Breaking the Billiard Ball

If the proper order is chosen, you can determine the breaking point with a maximum of 14 drops. Here's how to do it:

First drop the first ball from the 14th floor. If it breaks you can determine the exact breaking point with the other ball in at most 13 more droppings, starting at the bottom and going up one floor at a time.
If the first ball survives the 14 floor drop then drop it again from the 27th (14+13) floor. If it breaks you can determine the exact breaking point with at most 12 more droppings.
If the first ball survives the 27 floor drop then drop it again from the 39th (14+13+12) floor. If it breaks you can determine the exact breaking point with at most 11 more droppings.
Keep repeating this process always going up one less floor than the last dropping until the first ball breaks. If it breaks on the xth dropping you will only need at most 14-x more droppings with the second ball to find the breaking point. By the 11th dropping of the first ball, if you get that far, you will have reached the 99th floor.

Comments

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.

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...

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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