<t>QuoteOriginally posted by: JerryGarciathe answer to the original problem (43,948,843 bet cards) was quoted already several times assuming that positioning of right numbers is irrelevant:- there is eactly one bet card choice with 5 right numbers and - there are 5 x (90-5) = 425 bet card choices wi...

Certainly, as the algorithm would terminate (at 1) for an odd n if it does so for 2n...

True for example is b= a^3, in which case the ratio is a^2.

<t>e^pi > pi^e <==> pi > e Log(pi) <==> pi/e > Log(pi) <==> pi/e > Log(e(pi/e)) <==> pi/e - Log(pi/e) > 1 We only need to show that if 0<x<1 then (1+x) - Log(1+x) > 1 that follows from the development Log(1+x) = x - x^2/2 + y^3/3... for 0<x<1 </t>

Get the feather electrically charged and attach it to oneend of a long strip of cardboard you have prepared out of the box... Edit: Just noticed MattF preceded me in this...

<t>There is some proposed method, for all k. I will concede that the search required is not computationallypractical for even moderate k, to find such a linear transformation that cuts away some choice of 1/4th ofthe points (as you must get all the solutions, and then pick an appropriate 1/4th subse...

<t>First of all, this is not my approach as you say, I am only filling some gaps in the statements made bywannabequantie's friend so we can all understand them better. I am not taking ownership of anybody's method in doing so.My purpose is one of clarification, a break in the skepticism we perhaps a...

<t>hi wannabequantie, I think I can justify pretty much all of what you stated so far as follows:Do the following transformation of variables (you can think of it as a congruenceequation in the Gaussian integers, to be solved for (u,v) as stated x+iy = 1+0i + (u+iv)*(2+2i))x= 1 + 2(u-v)y= 2(u+v)Now ...

<t>QuoteOriginally posted by: pk14Please notice that once you wroteX^2+y^2=somethingthen the circle is symmetric with respect to the integer pt i.e. if (a,b) is on the circle then so does (-a,-b)Thus the # of interge pts on the circle must be even.The problem with the suggested solution is that, if ...

12^1000000 = 1762...

2 log(2) - 1 (twice the integral of (1/2-a)/(1/2+a) between 0 & 1/2, 1/2-a is the length of the short side...) Edit: I misunderstood the question. I went for the ratio of the short to the long side...

<t>Probably a good place to stop this discussion about nuances...This problem begs the question as to whether one can get the complete p-power expansion of (n, p), in closed form (based on that of n).Or if not, at the very least, a few more terms. Edit: no p-adic expansion per se, as this would put ...

I am not saying that you are using Wilson in the numerator, I am saying that I am,as I take the product of an equivalent set to 1,...,p-1 and replace it by -1 mod p in the numerator.Just a comment, not a criticism...

<t>Here are the details that justify 'Wilson' literally in my argument, if you are curiousAfter you cancel the p's, (n,p) maps to ( -1* [n/p] )/(-1) = [n/p] in the finite field Fpby the canonical map Z->Fp. The -1's come from Wilson.So since the difference of two integers maps to 0, it is divisible ...