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Old 2010-02-18, 07:56   #5
Joshua2
 
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Sep 2004

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Quote:
Originally Posted by Wacky View Post
Rather than "trying" 3, 8, 13, etc., do it algebraically.

If x = 3 mod 5, then x= 5A+3 for some integer A.
Substituting in the other constraints
5A+3 = 1 mod 3
5A+3 = 2 mod 4

But this is the same as
2A = 1 mod 3
A = 3 mod 4

Thus, we have reduced the number of constraints.
When we find permissible values for A, we can substitute and we will have the permissible values for x
This way makes a ton of sense. Is this the CRT as well? Its seems we can't continue with 2A = 1 + 3A and A = 3 + 4A? I think I did that wrong. How about A = 3 + 4B and 2A = 1 + 3 B with two equations and two unknowns? I'll look at the other posts more later.
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