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Topic: (X-A)*(X-B), etc. (Read 504 times) |
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evergreena3
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(X-A)*(X-B), etc.
« on: May 13th, 2004, 2:11pm » |
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Solve the following:
(X-A)*(X-B)*(X-C)*(X-D)...(X-Z)
EverGreenA3
(p.s. I did a quick search for this, and didn't find it listed anywhere. Please forgive me if it is a duplicate.)
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towr
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Re: (X-A)*(X-B), etc.
« Reply #1 on: May 13th, 2004, 3:08pm » |
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on May 13th, 2004, 2:11pm, evergreena3 wrote:| (p.s. I did a quick search for this, and didn't find it listed anywhere. Please forgive me if it is a duplicate.) |
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You're forgiven 
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Wikipedia, Google, Mathworld, Integer sequence DB
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ThudnBlunder
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Re: (X-A)*(X-B), etc.
« Reply #2 on: May 13th, 2004, 3:31pm » |
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Quote:
No, he's not. 
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| « Last Edit: May 13th, 2004, 3:32pm by ThudnBlunder » |
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Icarus
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Re: (X-A)*(X-B), etc.
« Reply #3 on: May 13th, 2004, 4:36pm » |
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Besides which, there is nothing to "solve". This is an expression, not an equation! "Simplify" would be the most appropriate instruction, though "evaluate" does also work in this case.
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I wonder: Which is larger
When their digits are reversed? " - Anonymous
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kellys
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Re: (X-A)*(X-B), etc.
« Reply #4 on: May 13th, 2004, 11:40pm » |
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Or if you really want to be mean, I like 'expand'
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Benoit_Mandelbrot
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Re: (X-A)*(X-B), etc.
« Reply #5 on: May 14th, 2004, 9:29am » |
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Expanded is :: 0!!!!! This is because you will eventually get to ...(X-X)..., which is zero, and anything times zero is zero! The answer is zero. ::
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| « Last Edit: May 14th, 2004, 9:29am by Benoit_Mandelbrot » |
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Because of modulo, different bases, and significant digits, all numbers equal each other!
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