Project Euler — Problem 37 Solution

Problem

The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.

Find the sum of the only eleven primes that are both truncatable from left to right and right to left.

NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.

Solution

[code lang=”fsharp”]
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 primeSequence = Seq.unfold (fun state -> Some(state, (state+1I))) 1I |> Seq.filter isPrime

let rec recTruncatable (predicate:bigint -> bool) (next:bigint -> bigint) (n:bigint) =
if predicate(n) then
let len = n.ToString().Length
if len = 1 then true else recTruncatable predicate next (next n)
else false

let leftTruncatable = recTruncatable isPrime (fun x -> bigint.Parse(x.ToString().Substring(1)))
let rightTruncatable = recTruncatable isPrime (fun x -> bigint.Parse(x.ToString().Substring(0, x.ToString().Length-1)))
let sum =
primeSequence
|> Seq.filter (fun n -> n > 7I)
|> Seq.filter (fun n -> leftTruncatable n && rightTruncatable n)
|> Seq.take 11
|> Seq.sum
[/code]