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A remarkable multiplication (Posted on 2008-11-09) Difficulty: 3 of 5
In the multiplication below, each digit from 0-9 appears exactly 2 times. Decode the multiplication.
       * * *
       * * *
     -------
       * * *
     * * *
   * * *
   ---------
   * * * * *
Note1: No leading zeros in any number.
Note2: For a long time, mathematicians thought that this problem could only be solved using a computer. A few years ago, it was solved analytically, but its solution itīs far from being simple (and short). So, welcome computers.

See The Solution Submitted by pcbouhid    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution, with some thoughts | Comment 2 of 3 |
  179
  224
  716
 358
358  
40096

Though the product of the long multiplication would be the same, using 224 as the multiplicand and 179 as the multiplier would result in a different set of intermediate products of the multiplicand and the individual digits of the multiplier. (The number of each individual digit would be, in general, a different quantity, i.e., only the number of 9's remain the same). Therefore, 179 as the multiplicand and 224 as the multiplier is required for the solution.

Having not solved this solely by analysis, I do offer these observations. At the initial point of analysis, before excluding such for further findings, the only position in the multiplicand or multiplier that may have a digit 0 is the middle digit of the multiplicand.  Any other position would cause the existence of more than the two minimum 0 digits.  Only a single 1 digit may also exist in the multiplicand or multiplier, any additional 1 digit would cause more than two of one or more digits in the multiplier to exist in the long multiplication. 

Edited on November 10, 2008, 10:17 am
  Posted by Dej Mar on 2008-11-10 07:58:01

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