Crashing Light Bulbs

You are living in a 100-floor apartment block. You know that there is one floor in the block, such that if you drop a light bulb from there or anywhere higher, it will crash upon hitting the ground. If you drop a light bulb from any floor underneath it however, the light bulb will remain intact. If you have two light bulbs at your disposal, how many drop attempts do you need such that you can surely find which the floor in question is?

The answer is 14 drops. You can do this by throwing the first bulb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100 (notice that the difference decreases always by 1) until it crashes and then start throwing the second bulb from the floors in between. For example, if the first bulb crashes at floor 69, you start throwing the second bulb from floors 61, 62, 63, etc. This way the total number of throws would be always at most 14.

Proving that 14 is optimal is done using the same logic. In order to use at most 13 throws, the first throw should be made from floor 13 or lower. The second throw should be made from floor 13+12 or lower, the third throw should be made from floor 13+12+11 or lower, etc. Continuing with the same argument, we conclude that the 13th drop should be made from floor 13+12+…+2+1=91 or lower. However, if the first light bulb does not crash after the last throw, you will not be able to find out which number among 92-100 is X.

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  1. It’s cool that you made a puzzle dude, but can’t this be done in only 7 throws by using something analogous to a quick sort algorithm? Throw the bulb at floor 50, if it breaks, go down 25 floors, if not, go up 25 floors. Then keep halving the number of floors like that, going either up or down until you finally increment by 1 floor and determine from that throw which one it was.

    Side note: Why did you say you have 2 bulbs? It serves to only confuse the phrasing of the question, does 2 bulbs mean I only get 2 attempts? Does it mean I get 2 bulbs per attempt? Why do I have 2 of them?

    1. Hi Kevin,
      The issue is that once your lightbulb breaks, you can’t use it for testing anymore, i.e. it can’t “break” twice. If you have an unlimited number of lightbulbs, then indeed a binary search algorithm would be fastest. With the current formulation however, once your two lightbulbs break, your experiment must end.

      1. I believe that is the problem. Your puzzle says that you should only have 14 attempts at dropping your light bulb. But if you only have 2, and they both break on each of their first go around, then you have only done 2 tests, and therefore cannot continue because you have no more light bulbs.

        If the answer to your puzzle says you need to do 14 tests to find the mystery floor, then you should have at least 14 bulbs at your disposal in order to complete the testing, should you not?

        1. Hi Leslie. If the first lightbulb crashes on the first throw, then the floor in question must be between 1 and 14. Then, if the second lightbulb crashes on the second throw, then the answer must be 1. You simply don’t need any more lightbulbs after that.