Quote:
Originally Posted by Wacky
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.