Yan Cui

I help clients go faster for less using serverless technologies.

#### Problem

Take the number 192 and multiply it by each of 1, 2, and 3:

192 x 1 = 192

192 x 2 = 384

192 x 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, … ,n) wheren> 1?

#### Solution

// define function which checks if a given number is 1-9 pandigital let isPandigital n = let str = n.ToString() [1..9] |> List.map string |> List.forall (fun x -> str.Contains(x) && str.IndexOf(x) = str.LastIndexOf(x)) // define function to generate the (possible) pandigital sequence let getConcatProduct n = // define recursive inner function let rec genSeq n' (digits:string) n= let digits' = digits + (n' * n).ToString() if digits'.Length > 9 then digits else genSeq (n'+1) digits' n genSeq 1 "" n let answer = [1..10000] |> List.map getConcatProduct |> List.filter (fun d -> d.Length = 9 && isPandigital d) |> List.maxBy int

Seeing as *n* must be greater than 1 as specified in the problem brief, we only needs to check numbers up to 10000 as the concatenated form of two 5-digit numbers will be longer than 9 digits.

The *getConcatProduct* function does most of the interesting work and is responsible for taking a number and returning the concatenated product of that number and 1, 2.. up to 9. The last section of the code checks the outputs and only look for concatenated products that are 9 digits long and pandigital from 1 to 9, finding the largest number of this kind.

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