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There's a common category of mathematical puzzle which involves determining missing numbers in a long multiplication problem.

As an example from this site (problem 10):

       1,7__ | (    1736)
 x     _,_43 | (    5843)
 ----------- | 
       5,20_ | (    5208)
      69,440 | (   69440)
   1,38_,800 | ( 1388800)
 + _,680,0_0 | ( 8680000)
 ----------- | 
  1_,14_,448 | (10143448)

It's possible without too much difficulty to deduce the original multiplication problem. However, it's fairly easy to see that not all of these numbers are required to achieve the same result. Trivially: 

       1,7__
 x     _,_43
 -----------
       5,20_
      _9,44_
   1,38_,8__
 + _,680,___
 -----------
  1_,14_,448

gives the same result. For a given multiplication problem, how many numbers can I maximally remove to still allow only one unique solution that can determined without brute-force guessing?

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1
  • $\begingroup$ The 0's at the end of each line in the addition part are extraneous. $\endgroup$
    – user88
    May 29, 2014 at 0:10

3 Answers 3

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10
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Kind of cheating, but...

       1,736 | (    1736)
 x     5,843 | (    5843)
 ----------- | 
       _,___ | (    5208)
      __,___ | (   69440)
   _,___,___ | ( 1388800)
 + _,___,___ | ( 8680000)
 ----------- | 
  __,___,___ | (10143448)

Answers the question in its literal form. ;-)

Also, note that with lots of zeroes in the multiplicands this works:

      _,___ | (   1000)
x     _,___ | (   1000)
-----------
  _,___,___ | (1000000)
-----------
  1,___,___ | (1000000)
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5
  • $\begingroup$ In the same vein, requiring that at least one number be removed from the multiplicands doesn't save you either. Just remove the 6 and include the 8 in 5208. $\endgroup$
    – user88
    May 29, 2014 at 0:12
  • $\begingroup$ Actually, believe it or not, I've seen a lot of puzzles for which this isn't an optimal solution. I saw a [4]x[4] puzzle in a book once that held six clues and was fully solvable. $\endgroup$
    – user20
    May 29, 2014 at 0:16
  • $\begingroup$ What's also cheating is using a semiprime like 2660443 as the product. That way you only have to reveal seven digits! $\endgroup$
    – user88
    May 29, 2014 at 2:44
  • $\begingroup$ @JoeZ. Heh, you just gave me an evil yet brilliant idea! edited $\endgroup$
    – Doorknob
    May 29, 2014 at 2:49
  • $\begingroup$ Well. I don't think we're going to get much better than that. $\endgroup$
    – user88
    May 29, 2014 at 2:51
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Doorknob's

      _,___ | (   1000)
x     _,___ | (   1000)
-----------
  _,___,___ | (1000000)
-----------
  1,___,___ | (1000000)

is not uniquely solvable; how about e.g.

      1,234
x     1,000
-----------
  1,234,000
-----------
  1,234,000

On the other hand, isn't the following uniquely solvable?

     _,___
x    _,___
----------
     _,___
+_,___,___
----------
__,__3,1__
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2
  • $\begingroup$ No, his is solvable. In your case, it would be expanded to 4000+30000+200000+1000000 $\endgroup$
    – user17008
    Jun 14, 2016 at 10:58
  • 3
    $\begingroup$ No, each of the partial products (the lines between the horizontal rules, which are added to make the full product) comes from multiplying the first number by one non-zero digit of the second number, e.g. 1736*3=5208 in another answer. $\endgroup$
    – Rosie F
    Jun 14, 2016 at 12:06
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In a $4 \times 4 $ multiply there are $ 8 $ single digit multiplies and $ 7 $ column additions. In the sense of simultaneous equations, you would expect to be able to delete $ 15 $ digits and find your way home. The fact that you know where digits belong, as well as pattern in arithmetic (like a product $5$ must come from $5$ times odd) gives you further information. I have seen problems like this with very few given digits that could be solved uniquely.

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