Improve performance and style of solution to Project Euler #32

Question

Since I'm fairly new to Mathematica, I'm trying to learn better ways to improve my coding skills so I've turned to Project Euler and this site to speed up my learning pace. Anyways, I was trying to solve problem 32 on the project Euler forum and came up with the following code

PanDigital[n_, m_] := Sort[Flatten[IntegerDigits /@ {n, m, n m}]] == Range[9];

Module[{k}, k = Select[Flatten[Table[{i j, PanDigital[i, j]}, {i, 2, 9}, {j, 1234, 
      9876}] ~Join~ Table[{i j, PanDigital[i, j]}, {i, 12, 98}, {j, 123, 987}], 
     1], #[[2]] == True &]; k = Union@Select[Flatten[k], IntegerQ] // Total] // Timing

{2.620817, 45228}

By the way here is the question:

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 × 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.

My question is, how can I make this code better, in terms of style and speed. I imagine using things like Reap and Sow could improve the readability and also speed. Also while my code run in under 3 seconds, I saw other people claiming less than 10 milliseconds time. Of course, using ParallelTable decreased the time to about 0.6 seconds on my PC but this is still not comparable to those times. Any advise is greatly appreciated.

Answer 1

Union @@ Table[
    If[a*b <= 9876 && 
      Union[IntegerDigits[a], IntegerDigits[b], IntegerDigits[a*b]] ==
        Range[9], a*b, 0], {a, 123, 1987}, {b, 2, 98}] // Tr // Timing

(*v7*)(*{0.686, 45228}*)
(*v8*)(*{0.078, 45228}*)

Answer 2

I'm really uncomfortable with hosting solutions to Project Euler problems here, but apparently the community feels otherwise.

I'll remark that it is often best to find the smallest set that encompasses the problem and test those cases. For example, you could consider only the numbers that might be products and then test those to see if they can be formed as the product of the remaining digits. This allows for about an order of magnitude improvement over your code I believe.

---

In light of the argument presented in the comments I shall go against my usual policy and post my solution using the principle described above.

test =
 MemberQ[
   Union @@@ Table[#[[{i, -i}]], {i, 2, Length@#/2}] &@IntegerDigits@Divisors@FromDigits@#,
   Range@9 ~Complement~ #
 ] &;

FromDigits /@ Select[
  TakeWhile[Reverse@Range@9 ~Permutations~ {4}, # =!= {3, 1, 8, 5} &],
  test
] // Tr