$$23d \equiv 1 \pmod{40}$$ $$73d \equiv 1 \pmod{102}$$
How can I solve this? $40$ and $102$ are not coprime, so I figured I can factor the moduli:
$$23d \equiv 25 \pmod 8$$ $$23d \equiv 26 \pmod 5$$ $$73d \equiv 25 \pmod 2$$ $$73d \equiv 25 \pmod 3$$ $$73d \equiv 18 \pmod{17}$$
All right, now - if I have both $23d \equiv 25 \pmod 8$ and the $23d \equiv 25 \pmod 2$, can I throw the first one out?
If so, I later have:
$$23d \equiv 26 \pmod 5$$ $$73d \equiv 25 \pmod 2$$ $$73d \equiv 25 \pmod 3$$ $$73d \equiv 18 \pmod{17}$$
What should I do next if I wanna use Chinese reminder theorem to solve this? I mean, how can I get rid of the $23$ and $73$?
For example, let's take $23d \equiv 26 \pmod 5$. I'd have to add $5$ to $26$ until I get the number that $23$ is divisible by. Is there a way to evaluate this quickly?
The way you express your congruences is rather unconventional. Given that $$23d\equiv1\pmod{40},$$ $$73d\equiv1\pmod{102},$$ and that $40=2^3\times5$ and $102=2\times3\times17$, it follows that $$23d\equiv1\pmod5,$$ $$23d\equiv1\pmod8,$$ $$73d\equiv1\pmod2,$$ $$73d\equiv1\pmod3,$$ $$73d\equiv1\pmod{17}.$$ It is common practice to express a congruence like $23d\equiv25\pmod8$ as $23d\equiv1\pmod8$, or even $-d\equiv1\pmod8$. It makes the modular arithmetic easier to do by heart.
Each of these congruences can further be reduced. For the first, note that $$23d\equiv3d\pmod5,$$ and that $3d\equiv1\pmod5$ implies $d\equiv2\pmod5$. You can do the same for the other moduli, and then use the Chinese remainder theorem to find all possible values of $d$.
As for your remark; if you have $23d\equiv25\pmod8$ then it follows that $23d\equiv25\pmod2$, but not conversely. So you cannot throw the first one out, but you can throw the second one out.