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18 digits (Posted on 2012-01-12) Difficulty: 4 of 5
Given 5 twos, 4 threes, 6 fives and 3 sevens create a multiplication problem , partial products included. I believe the solution is unique.

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.0000 (2 votes)

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Solution re(2): a solution | Comment 3 of 5 |
(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.

x  33

Besides the offer in my first post, there is no other solution where each number is in bsse-10 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 2012-01-15 00:15:09

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