# Project Euler – Problem 23 Solution

#### Problem

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

#### Solution

```let findDivisors(n) =
let upperBound = int32(sqrt(double(n)))

[1..upperBound]
|> Seq.filter (fun x -> n % x = 0)
|> Seq.collect (fun x -> [x; n/x])
|> Seq.filter (fun x -> x <> n)
|> Seq.distinct

let isAbundantNumber(n) = (findDivisors(n) |> Seq.sum) > n

let abundantNumbers =
Seq.unfold (fun x -> if x < 28123 then Some(x, x+1) else None) 1
|> Seq.filter isAbundantNumber
|> Seq.toList

abundantNumbers
|> Seq.collect (fun n -> abundantNumbers |> List.map (fun m -> n + m))
|> Seq.filter (fun n -> n <= 28123)
|> Seq.distinct
|> Seq.toList

```

I had to make a small modification to the findDivisors function which I had used previously to make sure we don’t have any duplicates in there as we now need to sum the divisors.

I first defined the isAbundantNumber function to check if the sum of the divisors of a number n is greater than n itself as per the definition given in the brief.

Then I generated the list of ALL abundant numbers from 1 to 28122 because though 28123 is the upper limit, if it it’s to be the sum of two POSITIVE integers then the two integers must be in the range of 1 to 28122.

The next list abundantNumbersSums is the one that will take up the bulk of the 15 seconds it takes this code to run, it generates all the unique sums less or equal to 28123 that you can generate using two abundant numbers.

The answer is then achieved by finding the difference between a) the sum of all numbers from 1 to 28123, and b) the sum of the abundantNumbersSum list.