145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
Find the sum of all numbers which are equal to the sum of the factorial of their digits.
Note: as 1! = 1 and 2! = 2 are not sums they are not included.
open System.Collections.Generic // factorial function let factorial n = if (n = 0) then 1 else [1..n] |> List.reduce (fun acc x -> acc * x) // create a cache of sorts to help improve performance let factorials = new Dictionary<int, int>() let start = [0..9] |> List.iter (fun n -> factorials.Add(n, (factorial n))) // get the digits of a number into an array let getDigits (n:bigint) = n.ToString().ToCharArray() |> Array.map (fun c -> int(c.ToString())) // get the sum of the factorials of a number's digits let getDigitsToFactSum (n:bigint) = bigint(getDigits(n) |> Array.map (fun n -> factorials.[n]) |> Array.sum) // get the max sum achievable by a number of the given number of digits let upperBound (digits:int) = bigint(digits) * bigint(factorial 9) // find the last number of digits n, where the max sum achievable by the factorials of the // digits of a number of n digits is greater than the smallest number of n digits // any number with more than n digits do not need to be checked let digitsToCheck = let n = Seq.unfold (fun state -> Some(state, (state+1))) 1 |> Seq.filter (fun x -> (upperBound x).ToString().ToCharArray().Length < x) |> Seq.head n-1 // get the next number with the given number of digits let maxNumber digits = [1..digits] |> List.map (fun x -> 9I * pown 10I (x-1)) |> List.sum let answer = let max = maxNumber digitsToCheck [3I..max] |> List.filter (fun n -> n = getDigitsToFactSum n) |> List.sum
This solution is very similar to my solution for problem 30, with the main difference being that the sum is calculated using the factorials of the digits rather the fifth power.