Inverse modular arithmetic method

Question

We have that if $nd\equiv k \mod p$, then $\dfrac{k}{n}\equiv d \mod p$. This is useful to solve small versions of CRT, for example:

$$2\cdot 7\equiv3\mod11$$

so

$$\dfrac{3}{7}\equiv 2\mod11$$.

It is easy to find $3|(2\mod11)(=24)$ so we have:

$$\dfrac{1}{7}\equiv 8\mod11$$, which gives us $0\mod7, 1\mod11$as $56$.

Does my method extend easily into larger $p$ or general $n$?

ADDENDUM

So far I have if we wish to find $x=(0,1)\mod(p,q)$ with $p,q$ primes. We can use this to find $y=(1,0)\mod(p,q)$ and using the linear properties of the CRT $(a,b)\mod(p,q)=ax+by\mod pq$.

Given $q\gt p$, let $q=\alpha p-\beta$, so that $\alpha p\equiv\beta\mod q$, and $\dfrac{\beta}{p}\equiv\alpha\mod q$. As we know $\alpha\mod\beta$ and $q\mod\beta$, we have a simpler solution to find a RHS divisible by $\beta$, say $k\beta$ which gives us $\dfrac {1}{p}\equiv k\mod q$, and hence $pk$ is $(0,1)\mod(p,q)$.

Answer 1

If you're writing $q=\alpha p-\beta$, and then considering $q,p$ modulo $\beta$, then you're basically doing the beginning of the Euclidean algorithm—which is the standard way to find modular inverses and hence to implement the Chinese remainder theorem. As far as I can tell, your method is basically: do the standard method, but skip a couple of steps at the end when you can identify a modular inverse by mental arithmetic. Nothing wrong with this (and it has nothing to do with $q,p$ being prime), but as the numbers get larger, you'll have to do most of the Euclidean algorithm before mental arithmetic will kick in. So I don't see much advantage to this method.