Solving systems of basic congruences

Question

I'm having some difficulty with a problem and I was hoping I could find some help here.

We've been covering congruences in my Discrete Math class, and, while I understand what they mean, I can't seem to solve systems of congruences greater than 2 equations in size.

What I mean is, I can solve problems that look like this:

$x \equiv 4 \mod 5$

$x \equiv 7 \mod 8$

I understand that I can solve this by doing something along the lines of:

$x = 5k + 4$

$5k + 4 \equiv 7 \mod 8$

$5k \equiv 3 \mod 8$

$k = 7$

$x = 5*7 + 4$

therefore $x = 39$

But I can't seem to figure out how to expand this to three (or more) equations, like so:

$x \equiv 4 \mod 5$

$x \equiv 7 \mod 8$

$x \equiv 3 \mod 6$

Disclaimer: I made these numbers up and I'm assuming there's always an answer. If this system doesn't work, any example problem will do. I'm just confused about the process.

Thank you!

Answer 1

The system

$x \equiv 4 \mod 5$

$x \equiv 7 \mod 8$

$x \equiv 3 \mod 6$

should be prepared more before solving.

The reason for this is that two of the modulus share a common factor: 8 and 6 are both multiples of 2.

This could lead to a clash where the two different equations demand contradictory properties mod 2, in this case it's actually fine though:the mod 2 part of both these equations agree so we should cast the 2 part out of one of the equations (this loses no information)

$x \equiv 4 \mod 5$

$x \equiv 7 \mod 8$

$x \equiv 3 \mod 3$

Answer 2

To construct an $x$ equal to $x_1 \mod m_1$, $x_2 \mod m_1$ and $x_3 \mod m_3$.

we would like some expression $x = A + B + C$ such that mod $m_1$ kills $B$ and $C$, mod $m_2$ kills $A$ and $C$ etc.. and gives the correct values ($x_1$, $x_2$, ..)

For the things to get killed in the right we we should have $x = m_2 m_3 A' + m_1 m_3 B' + m_1 m_2 C'$ so e.g. reduction mod $m_1$ gives $m_2 m_3 A'$ so we should set $A' = x_1 \cdot \text{inverse of (}m_2 m_3\text{) mod }m_1$ and similarly with the other two.