Source: Australian Mathematical Society Gazette - Puzzle Corner 28 Problem: The Mad Hatter is holding a hat party, where every guest must bring his or her own hat. At the party, whenever two guests greet each other, they have to swap their hats. In order to save time, each pair of guests is only allowed to greet each other at most once. After a plethora of greetings, the Mad Hatter notices that it is no longer possible to return all hats to their respective owners through more greetings. To sensibly resolve this maddening conundrum, he decides to bring in even more hat wearing guests, to allow for even more greetings and hat swappings. How many extra guests are needed to return all hats (including the extra ones) to their rightful owners? Other "Hat" Problems on the blog: Puzzle: What's the number on my Hat? Another Hat Puzzle Fair Hat Game Another Hat Problem Hats in a circle Hats and Rooms Number of Rounds of Derangem
Showing posts from February, 2013
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Source: http://www.ocf.berkeley.edu/~wwu/riddles/medium.shtml Problem: My fancy new digital alarm clock is broken! The time 'jumps' around. When I reset it, it reads 12:00:00. Then it runs as it should, but after 12:04:15 it resets back to 12:00:00. It counts up to 12:04:15 again and then it jumps to ... 12:08:32 ! Weird stuff. Do you know what's wrong with my alarm clock? Update (12/02/2013) Solution posted by Saumya Gupta, Abhimanyu Dhamija (CSE IITB 2011 Alumnus, Citibank Analyst) and Naga Vamsi Krishna in comments!
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Source: Counter-intuitive Conundrums Problem: Someone hands you a deck of cards which you thoroughly shuffle. Next, you start to deal them, face-up, counting the cards as you go. “One, Two, Three …” The aim is to predict what the count will be when you encounter the second black Ace in the deck. If you had to select one position before you started to deal, what number would you select that maximizes your chance of guessing the location of the second black Ace?