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Smudge Sludge (Posted on 2004-04-28) Difficulty: 3 of 5
   ??7
 x 3??
  ?0?3
  ?1?
 ?5?
 ?7??3
The above equation was written on a chalkboard to represent an example of a correctly answered multiplication problem. However, the class prankster went when the teacher wasn't looking and smudged many of the numbers. What did the original problem read?

  Submitted by Gamer    
Rating: 3.6667 (3 votes)
Solution: (Hide)
Since the only digit that can end in 3 when multiplied by 7 is 9, that must be the first digit of the bottom multiplicand. Also, since 3 times 7 is 21, the last digit on the row second from the bottom must have a 1 in the ones place. Since the only way ?7 times 3 will give 51 is 17, that must be the next number.
   ?17
 x 3?9
  ?0?3
  ?1?
 ?51
 ?7??3
Now we can fill in the ?0?3 with 9 times 17 or 53. Also, since 9 times ?17 needs to have an extra 900, the only ?00 times 9 to end in 900 is 1. This means the top multiplicand is 117 and the third line down is 1013 and the fifth line down is 351.
   117
 x 3?9
  1013
  ?1?
 351
 ?7??3
The second column adds up to 7 with no carrying possibility, so the ? in that column must be 1. This means the tens digit in the lower multiplicand must be 1 because it's the only thing that can result in 11? when multiplied by 117, and the ? must also be 7. The bottom three numbers can be added together to get the final solution.
   117
 x 319
  1013
  117
 351
 37283

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionPuzzle Solution With ExplanationK Sengupta2009-01-22 06:02:48
re: Solution incorrect. FK is correct.K Sengupta2009-01-22 06:00:51
re: AnswerJayDeeKay2009-01-21 13:37:16
Solution incorrect. FK is correct.jduval2007-09-16 19:57:16
AnswerK Sengupta2007-08-17 12:30:19
No Subjectvaradarajan2005-03-03 12:08:38
SolutionsolutionSteve Royer2004-04-28 21:53:13
SolutionReasoned solutionFederico Kereki2004-04-28 17:58:06
SolutionNo SubjectSaka Devi2004-04-28 14:11:31
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