Weigh each bag once.
(not yet completed)
Solution provided by Alan Fung.
Let's name the five bags of gold as A, B, C, D and E. One of them is odd and we don't know if the odd bag is lighter or heavier.
The first two weights are like these:A + B + C = 3X ----------------(weight #1) A + D = 2Y ----------------(weight #2)
X and Y are average unit weight of the combination. That is, they are roughly the weight of each bag of gold.
If X = Y, it means the weight ratio among A, B, C and D are the same. (You may check it by adjusting one of them as lighter or heavier, X will not equal to Y). Because A, B, C and D are the same, E must be odd. So the final weight will be to weigh E and that will be the answer.
If X < Y, it is because "B or C is lighter" or "A or D is heavier".
If X > Y, it is because "B or C is heavier" or "A or D is lighter". The final weight will be:C + D + E = 3Z ----------------(weight #3)
The reasons behind picking C, D and E is because now we know E has to be normal, we need to pick E as reference and two bags from the lighter and heavier possibilities. I tried to pick A but it wouldn't work because I shouldn't pick the same thing with the same coefficient that appears in all three equations. By the same token, C and B are interchangeable.
Now to make the solution easier to understand, I will start with the solution and work out the weight comparisons.
If X < Y
If B is lighter, X < Z and Y = Z
If C is lighter, X = Z and Y > Z
If A is heavier, X > Z and Y > Z
If D is heavier, X < Z and Y > Z
If X > Y
If B is heavier, X > Z and Y = Z
If C is heavier, X = Z and Y < Z
If A is lighter, X < Z and Y < Z
If D is lighter, X > Z and Y < Z
Because all results from weights comparisons are unique, we covered all possible solutions.
For the second part of the question, to know how exactly does the odd bag weigh, we can use simple substitution to work it out. For example, if we found B is odd, B = 3X - 2Y.