Hello, booniesgirl_29!
This is a classic problem.
(And I assume that the "scale" is a balance scale.)
The king has five bags of gold. \(\displaystyle \;\) He gave each of his most trusted people one bag.
Now one contains counterfeit gold. \(\displaystyle \;\)The king has a balance scale.
We knows for sure the counterfeit bag weighs less.
The king wants to know which bag weighs a different amount.
The king wants to use the scale only three times to figure this out.
Is this feasible? . . . Yes, but only two weighings are needed!
Label the bags: A, B, C, D, E.
First weighing: \(\displaystyle \;\) Balance A and B against C and D.
There are three possible outcomes.
\(\displaystyle [1]]\;AB\,=\,CD.\;\) Then \(\displaystyle E\) is the light bag.
\(\displaystyle [2]\;AB\,>\,CD.\;\) Then either \(\displaystyle C\) or \(\displaystyle D\) is light.
\(\displaystyle \;\;\;\)Second weighing: \(\displaystyle \;\)Balance \(\displaystyle C\) against \(\displaystyle D\) to determine the light bag.
\(\displaystyle [3]\;AB\,<\,CD.\;\) Then either \(\displaystyle A\) or \(\displaystyle B\)is light.
\(\displaystyle \;\;\;\)Second weighing: \(\displaystyle \;\)Balance \(\displaystyle A\) against \(\displaystyle B\) to determine the light bag.
