Simple Addition!

You can simplify the modular result even further before computing the table.

$18a\equiv 9c\pmod{24}\implies 9(2a-c)\equiv 0\pmod{24}\implies 3(2a-c)\equiv 0\pmod{8}$

Since $\gcd(3,8)=1$ , it follows that $c\equiv 2a\pmod{8}$ . Computing the table now uses this modular result along with the fact that $a,b,c$ are non-negative single digit integers and $a\neq 0$ .

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Also, there's actually a simple way to show beforehand that $b=0$ .

$\begin{aligned}\overline{abc}=34(a+b+c)&\implies 100a+10b+c=34(a+b+c)\\&\implies 9(11a+b)+(a+b+c)=34(a+b+c)\\&\implies 9(11a+b)=33(a+b+c)\\&\implies 3(11a+b)=11(a+b+c)\end{aligned}$

Use Euclid's lemma along with the fact that $11$ is a prime and $\gcd(3,11)=1$ to get that $11a+b\equiv 0\pmod{11}$ , i.e., $b\equiv 0\pmod{11}$ . It easily follows that $b=0$ since it's the only non-negative single digit integer less that satisfies the modular result.