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My solution is below. You can also look at Brian Wainscott's
solution.
Divide the coins into six pairs. Compare five of the pairs with the first five
weighings.
Case 1: There are five unequal weighings
The ten coins weighed have been identified. Weigh the last pair of coins to
determine which is heavier or lighter.
Case 1 total weighings:6
Case 2: There are four unequal weighings
The eight coins in the unequal weighings have been identified. The last pair
of coins are the same weight. Compare the equal pairs.
Case 2 total weighings:6
Case 3: There are three unequal weighings
The last pair are different, so compare them and also compare the two pairs
from the equal weighings.
Case 3 total weighings:7
Case 4: There are two unequal weighings
The last pair is equal. The four equal pairs can be compared in two
weighings. Call the pairs A, B, C, and D. Weigh A vs B and A vs C. One of those
weighings will be equal and the other will be unequal. Example: If A=B, A>C then A
and B are heavy and C and D are light.
Case 4 total weighings:7
Case 5: There is one unequal weighing
The last pair is unequal. Compare the two coins in the last pair and compare
the four equal pairs like in Case 4.
Case 5 total weighings:8
Case 6: There are zero unequal weighings
All the pairs are equal. Group the pairs into groups of two. Compare two of
the groups.
- Subcase 1: Both weighings are unequal: Those two groups have been
determined, compare the pairs from the last grouping.
- Subcase 2: One weighing is unequal: The unequal group has been determined,
and the last grouping is all equal. Compare the two equal groups.
- Subcase 3: Both weighings are equal: Compare the pairs from the last
grouping and compare the two equal groups against each other.
Case 6 total weighings:8
The maximum number of weighings needed is 8, in cases 5 and 6. |
