Catching a Coin Shaver

by Andrew Moskalik

This puzzle was inspired by a puzzle in VSP posted by Mohaamadlaeeg. This puzzle in turn apparently came from this site.

The minimum number of weighings required is two. Since you don't know the actual weight difference between real and underweight coins, you need to rely on the ratio of weight differences found during the two weighings. Here's how to do it:

Split the 81 bags into 19 groups of four, with five left over.
Label each of the 19 groups with one of the following sets of numbers, representing the number of coins to place in the scale in each weighing: 1/1, 1/2, 1/3, 1/4, 1/5, 2/1, 2/3, 2/5, 3/1, 3/2, 3/4, 3/5, 4/1, 4/3, 4/5, 5/1, 5/2, 5/3, and 5/4.
Label the four bags within each group with the following letters, representing which side of the balance to place the coins on: L/L, L/R, R/L, and R/R.
Label the last five bags as follows: (0/1, X/L), (0/1, X/R), (1/0, L/X), (1/0, R/X), and (0/0, X/X), where the 0 and X indicate no coins are to be weighed.
For the first weighing, take the number of coins indicated by the first number from each bag and place them on the side indicated by the first letter. For example, from the bag labeled (3/2, R/L), place three coins onto the right side of the balance. For bags that have a 0 in the first position, do not weigh any coins.
Note that for every bag labeled R, there is one corresponding bag labeled L in the same group. Thus the same number of coins end up on each side, and any weight difference is due solely to the underweight coins. After placing all coins in the correct pan, record the difference in weight between the two sides.
Repeat the procedure for the second weighing, using the second number and letter. Again record the difference in weight between the two sides.
There are three possible outcomes. First, if both weighings showed the two sides equal in weight, then the bag labeled (0/0, X/X) contains the underweight coins.
Second, if one, and only one, weighing showed the two sides equal in weight, then the underweight bag is the one whose coins were not weighed during the equal weighting, and on the side with lower weight during the other weighing.
Finally, if both weighings show a discrepancy, the underweight bag is one of the 19 bags whose coins were on the side with lower weight during both weighings. To find which of the 19 this is, divide the difference in weight from the first weighing by the difference in weight from the second weighing. Since each underweight coin weighs the same, the ratio of weight differences is equal to the ratio of coins weighed. Thus, the weight ratio will match the ratio of one of the sets numbers listed in step two...all 19 of which are unique. The bag with that number set contains the underweight coins. For example, if the left pan weighed 10g less during the first weighing, and the right pan weighed six grams less during the second weighing, the underweight bag is the one labeled (5/3, L/R).

If there were more than 81 bags, three weighings would be required. However, using the same procedure, where the ratios of coins in all weighings are unique, a single underweight bag can be found from a total of as many as 1155 bags in only three weighings. In four weighings, one underweight bag can be found in a whopping 13857 bags total.

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