I find this tricky for some reason. Don't really know how to approach it. I found the answer, but it was by luck.
So first off, I write down:
$$a = 9x +5$$
$$a = 18y + 14$$
$$a = 24z +20$$
Not really sure what I can do with this. What I did figure out (purely intuitively), is that "$z$" has to be a multiple of two (so that the remainders work out). I tested for $z = 2$ and it turns out to work. But I want a solid approach to this problem.
The least common multiple of $9$, $18$, and $24$ is $72$. Notice that one way of interpreting the clue is "$a$ is $4$ less than a multiple of $9$, $a$ is $4$ less than a multiple of $18$, and $a$ is $4$ less than a multiple $24$." All of these can be simultaneously satisfied if $a$ is $4$ less than a multiple of $72$. Namely, $a=68$ works, but so does $a=72\cdot n -4$ for all $n$.