Greatest common divisor (gcd) and lowest common factor (lcm) of numbers $a$ and $b$ are related by
$$a\times b= \mathrm{gcd}(a,b) \times \mathrm{\ell cm}(a,b)$$
The gcd function (which takes two inputs) has the following properties:
$$\mathrm{gcd}(a,b) = \mathrm{gcd}(b,a)$$ $$\mathrm{gcd}(a,b) = \mathrm{gcd}(a,b-a)$$ $$\mathrm{gcd}(a,0) = a $$
By repeated application of the second rule, we have
$$\mathrm{gcd}(a,b) = \mathrm{gcd}(a,b \bmod a)$$
Example
$$\mathrm{gcd}(300,144) = \mathrm{gcd}(144,300 - 2\times 144) = \mathrm{gcd}(144,12) = \mathrm{gcd}(12, 144 - 12\times12) = 12 $$
Another way to express the above computation is
$$\mathrm{gcd}(300,144) = \mathrm{gcd}(144, 300 \bmod 144) = \mathrm{gcd}(144,12) = \mathrm{gcd}(12, 144 \bmod 12) = 12 $$
gcd(300,144) # Take two numbers as input
gcd([300,144]) # Take a single input which is a list of two numbers
lcm(300,144)
lcm([300,144])
The list version is useful when more than two numbers are considered.
It turns out the algorithm produces a pair of integers $m,n$ such that $$300m + 144n = 12$$
It is always true that one can get a pair of integers $m,n$ such that $$ma + nb = \mathrm{gcd}(a,b)$$
Do you still remember the coin exchange problem in the test? This is the key to solve it.
To solve the equation $3x+3 = 4 \pmod{13}$ by hand as usual to get $ 3x = 1 \pmod{13} $.
By trial and error, we know that $3\times9=27=1\pmod{13}$.
Hence, $x=9 \pmod{13}$ is a possible solution.
Alternatively, note that $\mathrm{gcd}(3,13) = 1$. We can find a pair $m,n$ such that $$3m+13n = 1$$
The pair and the gcd can be computed at the same time in sage:
d,m,n = xgcd(1,3,13)
Giving $$(-4)\times 3 + (1) \times 13 = 1$$
by taking modulo 13, we have $$(-4)\times 3 = 1 \pmod{13}$$