k-man

I did it pretty much how you described, but there is a digit a there that is a part of the equation. I wasn't sure it's 1. I found this number assuming a=1, but I wasn't sure that for some other a there isn't a smaller number.

It took me about 5 minutes, but I'm not sure if this is the smallest number. This is what I came up with...

Can you prove this?

Suppose you give me a sheet pre-filled with all the numbers. I will start with an empty grid and will place the number 1 in the same square as on your sheet. Then I will place the number 2 in the same square as your sheet has it, and so on. It doesn't matter to me where the numbers on your sheet are located, I can "reproduce" any sheet with any number arrangement by placing numbers sequentially.

1. Some have already posted the explanation of the O() notation, but if you want to read some more check here http://en.wikipedia.org/wiki/Big_O_notation Usually, the notation is used to estimate the complexity of an algorithm, but it can also be used as a general measure of a magnitude of a function. 2. The proof doesn't imply any particular method of filling the sheet. It does imply that we start from the empty grid and add numbers sequentially beginning with 1. Even if the grid is already filled and we don't know what algorithm was used to fill it, we can still repeat the process by placing all numbers in their cells sequentially. However, I agree that my proof requires a little bit of the clarification, so here it is...

At which point does the reasoning break down for you? Is it the last sentence that you think requires more explanation?

Here is my attempt at this...

For every problem there is usually only one correct answer and an unlimited number of incorrect answers. The fact that you found 2 different mutually exclusive answers should tell you that at least one of them is wrong. I will try to explain why both of them are wrong using plain logic and no math. The first answer is assuming that every second the bug is moving by 1.01 meters (1 meter with the band and .01 meter on its own). To help see the error in this assumption let's imagine that the rubber band is originally 1 meter long and has markings at each centimeter. The mark 0 is tied to the wall and the mark 100 is tied to the scooter. The bug is sitting on the mark 0. During the first second assuming the band is stretching evenly the marks on the rubber band are moving with different speeds - mark 0 is not moving at all (it's tied to the wall), mark 1 is moving at 1 cm/sec, mark 2 is moving at 2 cm/sec, and so on. After the first second the total length of the rubber band will be 2 meters and the distance between each mark will also increase from 1 cm to 2 cm. For the bug to be 1.01 meters away from the wall it should be between the marks 50 and 51. How is that possible given that the bug's speed is 1 cm/second? In reality, the bug will be somewhere between marks 0 and 1. It will be more than 1 cm away from the wall, but less than 2 cm away. Why will it not reach the mark 1 that is now 2 cm away from the wall? Because it's moving continuosly during the second and not in one leap. If we divide the second into smaller time periods you will see that each fraction of the second the bug moves slower than the mark 1. Therefore at the end of the first second the bug will not reach mark 1. The second answer states that the bug will always be at 51% of the length of the rubber band. This would be true if the bug was initially placed in this spot (mark 51) and was not moving on its own. But it's moving, so once it gets to the mark 51 it will keep moving toward mark 52. At that time it will be many miles away, but the bug will eventually get there. Unfortunately to mathematically prove that the bug will eventually catch the Prof. you need to use some math. And to calculate the exact time it will take you need to use differential equations. I came to the same equation that the foolonthehill posted, but was too lazy to actually solve it.

Hint: I think the key here is that all people wearing black hats must be consistent in their answers and all say either "red" or "black". The red hat wearers must say the opposite.