Breaking the Billiard Ball
Logic puzzles require you to think. You will have to be logical in your reasoning.
It is your task to determine how high you can drop a billiard ball without it breaking. There is a 100 story building and you must determine which is the highest floor you can drop a ball from without it breaking. You have only two billiard balls to use as test objects. If both of them break before you determine the answer then you have failed at your task. What is the order of floors that you should drop the balls from to minimize the number of droppings that you will have to make to determine the answer?
Assume that if a ball doesn't break you can reuse it without worrying about it being weakened.
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.
Dec 17, 2002
|I don't know how someone can start with 14 and start from there. This was a very confusing and, yet, hard to guess problem.|
Jan 04, 2003
|Try working backwards from the top (there|
are several slightly different solutions).
E.g., if you still have 2 balls left and
have 5 floors to cover, you can do that in
three tosses by starting from floor 98.
Whenever you break the first ball, you will have
to cover whatever remaining floors are between
it and the last successful drop by working
up with the remaining ball one floor at a time.
So if the last successful ball #1 drop was from
floor #95 (to leave 5 floors/3 moves left if it succeeds),
the previous drop must have been from floor
91 or higher, so that there are only 3 moves left
if the drop from #95 breaks instead of works.
Hence the previous drop was from #91, etc.
Mar 12, 2003
|Aah! I see! Effectively we need minimum n such that:|
1 + 2 + 3 + ... + n >= 100
Jun 18, 2003
|This isn't logic, its math. |
Aug 14, 2003
|why not start from floor one?|
If it doesn't break then go up to floor 2, if it doesn't break then go up to floor 3 and so on. If it breaks on floor x, you hav egone one level too high so you go back down one and drop unbroken billiard balls to your hearts delight. Perhaps I'm missing something crucial because this seems too easy.
Aug 14, 2003
|Jessicag, I also figured your could start at floor one and work your way up. But the puzzle was to do it in the least # of steps. Doing it out way might be OK if the answer was less than 14--but if it was floor 99 then that certainly is not the least number of steps.|
Aug 14, 2003
|Nice one - but I have problem with the solution - given the rules laid down. It relies on both balls breaking at some point to determine the answer but the rules state that you have failed if both balls break before the answer is determined|
Aug 16, 2003
|Not quite - the rules say that if both balls|
break *before* you have determined the
answer, you fail. In the solutions
given, you know the answer, at the latest,
the instant the second ball breaks. In
some cases, the second ball never breaks.
Aug 20, 2003
|Just wanted to respond to the comment that this isn't a logic but rather a math problem. |
It may contain elements of basic math, but those are trivial. The solution can best be developed through logical reasoning.
Aug 15, 2004
|If someone really wants to be a wise guy they could figure out the terminal velocity of the billiard ball, by knowing its mass and air resistance. Factor in your 9.8m/s/s for gravity to determine the height at which that velocity is reached and deduct those floors. I'm too lazy to do the math, so TV might not be reached within 100 floors, blowing all this out of the water. Great teaser though. |
Dec 18, 2004
|One of the best ones I have seen on this webpage... keep it up! |
Apr 17, 2008
|There is a slightly better solution that leaves the max at 14 but has a better average.|
The principal is exactly the same but we use the fact that with the 11 drop of the first ball we are already at 99.
The new solution follows.
The columns are
the difference is that we start with 13 (instead of 14) but repeat the 9-floor step.
Thus below floor 55 we only need 13 max, above it is still 14.
This however gives a lower average.
I am not sure it this is the solution with the lowest average.
Apr 17, 2008
the average drops from 9.48 to 9.45 (wow, that is 945 drops instead of 948 if we cover all possiblities).
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