Given 5 twos, 4 threes, 6 fives and 3 sevens create a multiplication problem , partial products included. I believe the solution is unique.

(In reply to a solution by Dej Mar)

Amazing!

Your solution fits the 18 digits' distribution and is true in sense of the result, however there are no "partial products".

  My quest was for 3 digits multiplicand and 3 digits multiplier, three 4 digits  "partial products"  and a 6 digits result.

I still believe there vis only way to do it.

So - try to solve it as desribed.

Rem: only prime digits participate!!.

My quest was for 3 digits multiplicand and 3 digits

Posted by Ady TZIDON on 2012-01-14 16:30:20