A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.
Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.
How many starting numbers below ten million will arrive at 89?
let max = 10000000 // build up a cache for all the numbers from 1 to 10 million let cache = Array.init (max+1) (fun n -> match n with | 0 | 1 -> Some(false) | 89 -> Some(true) | _ -> None) // define function to add the square of the digits in a number let addSquaredDigits n = n.ToString().ToCharArray() |> Array.sumBy (fun c -> pown (int(c.ToString())) 2) // define function to take an initial number n and generate its number chain until // it gets to a number whose subsequent chain ends with 1 or 89, which means that // all previous numbers will also end in the same number let processChain n = let rec processChainRec n (list: int list) = if cache.[n] = None then processChainRec (addSquaredDigits n) (list@[n]) else list |> List.iter (fun n' -> cache.[n'] <- cache.[n]) processChainRec n  // go through all the numbers from 2 to 10 million using the above function [2..10000000] |> List.iter processChain // how many numbers whose number chain ends with 89? let answer = cache |> Array.filter (fun n -> n = Some(true)) |> Array.length
This is actually a fairly simple problem, with the main challenge being how to make it run fast enough to obey with the one minute rule which is where the cache comes in.
Using the property that if a chain starting with the number n, i.e. n, n1, n2, …, ends in 89 then the chain starting with any of the numbers n1, n2, … will also end in 89 we can minimise the amount of computation required by caching previous results. Note I didn’t have to mark the cache as mutable in this case because the Array construct in F# is a fixed-sized, zero-index, mutable collections of elements.