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By gandalf

Question: Everyday a peasant must pay the king one pound of gold and leave it on a collection plate in front of his house. Every morning a guard comes by to make sure he has put a pound of gold on the plate. The king collects the gold every six days from the plate. If the peasant only has one six pound block of gold, how can he make only two parallel cuts and still follow the kings rules each day?

Answer: He can cut one pound off and two pounds off. This would leave a one pound, two pound, and three pound block. On the first day he leaves the one pound block. On the second day he leaves the two pound block. On the third day he leaves the three pound block. On the fourth day he leaves the three pound and one pound blocks. On the fifth day he leaves the three pound and two pound blocks. On the last day he leaves all of them.

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Question: A man taking the census walks up to the apartment of a mathematician and asks him if he has any children and how old they are. The mathematician says "I have three daughters and the product of their ages is 72." The man tells the mathematician that he needs more information, so the mathematician tells him "The sum of their ages is equal to our apartment number." The man still needs more information so the mathematician tells him "My oldest daughter has her own bed and the other two share bunk beds."

How old are his daughters?

Answer: His daughters are 8, 3, and 3. The prime factorization of 72 is 2 * 2 * 2 * 3 * 3, so the possible ages are 2, 3, 4, 6, 8, 9, 12, and 18. Using the prime factorization and these numbers the only combinations of numbers that work for the first clue are:

18, 2 and 2.
9, 4 and 2.
6, 6 and 2.
6, 4 and 3.
8, 3, and 3.

Since he doesn't know the ages after this piece of information the sum of the three numbers must not be unique. The sum of 8, 3, and 3; and 6, 6, and 2 are the same. Now the final clue comes in handy. Since we know that the oldest daughter has her own bed it is likely that she has the bed to herself and is older than the other two so there ages are 8, 3, and 3 rather than 2, 6 and 6.

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By gandalf

Question: Two men find an old gold coin and want to have a coin toss with it to decide who gets it. The only problem is the coin is heavier on one side so it comes up heads more than tails. What is a fair way for the men to toss the coin and decide who gets the coin?

Answer: They just have to flip it twice. They call the first toss either heads or tails, then the next toss they automatically pick the opposite (ie if one man calls heads on the first flip, he automatically picks tails on the second and vice versa). If they both win one toss (a tie) out of the two, they just have to repeat until one of them wins both tosses.

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Question: If a mountain climber climbed up a mountain one day from midnight to midnight the next day and does the same thing the following day when coming down the mountain, would there be a time at which he was at the same position on the mountain both day?

Answer: Yes. Since the mountain climber would probably have an irregular rate it would be impossible to know when but he would be at the same spot at some point. If he could walk up and down the mountain at the same time he would have to cross himself at some point, and this would be the time at which he was in the same position.

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Question: A swan is in the center of a circular lake but he cannot take flight from the water, only on land. On the parameter of the lake there is a hunting dog that desperately wants the swan but cannot swim. So the swan must make it to the land before taking off and must do so before the dog makes it to him. The dog is almost 4 times faster than the swan and always runs to the point around the lake closest to the swan.

How can the swan get out of the lake and take flight before the dog gets him?

Answer: The swan can travel 1/4 of the way to land then swim in a circular path around the center of the lake (the swan will be moving slightly faster around than the dog in their circles). Once the swan is as far as he can get away from the dog in his circle he can swim the remaining 3/4 of the way to shore. The dog must travel the radius of the lake time pi (radius * π) while the swan only has to travel 3/4 the radius four times slower (3/4 * radius * 4). So the swan will make it to the shore and fly before the dog reaches it.

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