Project Euler - Problem #7

Description

• By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
• What is the 10001st prime number?

Solution (Improved)

My original solution was slow, and the code a bit ugly.  The new solution is more idiomatic and dynamic than the original.

let isPrime n =
    let upperBound = int32(sqrt(float(n)))
    let allNumbers = [2..upperBound]
    let divisors = allNumbers |> List.filter (fun div -> n % div = 0 && n <> div)
    if List.length divisors = 0 then
        true
    else
        false

let rec recursePrimes num count max =
    if count = max then
        num-1
    else
        if isPrime num then
            recursePrimes (num+1) (count+1) max
        else
            recursePrimes (num+1) (count) max

let UT_recursePrimes = recursePrimes 2 0 10001

Solution (Original)

open System
let primeFind find =
    let x = (find /10) * (find /10)
    let upperBound = Convert.ToInt32(System.Math.Sqrt(Convert.ToDouble(x)))
    let allNumbers = ref [2..x] in
    for div = 2 to upperBound do
        allNumbers := List.filter(fun num -> (num % div = 0 || div >= num)) !allNumbers
    !allNumbers

let resPrimes = primeFind(10001).[10000]