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)
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