The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.
Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:
- d2d3d4=406 is divisible by 2
- d3d4d5=063 is divisible by 3
- d4d5d6=635 is divisible by 5
- d5d6d7=357 is divisible by 7
- d6d7d8=572 is divisible by 11
- d7d8d9=728 is divisible by 13
- d8d9d10=289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
let rec distribute e = function |  -> [[e]] | x::xs' as xs -> (e::xs)::[for xs in distribute e xs' -> x::xs] let rec permute = function |  -> [] | e::xs -> List.collect (distribute e) (permute xs) // generate the 0 to 9 pandigitals let numbers = permute [0..9] |> List.map (fun l -> l |> List.map string |> List.reduce (+)) // the corresponding prime divisors and digits let primes = [2; 3; 5; 7; 11; 13; 17] let ns = [2..10] |> Seq.windowed 3 |> Seq.toList // returns the number retrieved from taking the digits at the supplied positions let d ns (numberStr:string) = int(ns |> Array.map (fun n -> numberStr.[n-1].ToString()) |> Array.reduce (+)) // predicate which identifies pandigital numbers with the desired property let predicate number = List.forall2 (fun n p -> (d n number) % p = 0) ns primes let answer = numbers |> List.filter predicate |> List.sumBy int64
One thing you might notice in the creation of the permutations of all 0 to 9 pandigital numbers is the use of (+) in the List.reduce function call. This is actually just the shortened form of:
List.reduce (fun acc item –> acc + item)
Another List function of interest here is the List.forAll2 function, which tests if all corresponding elements in both lists satisfy the given predicate pairwise. In this case, I’m using it to pair up the digits list ([[2;3;4]; [3;4;5];…]) with the primes list ([2;3;5;..17]) to test if a given number has the desired property as described in the problem brief.
We arrive at the answer by simply adding up all the pandigital numbers which matches the predicate condition.
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I’m an AWS Serverless Hero and the author of Production-Ready Serverless. I have run production workload at scale in AWS for nearly 10 years and I have been an architect or principal engineer with a variety of industries ranging from banking, e-commerce, sports streaming to mobile gaming. I currently work as an independent consultant focused on AWS and serverless.
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