So I know the answer for this is $2$, but based on my own work, I can't get to that solution. I haven't done a gcd before where $b>a$. I thought I could just flip the numbers and use the same method but that didn't seem to work. Here's what I have so far, what I am doing wrong?
$$\begin{align} \mathrm{gcd}(1714, 1814) &= \mathrm{gcd}(1814, 1714) \\ \mathrm{gcd}(1814, 1714) &= (1714, 100)\\ &= (100, 14)\\ &= (14, 9)\\ &= (9, 5)\\ &= (5, 4)\\ &= (4, 1)\\ &= (1, 0)\\ &= 1\\ \end{align} $$
I basically tried using the Euclidian algorithm method where you keep doing long division into each number to get the remainder and continue with that process.
\begin{align} GCD(1814,1714)&=(1714,100)\\ &=(100,14)\\ &=(14,2)\\ \end{align} So, $2$ is the $GCD$