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Find all positive integers
and which satisfyLast edited by tony123 (2008-08-20 09:38:04)
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Since all factorials are positive, x must be strictly greater than the other four letters.
(Otherwise one of the letters would be greater than or equal to x, and so you'd get x! = x! + k or x! = (x+n)! + k, neither of which are possible)
So then the maximum of all the RHS variables is x-1, meaning the RHS's maximum value is 4(x-1)!.
Since x! = x(x-1)!, this means that x≤4.
x is now between 1 and 4. The other variables are all between 1 and x, which means there are a finite amount of combinations to try. It's now fairly easy to see that the only solutions are:
x = 3; {f,g,y,z} = {1,1,2,2}
x = 4; {f,g,y,z} = {3,3,3,3}
There are 6 ways that the first way can happen, which gives 7 solutions in all.
If the positive restriction on the integers was dropped, the set {1,1,2,2} could have one of both of its 1's replaced with zero(es), which would give two more solution sets and 18 more solutions.
Why did the vector cross the road?
It wanted to be normal.
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