IMC 2009 Q25
25. One coin among \(N\) identical-looking coins is a fake and is slightly heavier than the others, which all have the same weight. To compare two groups of coins you are allowed to use a set of scales with two pans which balance exactly when the weight in each pan is the same. What is the largest value of \(N\) for which the fake coin can be identified using a maximum of two such comparisons?
- A 4
- B 6
- C 7
- D 8
- E 9
Intermediate Mathematical Challenge 2009, Question 25. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: E 9
As it is known that the fake coin is heavier than all of the others, it is possible in one comparison to identify which, if any, is the fake in a group of three coins: simply compare any two of the three coins – if they do not balance then the heavier coin is the fake, whereas if they do balance then the third coin is the fake. This means that it is possible to find the fake coin when \(N = 9\) using two comparisons: the coins are divided into three groups of three and, using the same reasoning as for three individual coins, the first comparison identifies which group of three coins contains the fake. The second comparison then identifies which of these three coins is the fake. However, it is not possible to identify the fake coin in a group of four coins in one comparison only, so it is not always possible to identify the fake coin using two comparisons when \(N = 10\). If less than four are put on each side for the first comparison and they balance, then there are more than three left and the fake coin amongst these cannot be identified in one further comparison. Alternatively, if more than three are put on each side for the first comparison and they do not balance, then the fake coin in the heavier group cannot be identified in one further comparison.
Solution © UK Mathematics Trust, from the IMC 2009 solutions.