Math Is Fun Forum

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If 65 67 83 is ascii, it would be ACS. If we relate that to time, the closest I get is the Australian Central Standard time. That would be UTC/GMT + 9:30h. Just throwing something out there... does Australia say anything at all?

Also, I was thinking that 65 67 83 could be rows or something in Pascal's triangle. 65 would be row 6, number 5 in the sequence. But 67 doesn't quite work out.

The one thing that comes to mind is that they represent degrees, minutes and seconds like some sort of location. As with 9 000, I'm clueless.

check this out:
http://www.research.att.com/~njas/seque … &go=Search

The only matching (integer) sequence would be the first of the three, none of the other are followed by either 11,12,13 or 14. But then, the rest are still wrong... If it really is a sequence (can't see it myself), nice find.

That symbol, uppercase phi, is the reciprocal of the golden ratio. I think he missed something there

Anthony, take a look at this:

   1.00000000...
-  0.99999999...
= 0.00000000...

Since the number of zero's is infinite, it is the same as writing 0 as there can't be anything after the infinitely many zero's (just as 1.000... is the same as 1, or 5.5999... = 5.6). Thus, there is no difference between 1 and 0.999... they are the same thing!

As pointed out already, there is no such thing as 0.0000...1. The only thing it proofs is that you don't understand the concept of infinity.

If you still don't believe it, prove us wrong. Otherwise none will take you seriously.

It doesn't seem to work in Firefox, try IE.

Sure thing. I'll e-mail my solution soon, hoping that, in return, you'll e-mail me yours Hopefully you can confirm it's correct.

If it's a multiple of 3, it's between 50-59, but not all numbers between 50-59 are multiples of 3. Cancel out those numbers. For the rest between 50-59, how many are multiples of 4? Cancel out those who are not multiples of 4. You following so far?

Good. So what can you conclude now? Exactly, the number is not a multiple of 3 because none of the numbers between 50-59 remains. Cancel out all remaining numbers that are multiples of 3.

Remember, it has to be a number between 50-79, because it's the only definitions you've got.

Remove your second criteria, you've used it up already. Continue the same way with your other two criterias

Edit: Bah, Rod gave the answer away, lol.

So, basically, a prisoner must be absolutely sure that everyone has visited the central room at least once when he says he wants to go home. But they can only contact eachother through the switch... So they must be clever enough to figure out what n is, right? or perhaps be able to count to 100 through the switch? I guess that's two options.

We might have to group the prisoners somehow before they are being put into the cells.

Is the switch on or off from the start?

No problem : )

I don't know what L'hopitals rule is, but here's what I did (hoping that I didn't use it!):

I'm guessing you use ε on your y-axis, and δ on your x-axis (like we do over here). If so, consider this problem (same principle, I'm just using x² instead):

Show that x² --> 4 when x --> 2.
We need to show that, as soon as we've picked our ε>0, we can find a δ>0 so that |x²-4|<ε when 0<|x-2|<δ. Don't forget the 0 right there!

|x²-4| = |x+2|*|x-2|

Let ε>0. If we demand |x-2|<1 (namely that 1<x<3), we get |x+2|<5. If we also demand that |x-2|<ε/5 (notice why we picked 5 below, and not any other number), we get:

|x²-4| = |x+2|*|x-2|<5*ε/5=ε

Thus, if δ is the least of the numbers 1 and ε/5, we get that |x²-4|<ε when 0<|x-2|<δ. Maybe someone could pull out a graph of the definition to make things easier...

That's all there is, I think. I don't see where you got that ε/3=δ from though?