markr wrote:
Actually, the answer is A W.
Why, Y o'why. Oh gawd, will this damn answer haunt me thru time?
I agree with DP's simple weighing answer.
For anyone interested in this type of problem - the ultimate guide:
In the November 1997 College Mathematics Journal, Mario Martelli and Gerald Gannon of California State University, Fullerton address the following question: What is the largest number, m, of otherwise identical coins among which a single odd coin can be detected using a balance scale n times?
Suppose that the odd coin weighs less than the others. If there are three coins, you can find the light one by putting one coin on each side of the balance while holding the third. With four or more coins, one weighing isn't enough because two coins must be either held or set on both pans. Hence, the largest number of coins among which it is possible to detect the lighter one using the scale twice (n = 2) is 3^2. You place three coins on each pan and hold the remaining three. If neither pan rises, the coin is among the three in your hand. Otherwise, the pan that rises contains the light coin. In either case, a second round of weighing detects the counterfeit.
In general, 3^n is the largest number of coins among which a lighter one can be detected using the balance n times.
Martelli and Gannon go on to demonstrate that, in general, n weighings are needed to detect the counterfeit one among (3^n - 3)/2 coins. For example, four weighings will detect the odd one among 39 coins. You divide 39 by 3 to get 13. Setting aside 13 coins, you divide the remaining 26 equally between the two pans of the scale. If the pans remain level, the odd coin is among the 13 set aside.
You can then choose a test coin of the correct weight from the 26 coins that balanced. To find the odd one among the remaining 13 coins, set aside four, put five on the left pan and the other four together with the test coin on the right pan. Suppose the right side is lighter. Taking away the test coin, the odd coin is either among the five on the left (heavier) or the four on the right (lighter).
With two more tries, it's possible to detect the counterfeit coin and determine whether it is lighter or heavier than the others. Call the five coins on the left "red" and the four on the right "black." Set aside one red coin and two black coins, leaving four red and two black for the balance. Put two red and a black on each pan. If the right rises, the counterfeit coin is among the two red on the left or it is the black on the right.
With just three coins, two red and one black, just put a red on each pan and keep the black in your hand. Whatever the outcome of the weighing, you can pick out the counterfeit.
Variations of that method produce the required result in the other possible cases.
Martelli and Gannon note that their approach "demonstrates the strategic value of the old principle divide et impera: to solve a difficult problem, break it into simpler problems."
Talking of small numbers:
Alan, Bill and Chris dug up 9 nuggets. Their weights were 154, 16, 19, 101, 10, 17, 13, 46 and 22 kgs. They took 3 each. Alan's weighed twice as much as Bill's. How heavy were Chris's nuggets
The product of 3 brothers' ages is 175. Two are twins. How old is the other one
A man has 2 bankcards, each with a 4 digit number. The 1st number is 4 times the 2nd. The 1st number is the reverse of the 2nd. What is the first number
