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One equation with nine unknowns! (Posted on 2006-12-13) Difficulty: 4 of 5

Solve the given equation by using the integers from 1 to 9 inclusive exactly once each.


Note: Obviously this problem is easily susceptible to a brute force approach - however, it would be nice to see some arguments that don't use computer programs!

No Solution Yet Submitted by John Reid    
Rating: 4.2000 (5 votes)

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re(2): First little thought ... | Comment 6 of 7 |
(In reply to re: First little thought ... by Joel)

Thanks, Joel, for expanding on my thoughts.  I can take your work a little further. 

a) You have pointed out that 9*6 and 9*3 cannot be any denominators.   It is also true that 9, 6, and 3 cannot be in separate denominators, because the threes cancel out and and the remaining 3 (formerly part of the 9) is now relatively prime to the the other denominators.  So, if 3, 6 and 9 are in the denominators, then 3*6 must be together as one of the denominators.

b) You have also pointed out that 8*2, 8*4, and 8*6 are not going to work.  This means that if 8 is in the denominator, it must be 8*1, 8*3 or 8*9

c) If 1 is the third numerator (along with 5 and 7), then the denominators must be 3*6 from a, 8*9 from b, and 2*4 because that's all that's left.  As we have seen, this leads to a solution.

I haven't checked out other possible numerators, so can't say whether the solution is unique.

A little more work:

d) 4 cannot be the third numerator.  This would make the denominators 3*6 from a, 8*9 from b, and 2*1.  But 2*1 as a denominator causes a fraction greater than 1.

e) 2 cannot be the third numerator.  This would make the denominators 3*6,  8*9, and 4*1.  To avoid a fraction greater than 1, one of the fractions would need to be 2/(4*1) = 1/2.
There are two remaining combinations:
  5/(8*9) + 7/(3*6) = 33/72
  5/(3*6) + 7/(8*9) = 27/72
Neither adds to 1 when you add in 2/(4*1)

I still haven't ruled out 3,6,9 or 8 as the third numerator.

Edited on December 15, 2006, 7:29 pm
  Posted by Steve Herman on 2006-12-15 07:14:13

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