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Swimming Pool

 


18 days; Working backwords in 19 days it will be half of full i.e. 648 ft sq. In 18 days it will 324 ft sq.


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Guess the Number

  Sage had a 1, Rosemary had an 11, and Tim had a 2. From Sage's statement, he had to have an odd number. He knew that the sum of all three numbers was even, so if his number was odd, one of the other two numbers would have to be odd, and the other even, leaving them unequal. Rosemary could deduce this, too, so when she saw an 11, she knew that Sage had a 1 and Tim had a 2. If she had seen a 12, she would have known that the other two boys each had a 1, which is contradicted by her saying that she already knew they all had different numbers. If she saw any number other than an 11 (say 9, for example), she would not have know which odd number Sage held. (In our example, he could have had a 1 and Tim a 4, or he could have had a 3 and Tim a 2). Tim, understanding all of this, therefore knew their numbers.
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Greedy Pirates

If there are two pirates left (#4 & #5), #4 has no options. No matter what he proposes, pirate #5 will disagree, resulting in a 1-1 vote (no majority). #5 will kill #4 and will keep all of the gold. Now say there are 3 pirates left. #4 has to agree with whatever #3 decides, because if he doesn't #3 will be killed (because #5 won't vote for #3's proposal no matter what it is). #3 will just propose that he keep all of the gold and will get a 2-1 vote in his favor. Now if there are 4 pirates left: #3 won't vote for #2's proposal because if #2's fails, #3 will get all of the gold. #4 and #5 know that they will get nothing if the decision goes to #3, so they will vote for #2's proposal if he gives them one gold piece each. Therefore, #2 would keep 998 gold, and #4 and #5 would each get one gold. So let's wrap this up: Pirate #1 needs 2 other votes. He will not get a vote from #2 because #2 will get 998 gold if #1's plan fails. #1 offers #3 o...
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Backwards Clock

  Suppose that a second pair of hands turns together with the wrong pair of hands, but then in the correct way. When the wrong pair is in the same position as the correct pair, this means that the time is shown in the right way. First look at the hour-hands that are at the twelve. One turns with the correct speed, the other with a speed that is twelve times as small. These two hands are in the same position again when the 'slow' hand has progressed x minutes. The fast hand then has progressed 60+x minutes. For the time x that passed, then holds: (60+x)/12 = x. This means that x = 5 5/11 minutes. For the minutes-hands that start at six holds the same. The confused clock therefore shows the correct time again at 5 5/11 minutes past 7. 
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