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
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
aefhi=5(cefhidchigcef)
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 20061215 00:06:43 