#### Problem

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6^{th}prime is 13.

What is the 10001^{st}prime number?

#### Solution

open System let findFactorsOf(n) = let upperBound = int32(Math.Sqrt(double(n))) [2..upperBound] |> Seq.filter (fun x -> n % x = 0) let isPrime(n) = findFactorsOf(n) |> Seq.length = 0 let primeNumbers = Seq.unfold (fun x -> Some(x, x + 1)) 2 |> Seq.filter isPrime let p = primeNumbers |> Seq.nth(10000)

Here I borrowed the *findFactors* and *isPrime* functions I first used in the problem 3 solution, except this time they don’t have to be constrained to work *int64* types.

The rest of the solution’s pretty straight forward too, to find the 10001st prime number I first need to generate the sequence of all prime numbers. I did so by using Seq.unfold to generate the list of all natural numbers equal or greater than 2 (the first prime number), coupled with the *isPrime* predicate:

let primeNumbers = Seq.unfold (fun x -> Some(x, x + 1)) 2 |> Seq.filter isPrime

With the sequence of all prime numbers at hand, Seq.nth is used to find the element at the index position 10000 (sequences are zero-indexed like arrays, so this is actually the 10001st element in the sequence) to get our answer!