This one is pure math.. ;o I also brought the difficulty down a notch from the last one.

Edit: On #2, Circle A and Circle B do not share centers.
#1 was solved by Devourer
#2 by Azala
#3 by Vrael

Post has been edited 4 time(s), last time on Jun 26 2010, 4:15 pm by BeDazed.

#1: Not sure what you mean by 'most concise manner' but I think the answer is:



Edit: Woops.

Problem #1.


I don't know what you mean by most concise manner but I've got the exact number and an expression that represents it so if that isn't enough please explain.

Oh, fuck, I misread. I thought it said to find the sum of the 100th row. Woops.

Doesn't #2 imply #3? If a set has two consecutive intergers, then it automatically is not comprised entirely of inconsecutive (you mean non-consecutive) intergers.

Edit:
And for anyone who didn't know, (e^x) = lim n-> infinity of 1 + x/1! + (x^2)/2! + (x^3)/3! + . . . + (x^n)/n!
So essentially, (e^x)^(5/x). I have no idea why BeDazed expressed the problem using the taylor expansion, unless perhaps its convenient in some way to the solution of the problem.

Also, for the hyperbola in #2, do you mean the surface area of the triangle between the asymptotes and the x or y axis? Because the asymptotes themselves form a cross, not a triangle.

Post has been edited 2 time(s), last time on Jun 20 2010, 11:38 pm by Vrael.

For problem number 3, the conditions 2 and 3 imply each other- yes. But I wrote both of them just to clarify.

Asymptotes and a tangent line. It will form a triangle. And, I didn't intend that to be a Taylor expansion. It was supposed to be all high school material. It was supposed to express lim[n->inf](1+x/n)^n in binomials.


(1+x/n)^n*1/x*x = e^x = 1 + nC1(x/n)^1 + nC2(x/n)^2 + ..... nCn(x/n)^n = 1 + x/n*n!/1!(n-1!) + x^2/n^2*n!/2!(n-2)! + ... + x^n/n^n*n!/0!n!
And since n is going infinity, it basically approximates to a Taylor, or Laurent series. But you don't even have to get that far to know that it was a binomial expression for (1+x/n)^n.

Post has been edited 5 time(s), last time on Jun 21 2010, 7:24 am by BeDazed.

On Problem #1:
The sum of each line is the double of the line above + 2.
2*2^100 + 100*2 = 2.5353012 * 10 ³⁰



EDIT:
Your way is the right way, or not? I mean, test it with 2x2^3 + 2*3 = 22. 2+6+14 = 22. This is for row three. I think this is dynamic, no?
However, why did you chose 101?

(I forgot how to "read" that * 10 ³⁰ thing so I just wrote it down)
(Being at school with no tasks, haha)

I don't get task 3

Post has been edited 3 time(s), last time on Jun 21 2010, 7:14 am by Devourer.

Find the sum UP TO the 100th line.

Maybe I'm getting this wrong, however:
2*2^99 + 99*2 = 1.2676506 * 10 ³⁰

7897 is my answer for #3. I'm not satisfied with how I got it, but I think it's right.

It's wrong. There maybe combination you are overlooking...
Of course.. if you did it the hard way, the probability of the result being wrong is pretty high.

err, maybe 8625

and yeah, I think I did it the hard way

http://projecteuler.net/

Right. I have no knowledge of programming though. So this obviously doesn't fit me.

Correct <_<.

..Now for #2. Oh, just a hint though. The locus will probably have one more variable then x and y.

So, what was the easy way? I counted them all by hand lol.

Edit:
By hand I mean scratch paper + calculator for the multiplication.


And ummm... I doubt anyone has any idea wtf you're asking for in #2. You're putting circle B somewhere in circle A non-concentrically, but I have no idea where circle B's center is relative to circle A's. I don't think you gave us any information about where the A and B circles centers are located.

Post has been edited 2 time(s), last time on Jun 24 2010, 4:43 am by Vrael.

Oh right. Just place them wherever you want it. Well, that's two more variables. I guess you could just place the relative centers of those two circles on a single axis, then say it rotates to save you the trouble of angles That should make a total of 5 variables.

But in all seriousness... it would be a lot more fun if more people participated <_<

Well, the way I did it was to count all the set of integers that had partial or fully consecutive numbers, and subtracted that number from 19C5. It's probably way shorter than counting all the sets that aren't consecutive. And since numbers are meaningless, I made 19 slots and solved it by calculating the number of ways you can place 5 markers partial or fully consecutive. ;o I'm wondering if you did it similarly though.

Post has been edited 3 time(s), last time on Jun 24 2010, 7:59 am by BeDazed.