James33

No, Sherlock, but great investigative reporting! Oh that makes things different. In that case I think there are several solutions.

I don't think there are any integer solutions to this.

May be cheating but Avogadro's constant=R/K

By the logic above 97 is the oddest prime number because it is the only prime that has the property that it is not equal to 97.

I don't claim to have made this puzzle, nor do I have a solution. The is a circle of radius 3 meters with an ant in the centre. The ant picks a direction randomly, all angles have an equal probability. It walks in this direction for 1 meter before fogetting where it was going and chosing another direction randomly. This could go on forever, but on average how many meters has the ant travelled before it exists the circle? Computers are allowed, but I do want to see a mathematical solution.

I'd like to add that this whole area is bizarre, and yet it does seem (you've said so, and SP has shown an example) that one could do some sort of analysis for problems in this area. Thank you, James33, for a glimpse into a weird world! Thats very similar to the way I was expecting it to be solved, a lot of these problems cannot be solved perfectly analyticly but, as you did, can reduce the possibilities of x to a small, finite number of options which can be eliminated without much trouble.

Superprismatic is correct for the first, CaptainEd is correct for the second but there are much better ways of doing these than looking at the graph, also you have to prove there are no other solutions.

Yes that is what I meant.

For those that don't know, the floor function which I will denote by [x] is the largest integer less than or equal to x so for example [6]=6, [8.7]=8 and [-4.3]=5. Solving equations using this function can be very interesting and there are often loads of ways to do it. An easier one to start off: Solve -6x^2+[x^4]+19=0 This one is harder; Solve [x^3]-x^2+6[x]=23

Replace infinity with n and 0 with a. From the first one we have as n-> infinity a->0. 1/n=a so 1/a=n. In this case (equation is equivilent) as n->infinity a->0 so we have the second equation.

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