Challenging question

You, sir, are a bigger idiot than the guy who posted the wrong answer that you just voted as Best Answer.

Based on your own explanation of the problem, my explanation of the answer, and his explanation of the answer, his is clearly not answering the question that was asked.

Whatever.

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i think the answer is ZERO because when the process goes on till 100, the last man comes and offs all the bulbs because the multiple of 100 is 100, 200...so on.. and as there r only 100 bulbs then he offs every bulb and zero bulb is left glowing.... IM NOT SURE AND IF ITS WRONG THEN SORRY

Only 1 bulb having button No. 1
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First you will take intersection of multiples of 2s and 3s
Then 3s and 4s

..... and so on

no bulbs will be glowing...

Only the perfect squares will remain lit, and there are 10 perfect squares between 1 and 100 inclusive (1^2 through 10^2). So 10 bulbs will be glowing.

The reasoning behind this is that any bulb with an even number of distinct factors will be turned off, because each time a bulb N is turned "on" by one factor X, it's turned "off" by the corresponding factor N/X. Only perfect squares have an odd number of distinct factors, so they will get one less "flip" than non-squares, and therefore will remain on.

Example:

The factors of 80 are 1, 2, 4, 8, 10, 20, 40, and 80.
Bulb 80 is turned on initially, which counts as the first flip. Then it's turned off by the multiples-of-2 flip, on by the multiples-of-4, off by the multiples-of-8, on by the mutiples-of-10, off by the multiples-of-20, on by the multiples-of-40, and off by the multiples-of-80.