Thelema

Ahh guys this ain't so hard as it looks - a little bit of logic goes a long way. For instance, the 1% has GOT to swing the 98 weeks downwards. By how much? Proportionate to as often as they are sitting at the table next to each other. We already calculated this by calculating how often they sit at the same table. All you have to add now is the variant of them not sitting at adjacent tables or falling in love! &etc

The Café Lover's Problem, PART 2 Poor Marcello, can't sleep over the issue. He wants to increase dramatically the probability of the man and the woman meeting before 98 weeks have passed. He knows he cannot attract more than randomly 0-6 patrons at any one time. He enlists the help of a Mathematician who proves to him that variants of both his and David's solutions actually can't decrease the probability of them meeting. The Mathematician also shows him that if n=# of tables, then p = 1/(28n - 14) where n>3 and 1/98 if n<=3. The Mathematician also shows that subtracting a chair makes no difference to p if n>4, and p= (n-1)/49(2n-1) if n<=4. Therefore he shows them that 98 weeks is the optimum, given that their clientele won't vary - and that they needn't bother re-arranging the shop over the issue. Marcello wakes in the middle of the night and thinks he has a solution: offer half price coffee's to couples that come into the shop. Let's say for the sake of the problem that this will increase the number of couples by 25%, and that couples will occupy both chairs of a table. Let's also say that if the man and the woman are sitting not opposite from each other, this deal will increase their chances of talking and pretending to be a couple by 1% in order to take advantage of the deal, but not if they're both on outer tables - and only if either the man or the woman can sit opposite each other, either on their own tables or on a vacant table. Has Marcello hit on a way to reduce the 98 weeks? By how much? Hint: the distributions 0-6 patrons rather than being random are now calculated to be 0.1215, 0.1215, 0.1366, 0.1366, 0.1544, 0.1544, 0.1751. You may not need this to solve the problem. If you do, double check. For background to this problem, please read the original PART 1 thread. This has also been posted there on p2.

OK this is easy now, the hard part was essentially deriving the formula that Wameron gave. Now we have prove that (2x +1)2mn = nm(m^2 - n^2) is false for all integer values of x. Cancel nm from both (2x + 1)2 = (m^2 -n^2) Can you get it from there?

The Triangle Problem Suppose I construct a right angled triangle so that both sides and the hypotenuse are integers and all relatively prime to one another. Prove that the area of ANY and all such triangles is not an odd product of the longest non-hypotenuse side. edit: added all relatively prime to one another condition - forgot this. clarified that "longest side" refers to longest non-hypotenuse side.

sorry, let's change that I found the following, and leave this to you as a proof: a^2 + b^2 = c^4 for an infinite number of solutions a^2 + b^2 = c^8 for exactly one solution I have edited the previous. Strangely, I have also found a single solution for a^2 + b^2 = c^12 and I postulate that there is a single solution for all c^4n, n>1. In case you're wondering, the c^12 is 450117362^12. But I won't make it so easy for the c^8 solution ( it's farrrr simpler, I guarantee that c is less than 50) Sadly, I have now reached the limit of excel spread-sheets and most online computational tools. Some other interesting things came up.... 1) if not c^2 then must be a^2. 2) Also, if you map the progress of the infinite solutions of a with c where a^2 + b^2 = c^4 then strangely the ratio of successive c's and a's approach quite quickly (3 + √8). For Instance, the first few (a,c) sets are (7,5) (41,29) (239,169): 41/7= 5.857 and 29/5=5.8; 239/41=5.829 and 169/29=5.828 etc.......and (3 + √8) = 5.82843

Federal government's order to ban the national supply and sale of 19 synthetic drugs that closely resemble illegal substances for 120 days. In effect from 19/06/2103 White Revolver Ash Inferno Kyote K2 Black Widow Iceblaze Montana Madness Iblaze Galaxy Ultra Nova Skunk Kronic Vortex Inferno Herbal Incense Buddha Express Black Label Iblaze Tropic Thunder Sharman Slappa King Karma Circus Deluxe Last weekend Home Affairs Minister Jason Clare announced Australia would soon develop legislation to ban all synthetic drugs. A media release announced the ''onus of proof'' would be reversed, until authorities clear them as safe and legal. Hmmm...these are "products" not "drugs" for a start...and as for legislation banning ALL synthetic drugs! LOL. I assume they don't mean paracetamol!

now avast shows WIN32:evo-gen virus linked to forum page thru java platform

. Also, just because something is even and an even product of something does not mean it can't be the odd product of something else. eg 6 is even, and an even product, but it is still the odd product of 2. The question asks that the area cannot be represented as an odd number X 2mn. Also, any even number divided by 2 will not also be even, it can be odd.. The formula is right, so the question becomes: prove that (2x +1)2mn = nm(m^2 - n^2) is false for all integer values of x.

sorry, I should have mentioned, but I realised too late. All sides of the triangle have to be relatively prime to one another. And by "longest side" I mean the longest side that is NOT the hypotenuse. And examples won't do, I need a general proof. I've adjusted the original post and repost here: The Triangle Problem Suppose I construct a right angled triangle so that both sides and the hypotenuse are integers, and that these integers are all relatively prime to one another. Prove that the area of ANY and all such triangles is not an odd product of the longest side (not being the hypoteneuse).

The Café Lover's Problem, PART 2 Poor Marcello, can't sleep over the issue. He wants to increase dramatically the probability of them meeting before 98 weeks have passed. He knows he cannot attract more than randomly 0-6 patrons at any one time. He enlists the help of a Mathematician who proves to him that variants of both his and David's solutions actually can't decrease the probability of them meeting. The Mathematician also shows him that if n=# of tables, then p = 1/(28n - 14) where n>3 and 1/98 if n<=3. The Mathematician also shows that subtracting a chair makes no difference to p if n>4, and p= (n-1)/49(2n-1) if n<=4. Therefore he shows them that 98 weeks is the optimum, given that their clientele won't vary - and that they needn't bother re-arranging the shop over the issue. Marcello wakes in the middle of the night and thinks he has a solution: offer half price coffee's to couples that come into the shop. Let's say for the sake of the problem that this will increase the number of couples by 25%, and that couples will occupy both chairs of a table. Let's also say that if the man and the woman are sitting not opposite from each other, this deal will increase their chances of talking and pretending to be a couple by 1% in order to take advantage of the deal, but not if they're both on outer tables - and only if either the man or the woman can sit opposite each other, either on their own tables or on a vacant table. Has Marcello hit on a way to reduce the 98 weeks? By how much? Hint: the distributions 0-6 patrons rather than being random are now calculated to be 0.1215, 0.1215, 0.1366, 0.1366, 0.1544, 0.1544, 0.1751. You may not need this to solve the problem. If you do, double check it.

The Café Bella-Roma love-seats A man and a woman go to Café Bella Roma either during the whole morning or whole afternoon, on just one random day every week. The thing is, if this man and woman ever sit opposite each other, they will fall in love, and the only chance they get to do this is sitting opposite each other at the Café Bella-Roma. They choose where to sit randomly. The Café has 3 tables, each with 2 chairs opposite each other, and never has any other patrons sitting at the tables when either the man or woman are there. What is the average number of days it will take them to sit opposite each other and fall in love? Suppose now that the Café has up to 6 patrons at any one time, and there is equal probability that the seats in the Café are occupied with any of 0 -6 of them. Now what is the average number of days it will take them to sit opposite each other and fall in love? Marcello, one of the owners, (both of whom secretly know all of this), at the beginning, Marcello wants to make an even bet that they will have met by a certain number of weeks. What number should he bet on so that his chances of winning the bet are 50/50?" The other owner, David, is more civic-minded and wants to know what the best thing he can do is to increase the chances of the man and the woman falling in love. His idea is to put in an extra table. Marcello reckons they should take a chair away from one of the tables. For this part, assume that still randomly 0-6 patrons are occupying the tables at any one time. There is standing room at the bar for extra's. Which idea is the best? edit: part c and d have been reworded for greater clarity.

Yes, you're right, it's a case of bad wording. I should have said: "At the beginning, Marcello wants to make an even bet that they will have met by a certain number of weeks. What number should he bet on so that his chances of winning the bet are 50/50?" I got the counting wrong, CBL is right. The correct solution is: 2) for 0,1,2,3,4,5,6 strangers respectively, the probability that if the man and woman are also in the café that they sit opposite each other is (1/5, 1/5, 1/5, 1/5, 1/5, 0, 0) each x 1/14 = (1/70, 1/70, 1/70, 1/70, 1/70, 0, 0). Add, divide by 7, convert to a reciprocal = 1/98. So 98 weeks. 3) How many weeks should Marcello bet for them meeting to have an equally-weighted chance in scenario 2? When (97)^n / (98)^n = 0.5 ie n = log0.5/log(97/98) = 67.58 weeks 4) a) OK removing a seat, with the SAME DEAL concerning 0-6 strangers (another thing I should have specified): For 0,1,2,3,4,5,6 strangers respectively, p that if man and woman are in the café that they sit opposite each other is (1/5, 1/5, 1/5, 1/5, 0, 0, 0) = 122.5 weeks Adding a table gives (1/7, 1/7, 1/7, 1/7, 1/7, 1/7, 1/7) = 98 weeks. David's estimate is better than Marcello's, but no improvement. If we use Bal-sac’s method, that when removing a chair we have randomly 0-5 other patrons, and when adding a table we have 0-8 randomly other patrons: a) (1/5, 1/5, 1/5, 1/5, 0, 0) = 105 weeks (1/7, 1/7, 1/7, 1/7, 1/7, 1/7, 1/7, 0, 0) = 126 weeks. Then Marcello still comes out on top, but both are worse. Please note for the record of vanity that Balzac’s estimates are close but wrong. Next, The Café Lover's Problem, Part 2. Note: the original problem has been reworded slightly in light of the constructive insights of Balzac. PS Don't you find it interesting, that no matter how many people are sitting at the tables, as long as n < 2t-1, then the probability is exactly the same for the lover's meeting? One way to think of this is that the reduction in probability of a table being occupied is exactly balanced by the increase in probability of them being constrained to less options in order to not sit next to each other.

whoops, yes weeks, sorry. by even bet, I mean, the point at which the probability they have not yet met equals the probability that they have. i.e. the point at which his chance of doubling his money are equal to his chance of losing his money = even bet. 5/24 comes from: Case 5 free seats (1 other sitting) : 4/5 chance she will select an empty table. Guy then has 1/4 chance of sitting across from her. P = (4/5)*(1/4) = 4/20 Hmmm maybe I did my counting wrong, I'll double check it.

solution: 1) there's a 1/14 chance of them being in the café at the same time on any given shift within any given week. There's a 1/5 chance they'll sit opposite each other...thus 1/70. 70 days average. 2) for 0,1,2,3,4,5,6 strangers respectively, the probability that if the man and woman are also in the café that they sit opposite each other is (1/5, 5/24, 1/5, 3/10, 1/5, 0, 0) each x 1/14 = (1/70, 5/336, 1/70, 3/140, 1/70, 0, 0). Add, divide by 7, convert to a reciprocal = 1/88.421 Thus in this situation 88.4 days average. 3) How many days should Marcello bet for them meeting to have an equally-weighted chance in scenario 2? When (87.421)^n / (88.421)^n = 0.5 ie n = log0.5/log(87.421/88.421) = 60.94 days. 4) I'll leave the rest to you....

Pardon me for saying, but it sounds like you didn't get proper representation. Did you check out the warrant? etc

ahh the myth of willpower, hey? Substances can erode this "willpower", it must be remembered, rendering people unable to seek help.

There are hard-to-find variants of fentanyl on the marketplace, never heard of them in Oz tho'. I wonder if these guys are going to target natural products too?

yeah that's great input too.

please note original post edited to include current law prohibiting import of these plants.

I fear that Nabraxas might be, you know, :crux:

Hofstader is hypermanic. That is plain...

typical of hypermania. Have you ever tried to read one of Hoftstader's works? He's a bit the same...

sad to learn today that MXE has just been listed as a customs-prohibited drug.

Is this why the body produces mucus in response to bacterial infection of the mucosa?