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You are not logged in. #1 20060121 17:50:09
Find OneU have two numbers ..u must find a method to find a third number from this two numbers such that afterwards only the 3rd number is provided u should get back the first two number..One more thing to easy the work...the two numbers is in limit 0 to 255 .CAN ANY ONE????? #2 20060121 18:09:36
Re: Find OneC = A + B×256 "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #3 20060121 22:44:49
Re: Find OneHI..It was nice...But would u mind me putting one more constrain?? #4 20060121 23:10:44
Re: Find OneNot if they must be whole numbers (I assumed whole numbers by the nature of your question). "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #5 20060121 23:20:47
Re: Find OneNOOOOOOOOOOOOOOOOOOOOOOOOOOOOO...............The C being real number is not atall acceptable..U CAN DO ANY OPERATION LIKE IN BINARY OPERATIOBS AND CONVERTION ALL>..Can any one.....Its very challenging... #6 20060122 00:10:20
Re: Find OneI would say that with those conditions, it's impossible. Why did the vector cross the road? It wanted to be normal. #8 20060122 02:43:24
Re: Find OneDnt u have any idea like in puzzles some tricks are there to find out number by special operations all....any idea like that in this case? #9 20060122 08:52:23
Re: Find OneIf there are only 256 values for C, then you can only reconstruct 256 cases. If A and B can have 256 independent values each, then there are 65536 cases (as mathsy said) and I so it cannot be done. "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #10 20060122 16:11:52
Re: Find OneAlso discussed here "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #11 20060122 16:24:54
Re: Find OnePoor MathsIsFun has to go running around stitching together crossforum forays by posting links to far flung fragments of desperation. #12 20060122 19:45:43
Re: Find One
Nicely done. I'm assuming you, MathsIsFun, know why this works, but for an explanation on why it works, all you need to know is bit shifting. Bit shifting is the way all multiplication works. If you shift all the bits in a number:
I can prove through the use of the pigeon hole principle, that it is not possible if the function must be onto (valid for every value of C). Once you prove something impossible, there are no such tricks. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #13 20060122 20:01:07
Re: Find One
I may have created a loop! "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #14 20060122 23:29:33
Re: Find One
please proove it 