Solve the given equation by using the integers from 1 to 9 inclusive

**exactly once each**.

a/(b*c)+d/(e*f)+g/(h*i)=1

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!

(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**