Yan Cui
I help clients go faster for less using serverless technologies.
Problem
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.
The product 7254 is unusual, as the identity, 39 x 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
Solution
1 2 3 4 5 6 7 8 9 10 11 12 | let isPandigital(n) = let upperBound = int32(sqrt(double(n))) [2..upperBound] |> Seq.filter (fun x -> n % x = 0) |> Seq.map (fun x -> x.ToString()+(n/x).ToString()+n.ToString()) |> Seq.exists (fun str -> [1..9] |> List.map (fun n -> n.ToString()) |> List.forall (fun n -> str.Contains(n) && str.IndexOf(n) = str.LastIndexOf(n)))let answer = [1000..9999] |> List.filter isPandigital |> List.sum |
This problem here is somewhat open ended in terms of how far you have to go before you can be sure that you’ve checked all the relevant numbers, i.e. do I have to cover all 1-5 digits numbers, or 1-9, or somewhere in between?
To answer that, think of each of multiplicand, multiplier and product as a * (10 POW b) – e.g. 39 = 0.39 * (10 POW 2), 186 = 0.186 * (10 POW 3) and 7254 = 0.7254 * (10 POW 4) – then you end up with:
a1 * (10 POW b1) * a2 * (10 POW b2) = a3 * (10 POW b3) where b1 + b2 + b3 = 9 and b1 + b2 >= b3
As it turns out the only b3 which satisfies both criteria is 4, so we only need to check 4-digit products, hence you see in my solution that I’ve only considered the numbers from 1000 to 9999.
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