After the student's passing of the first test the mighty and
wise wizard from problem #15 decided for a second test. With
a magic spell he created diamonds with integer weights between
1 and 80 magic pounds. Plus he gave the student a standard
pair of non-magic scales.
Now the student's job was to create a set of counterweights.
With this set and the scales it should be possible to
determine the weight of each diamond (without using already
weighted diamonds).
What is the minimum number of counterweights (and which) that
the student had to create to pass the second test?
Correct answer: 4 (2 - 6 - 18 - 54)
Explanation:
You need only four counterweights: 2, 6, 18 and 54. Note, that you don't need the equilibrium to determine the diamond's weight. With the given set you will get the equilibrium for every even weight and the odd weights must be determined by lighter or heavier.
Examples:
D=14: 18/2 + D/6 (equilibrium)
D=16: 18 + D/2 (equilibrium)
D=15: 18/2 + D/6 (side with D is heavier => D is heavier than 14)
D=15: 18 + D/2 (side with D is lighter => D is lighter than 16)
=> D must be 15
