# 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

```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
```