Solving system of linear congruence equations

Solving system of linear congruence equations

$$23d \equiv 1 \pmod{40}$$ $$73d \equiv 1 \pmod{102}$$
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 those $23s$ and 473s$?
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?