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Old 2007-11-17, 04:08   #1
ShiningArcanine
 
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Dec 2005

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Default Chinese Remainder Problem

http://library.thinkquest.org/22494/...ic/ari_pr1.htm

According to the solutions page, the second problem has no solution. I do not quite understand how they arrived at the second to last sentence of their proof and a C program I wrote for solving such problems says that 59 is the answer. Doing some quick calculations, it would seem that:

59 mod 2 = 1
59 mod 3 = 2
59 mod 4 = 3
59 mod 5 = 4
59 mod 6 = 5

Which makes 59 satisfy the conditions for having solved the problem, which I believe makes 59 + 60k give solutions of the problem for all numbers in Z. Does anyone know how they arrived at their second to last sentence?

Last fiddled with by ShiningArcanine on 2007-11-17 at 04:10
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Old 2007-11-17, 05:10   #2
Peter Hackman
 
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They arrived at the soultion by a horrible mistake.

The solution form suggested assumes that the moduli are relatively prime in pairs; however, (4,6)=2. (And 2 divides 5-3=2, so the data are indeed compatible).
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Old 2007-11-17, 10:01   #3
Peter Hackman
 
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Quote:
Originally Posted by Peter Hackman View Post
They arrived at the soultion by a horrible mistake.

The solution form suggested assumes that the moduli are relatively prime in pairs; however, (4,6)=2. (And 2 divides 5-3=2, so the data are indeed compatible).
And, of course, the problem is perfectly trivial if you replace all right members by -1.
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