Here is another favorite:
If you are presented with 50 gold coins and told that 49 are plated but 1 is solid gold, and told that by using an old fashioned balance beam scale to determine which is the solid gold coin you can earn the coin, you can easily determine the solid coin by comparing the coins weight one against the other, 1 on each side of the scale until you find the heavy coin and claim the prize, right?
but what if the challenge says you can only use the scale 4 times? There are no weights available, just 50 coins and a scale, the coins look identical and are very close in weight, the plated ones all exactly the same and the solid one just slightly heavier. Can it be done with only 4 trips to the scale?
I guess this was overlooked, or just posted in the wrong place or didn' t get any traction- no problem.
If anyone cares, I want to provide the answer before going on vacation.
The scale can be used to eliminate coins at a faster rate per use than by dividing the fifty into two piles which would eliminate 25 coins. If you divide the fifty into 3 piles, 16,17 and 17, and weighing the two 17 coin piles against each other, 33 or 34 coins can be eliminated in 1 weighing.
So, divide into 17, 17, and 16. weigh the 17's against each other. If equal, the solid coin is in the 16 coin pile. If one 17 coin pile is heavier, thats the pile with the solid coin.
Next weighing is either 16 or 17 coins depending on the results of the weighing, divide it 5,5 and 6 or 6,6, and 5 and weigh the equal numbered piles. this will leave you with a pile of 5 or 6 that includes the solid coin after just 2 trips to the scale.
Divide the pile of 5 or 6 into 3 piles of 2 or 2,2, and 1. Weighing 2 coins on one side against 2 coins on the other side could result in an equal weight and if the third pile is a single coin you have found the solid coin in 3 trips, more likely you will end up with 2 coins and will need one more weighing.
It can be done, you just have to think outside the box a little bit.