Problem 8: Find the greatest product of five consecutive digits in the 1000-digit number.

First we get our 1000-digit number into a string suitable for digit traversal.  F# has multi-line strings and the backslash character may be used at the end of each line to prevent newline characters from being inserted.

let digits1000 = "73167176531330624919225119674426574742355349194934\
96983520312774506326239578318016984801869478851843\
85861560789112949495459501737958331952853208805511\
12540698747158523863050715693290963295227443043557\
66896648950445244523161731856403098711121722383113\
62229893423380308135336276614282806444486645238749\
30358907296290491560440772390713810515859307960866\
70172427121883998797908792274921901699720888093776\
65727333001053367881220235421809751254540594752243\
52584907711670556013604839586446706324415722155397\
53697817977846174064955149290862569321978468622482\
83972241375657056057490261407972968652414535100474\
82166370484403199890008895243450658541227588666881\
16427171479924442928230863465674813919123162824586\
17866458359124566529476545682848912883142607690042\
24219022671055626321111109370544217506941658960408\
07198403850962455444362981230987879927244284909188\
84580156166097919133875499200524063689912560717606\
05886116467109405077541002256983155200055935729725\
71636269561882670428252483600823257530420752963450"

While the algorithm required for solving this problem is straight forward, it gives us a nice opportunity to compare an iterative approach (a) with a pure functional approach (b).  Note: make sure to convert each char to a string before converting to an int!

let problem8a =
    let mutable max = 0
    for i in 0..(digits1000.Length-5) do
        let mutable next = 1
        for j in 0..4 do
            next <- next * (digits1000.[i+j] |> string |> int)
        if next > max then max <- next
    max
       
let problem8b =
    {0..digits1000.Length-5}
    |> Seq.map
        (fun i -> 
             {0..4} 
             |> Seq.map (fun j -> digits1000.[i+j] |> string |> int) 
             |> Seq.fold (*) 1)
    |> Seq.max

I personally prefer the descriptiveness of (b) and it’s performance is on par with (a) .