User talk:Jaigoda/Math Problem of the Whenever/Archive

I'll figure this out in a second. Testing links though. -- Jai .  -  04:00, May 22 2012 (UTC)
 * 2.  PermaSwag     04:22, 22 May 2012 (UTC)

05/21/2012: Flowers
quick question, when you say the amount he has doubles, do you mean the amount he has after he has laid flowers or before. If it's before it's pretty derp, if it's after then it's hard. Frostels 06:57, 22 May 2012 (UTC)

3.5, leaving 4 flowers. Frostels 07:07, 22 May 2012 (UTC)
 * You're supposed to leave 0 flowers. It's 7,8. 7->14 6->12 4->8 ->0. gg? Charr Centipede 07:24, 22 May 2012 (UTC)
 * I assume if you keep doubling our answers you will get 0 over and over again. Frostels 08:09, 22 May 2012 (UTC)
 * And I meant leave 4 flowers at each grave. Frostels 08:11, 22 May 2012 (UTC)
 * Oh, that was half a flower? And you're right. Charr Centipede 08:19, 22 May 2012 (UTC)
 * Why on earth did you use half a flower... But you are right again Frostels. Any linear combination of starting=3.5, leaving=4 will give you a correct answer, so you might as well pick whole numbers such as 7 and 8, but you could just as well choose 2289 and 2616. A new misery  08:38, 22 May 2012 (UTC)
 * No idea, I just thought 7/4 would work, and half of 7 = 3.5 D: Frostels 08:49, 22 May 2012 (UTC)
 * Well the slightly more rigorous way of doing it would be to consider x = starting flowers, y = flowers left at the graveside. You start with x flowers, double it each river and take away y each time you hit a grave. As such, your number of flowers goes x->2x->2x-y->4x-2y->4x-3y->8x-6y->8x-7y. You want zero at the end, so 8x-7y = 0, therefore x = 7/8.y and this will work for any value of y you fancy to choose, but you should probably choose it such that x and y are both whole numbers. The trivial and dumb answer that I'm surprised no one said is zero flowers, herpaderp. A new misery  08:57, 22 May 2012 (UTC)
 * Question should ask for the lowest number of flowers you can lay on each grave if you start with n numbers of flowers so that you end up with 0 flowers at the last grave, ignoring the trivial solution? -- Relyk 09:16, 22 May 2012 (UTC)
 * Wut. Your Wolfram link makes no sense and doesn't even solve the problem. A new misery  10:48, 22 May 2012 (UTC)

Frosty, Charr (hi minion), and Misery were all correct. Well, sort of for Frosty. I feel bad for the people that get just the stems of the flowers. :D -- Jai .  -  11:32, May 22 2012 (UTC)

Real men do physics in french, not shitty word problems.--TahiriVeila 11:48, 22 May 2012 (UTC)
 * I like how it's dated two days in the future. And if you'd like Jake, I could give you some nice Physics problems to solve. -- Jai .  -  12:37, May 22 2012 (UTC)
 * Its dated two days in the future because he found how to build a time machine with that sheet up there. The fact that they are problems about electrodynamic is completely pointless. [[Image:SilentSign.png|x19px|link=User:Silent]] 12:42, 22 May 2012 (UTC)
 * Or because it was assigned a week ago and is due thursday...--TahiriVeila 12:44, 22 May 2012 (UTC)
 * But that answer is much less cool. D: -- Jai .  -  12:45, May 22 2012 (UTC)
 * ^ And mine is the right one. I have prrrrooofs. [[Image:SilentSign.png|x19px|link=User:Silent]] 12:48, 22 May 2012 (UTC)
 * jake so french. Frostels 13:05, 22 May 2012 (UTC)
 * jake so swiss--TahiriVeila 13:29, 22 May 2012 (UTC)

05/22/2012: Cylinder and String
It must touch the bottom and the top or the center of the circumferences? And does the string make only 7 turns or it can go wherever you want after you completed 7 turns? However if it touches the center and it goes whenever you want its ***  17:09, 22 May 2012 (UTC)
 * You can do it in much less string. Shortest path is a straight line. Also, it only needs to touch an edge of the cylinder, it doesn't have to reach over the top/bottom to get to the center or anything. -- Jai .  -  17:17, May 22 2012 (UTC)

Meh, I don't want to put much thought into this but... answer. Assuming my first thought works with absolutely no sketching to check it makes sense ^___________^ A new misery  17:25, 22 May 2012 (UTC)

So, if I understood how to answer, i'd say *** --Sewa 17:27, 22 May 2012 (UTC)
 * I do not understand why you did the working that way... I just went like this o_O A new misery  17:31, 22 May 2012 (UTC)
 * The same, I just counted the lenght of every spiral and then multiplied by 7 --Sewa 17:34, 22 May 2012 (UTC)
 * I dont know the formula to calculate the spiral, so, whatever... [[Image:SilentSign.png|x19px|link=User:Silent]] 17:37, 22 May 2012 (UTC)
 * Just a rectangular triangle: hypotenuse = sqrt(c^2 + C^2)... C is the circumference and c is the height/7 --Sewa 17:42, 22 May 2012 (UTC)
 * Ah easy. I think those math problems will bring me 10 years forward than my class. [[Image:SilentSign.png|x19px|link=User:Silent]] 17:46, 22 May 2012 (UTC)

I'm probably thinking way too easy
but isn't the answer answer? Brandnew.  20:24, 22 May 2012 (UTC)
 * or maybe answer. idk. Brandnew.  20:26, 22 May 2012 (UTC)
 * Thats my same error. However, no, it seems you need to calculate the spiral with the pitagorean theorem. [[Image:SilentSign.png|x19px|link=User:Silent]] 11:52, 23 May 2012 (UTC)
 * yeah, it's sort of obvious actually, i was just being retarded. Brandnew.  14:36, 23 May 2012 (UTC)
 * It's easy if you open up the cylinder and flatten it: answer. Dzjudz sig.png talk 06:32, 24 May 2012 (UTC)