Quote:
Originally Posted by Joshua2
1.
2. I think that is what we did before, reduce by one.
A = 2  B = 3 + 4B so 2  3 = 5B or 5 = 5B so B = 1. So is that the right idea?

No. I think that you may be confusing the "algebraic equality" (as in X = 5 A + 3) with the "modular equality" (as in X = 3 mod 5).
They are somewhat different concepts. Therefore, for clarification, I will use "==" in the latter case.
We are reducing the number of constraints by one by making a substitution that causes one of the constraints to be met for all values of the unknown.
We started with:
Code:
Find the set of integers "X" such that
X == 1 mod 3
and
X == 2 mod 4
and
X == 3 mod 5
We made the substitution
X = 5 A + 3
This transformed the last constraint into
which is true for all integers A
That left us with the equivalent problem:
Code:
Find the set of integers "A" such that
5 A + 3 == 1 mod 3
and
5 A + 3 == 2 mod 4
or, equivalently:
Code:
Find the set of integers "A" such that
2 A == 1 mod 3
and
A == 3 mod 4
So, continuing this procedure:
We let A = 4 B + 3 which will meet the last constraint for all integers B, leaving only one constraint (mod 3).
Then we let B = 3 C + … (something), which will be true for all integers C.
Finally, we combine all of the substitutions to get one substitution
X = f(C), which meets all of the constraints for any integer C.
As you already know, from other posts, this will be
or equivalently,
However, you should work out the details to derive the answer yourself.
There are a couple of aspects of the derivation that might catch you unaware. You should also look to see similar coefficients in the CRT solution. Observing these similarities might give you a greater insight into "why the methods work"