Project Euler – Problem 7 Solution

Yan Cui

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Problem

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

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!

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1 thought on “Project Euler – Problem 7 Solution”

  1. here’s another solution

    let p7 q =
    let isPrime n =
    let rec check = function
    |i when i > n/2I -> true
    |i when n % i = 0I -> false
    |i -> check (i+1I)
    check 2I

    let rec loop nth = function
    |o when isPrime o -> //printfn “%A %A” o nth
    match nth with
    |x when x = q -> o
    |_ -> loop (nth + 1I) (o + 1I)
    |o -> loop nth (o+1I)
    loop 1I 2I

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