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You are not logged in. #476 20060302 02:40:37
Re: Problems and Solutionsmathsyperson, a good attempt! I shall post the proof after a few days (during the weekend, when I am free). Character is who you are when no one is looking. #477 20060303 17:44:03
Re: Problems and SolutionsProblem # k + 110 Character is who you are when no one is looking. #478 20060712 23:19:14
Re: Problems and Solutions
I think that you would have to add 5 liters of water to make the solution 20%. #479 20060713 00:34:31
Re: Problems and SolutionsDaisy  that is correct. 9liters(the new amount of water) is in fact 1/5 of 45liters(the new total) #480 20060723 20:25:17
Re: Problems and Solutions#k+110 IPBLE: Increasing Performance By Lowering Expectations. #481 20090105 10:03:07
Re: Problems and Solutions
The trick is to use complex numbers – or Gaussian integers (complex numbers with integer real and imaginary parts). Thus, factorizing in the ring of Gaussian integers, we have Since and we have Now we multiply the factors in a different order! And it is clear that if , both and are integers greater than 1. Hence is composite if ! Last edited by JaneFairfax (20090105 10:48:44) #482 20101022 12:00:42
Re: Problems and SolutionsHi ganesh; It is easy to spot a recurrence form! We solve this by standard means: Do not bother to simplity. Just substitute 5 for n. There are 5 guys remember. You get the fraction: Set it equal to y, ( I like x and y ). The step is justified because 1024 x  8404 is obviously a multiple of 3125. Rearrange to standard form for a linear diophantine equation. Solve by Brahmagupta's method, continued fraction, GCD reductions... Whatever you like. You just need 1 solution! I have a small answer found by trial and error of ( x =  4 , y =  4 ). Now if a linear diophantine equation has one solution it has an infinite number of them. Utilize Bezouts identity, which says if you have one answer (x,y) then you can get another by: Plug in x = 4, y =  4, a = 1024, b = 3125 Now it has been solved in terms of a parameter k. Substitute k = 1,2,3,4,5 ... to get all solutions. k = 1 yields (3121, 1020) which is the smallest positive solution. So there are 3121 coconuts in the original pile. It was not necessary to even know of Bezouts identity. From equation A you have the congruence: Once one answer of x = 4 was found you just have to add 3125 to get x = 3121. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 