#### Problem

The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.

How many circular primes are there below one million?

#### Solution

open System let hasDivisor(n:bigint) = let upperBound = bigint(sqrt(double(n))) [2I..upperBound] |> Seq.exists (fun x -> n % x = 0I) let isPrime(n:bigint) = if n = 1I then false else not(hasDivisor(n)) let rotate(n:bigint) = let charList =n.ToString().ToCharArray() |> Array.toList let len = List.length charList [0..(len-1)] |> List.map (fun r -> List.permute (fun i -> (i + r) % len) charList) |> List.map (fun l -> String.Join("", l |> List.toArray)) |> List.map bigint.Parse let isCircularPrime(n:bigint) = rotate n |> List.forall isPrime let answer = [2I..999999I] |> Seq.filter isPrime |> Seq.filter isCircularPrime |> Seq.length

The *rotate* function utilizes the List.permute function to rotate the elements in the char array representation of a given number, the rest is all pretty straight forward.

It’s worth noting that this is a brute force approach and does take quite a bit of time to run, but is also the most simple and easily understood solution. In the forum thread for the problem there are many suggestions for optimizations, and indeed many people have managed to write code which runs under a second! Also, you can improve on the performance of my solution hugely if you just cache some of the computed results (e.g. hasDivisor is a great place to start) with a dictionary.