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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: First little thought ... | Comment 5 of 7 |
(In reply to First little thought ... by Steve Herman)

In fact, this can be proven. 

if 5 (e.g.) is in the denominator of one of them (say b=5 wlog) then :
sum = (aefhi+5(dchi+gcef))/5cefhi
so for the sum to be 1
aefhi+5(dchi+gcef) = 5cefhi
i.e. aefhi is divisible by 5 which is not possible because 5 is relatively prime to all of those variables.

Similar reasoning can prove that 9 *6 and 9*3 cannot appear on the bottom of any fraction, nor can 8*2, 8*4, or 8*6.

  Posted by Joel on 2006-12-15 00:06:43

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