20071117, 04:08  #1 
Dec 2005
2^{2}×23 Posts 
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 20071117 at 04:10 
20071117, 05:10  #2 
Oct 2007
linköping, sweden
2^{2}×5 Posts 
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 53=2, so the data are indeed compatible). 
20071117, 10:01  #3 
Oct 2007
linköping, sweden
2^{2}·5 Posts 
And, of course, the problem is perfectly trivial if you replace all right members by 1.

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