reciprocal of the divisors, they always equal 2:
e.g. 1/1 + 1/2 + 1/3 + 1/6 = 2
2) The only even perfect number of the form x^3 + 1 is 28
3) The first 39 even perfect numbers are 2n-1(2n - 1) for
- n = 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917
(The numbers start getting quite large after 7)
Anyway here is the linq code that finds the numbers with n = 2,3,5,7. It loops through all numbers from 1-10000, gets each numbers divisors with LINQ and sums the divisors with LINQ. If the sum of the divisors equals the number itself, we found a perfect number.
private static void WriteOutPerfectNumbers()
{
List<int> allNumbers = new List<int>(); // list of range of numbers to check
List<int> perfectNumbers = new List<int>();
int numbersToSearchTo = 10000; // look for perfect numbers up to 10000
// get a list of all numbers from, 2 - 10000 (1 is not considered perfect)
var allNumbersVar = System.Query.Sequence.Range(2, numbersToSearchTo - 1);
allNumbers = allNumbersVar.ToList<int>();
// loop through all numbers and see if each one is perfect
for (int nextNumber = 1; nextNumber < allNumbers.Count; nextNumber++)
{
// first get all the unique divisors of the number
List<int> divisors = new List<int>();
var allPossibleFactors = System.Query.Sequence.Range(2, nextNumber);
var divisorsVar = from num in allPossibleFactors where nextNumber % num == 0 && (int) nextNumber > num select num;
divisors = divisorsVar.ToList<int>();
divisors.Insert(0, 1); // insert 1, because it counts as a divisor
// if the number is not prime, check it for 'perfection'
if (divisors.Count() > 1)
{
// get the sum of the divisors
int sumVal = divisors.Sum();
// if the sum of the divisors equals the number itself
// we found a perfect number!!
if (sumVal == nextNumber)
{
perfectNumbers.Add(nextNumber);
}
}
}
// write out all the perfect numbers we found
foreach (var num in perfectNumbers)
{
Console.Write("{0}, ", num);
}
Console.WriteLine();
Console.ReadLine(); // wait for enter key to exit
}
If your run this method in .NET 3.0, you should get the following output
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Happy LINQing!