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
re: a solution by Ady TZIDON)
'My quest was for 3 digits multiplicand and 3 digits multiplier, three 4 digits "partial products" and a 6 digits result.'
I believe your quest is in error. A 3 digit multicand and 3 digit multiplier, with three 4 digit "partial products" and a 6 digits result gives 24 digits. There is a solution, however, with a 3 digit multiplicand and 2 digit multiplier with two 4 digits "partial products" and a 5 digit result.
775
x 33
2325
2325
25575
Besides the offer in my first post, there is no other solution where each number is in bsse10 and no extraneous notation is used, such as using digits for exponentiation.
Edited on January 15, 2012, 12:20 am

Posted by Dej Mar
on 20120115 00:15:09 