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 2007-11-17, 04:08 #1 ShiningArcanine     Dec 2005 22·23 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 2007-11-17 at 04:10
 2007-11-17, 05:10 #2 Peter Hackman     Oct 2007 linköping, sweden 1416 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 5-3=2, so the data are indeed compatible).
2007-11-17, 10:01   #3
Peter Hackman

Oct 2007

2010 Posts

Quote:
 Originally Posted by Peter Hackman 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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