elementary number theory - solving and manipulating linear congruences

I need to find a multiple of $5$ call it $5k$ with the following properties:

1. $5k \equiv 3$ mod $6$


2. $5k \equiv 2$ mod $7$

My first instinct is to start with the Chinese Remainder Theorem, but I do not know how to start. Could I get a few hints? Please explain the modular arithmetic manipulations needed to solve this problem.

Answer

To make the computations easy to understand, we first consider general equations.

Let $a,b,n$ be integers.
Suppose gcd$(a, n) = 1$.
Consider the following equation.

$ax \equiv b$ (mod $n$)

Since gcd$(a, n) = 1$, we can solve $ay \equiv 1$ (mod $n$) by Euclid's algorithm.

Then $x = by$ (mod $n$) is the solution.

Let $a,b,n,m$ be integers.
Suppose gcd$(n, m) = 1$.
Consider the following equatiuons.

$x \equiv a$ (mod $n$)

$x \equiv b$ (mod $m$)

Since gcd$(n, m) = 1$, we can find, by Euclid's algorithm, integers $k, l$ such that

$mk \equiv 1$ (mod $n$)

$nl \equiv 1$ (mod $m$)

Then $x = amk + bnl$ (mod $nm$) is the solution.

Now let's solve the given equations

$5k \equiv 3$ (mod $6$)

$5k \equiv 2$ (mod $7$)

We get(by Euclid's algorithm or just by testing)

$k \equiv 3$ (mod $6$)

$k \equiv 6$ (mod $7$)

Then we can apply the above method to find $k$.
Since

$7 \equiv 1$ (mod $6$)

$-6 \equiv 1$ (mod $7$)

$k = 3\cdot7 -6\cdot6 = -15 \equiv 27$ (mod $42$)