i think that most of you people are smart enough to find the first part, a piece of cake. WELL, HOLD ON!!!! that was just the warming up session. this one is the real cracker. let's see how good are you at this one. You're sitting in your office, when your boss walks in, carrying a basket filled with bags. He picks up a bag and opens it, pouring five gold coins on your desk. "The Mint asked us to find a coin-shaver," he says. "The Mint has exactly 81 different suppliers; one of them is consistently supplying underweight coins, but we don't know which one. He gestures to the bags. "Here," he says, "we have 81 bags each containing five coins; one bag from each supplier. We know that all the coins are absolutely identical, except for the coins from the one underweight bag. We don't know how much underweight these coins are, but we know they're all underweight by the same amount. I need you to find out which bag contains the underweight coins. And, I need you to find out in the minimum number of weighings. Get to work." All you have to work with is your trusty two-pan scale. This fancy scale will report the exact difference in weight between masses placed in the two pans, as well as telling which side is heavier. That means you can tell exactly how much more or less one pile of coins weighs compared to another...but remember you don't know how much less the underweight coins weigh. What is the minimum number of weighings required to find the one bag of underweight coins, if each of the 81 bags contains five coins? You can take individual coins out of bags and label them if you desire.