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Topic: Divisible by 2003 (Read 468 times) |
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Aryabhatta
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Divisible by 2003
« on: Feb 24th, 2005, 10:39am » |
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Yeah, I know, two years late... 
Here is the problem.
s is not divisible by 2003.
There is some integer a such that pa^3 + qa^2 + ra + s is divisible by 2003.
Show that there is some integer b such that sb^3 + rb^2 + qb + p is divisible by 2003.
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| « Last Edit: Mar 6th, 2005, 9:32pm by Aryabhatta » |
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Barukh
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Re: Divisible by 2003
« Reply #1 on: Mar 6th, 2005, 11:29am » |
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Take b = a-1 mod 2003
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on Feb 24th, 2005, 10:39am, Aryabhatta wrote:| Yeah, I know, two years late... |
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…or you would need to wait another 6 years… 
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rmsgrey
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Re: Divisible by 2003
« Reply #2 on: Mar 6th, 2005, 5:19pm » |
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on Feb 24th, 2005, 10:39am, Aryabhatta wrote:| Show that there is some integer b such that sb^3 + rb^2 + qa + p is divisible by 2003. |
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Judging by Barukh's answer, shouldn't that penultimate term be "qb" not "qa"?
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Aryabhatta
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Re: Divisible by 2003
« Reply #3 on: Mar 6th, 2005, 9:32pm » |
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on Mar 6th, 2005, 5:19pm, rmsgrey wrote:
Judging by Barukh's answer, shouldn't that penultimate term be "qb" not "qa"? |
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Yes, thanks for pointing that out. I have modified the original post. I did not catch that even after reading it 2 times!
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Aryabhatta
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Re: Divisible by 2003
« Reply #4 on: Mar 6th, 2005, 9:33pm » |
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on Mar 6th, 2005, 11:29am, Barukh wrote:::
Take b = a-1 mod 2003
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…or you would need to wait another 6 years… |
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Well done!
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