Chinese Remainder Problem

[ShiningArcanine]

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?

[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).

[Peter Hackman]

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.