What is the largest positive integer that is not the sum of a positive integral multiple of 42 and a positive composite integer?
The solution listed says
The requested number $\mod {42}$ must be a prime number. Also, every number that is a multiple of 42 greater than that prime number must also be prime, except for the requested number itself. So we make a table, listing all the primes up to 42 and the numbers that are multiples of 42 greater than them, until they reach a composite number.

I understand the first sentence's reasoning. However I don't get why every multiple of 42 greater than that number also has to be prime except for the solution. If it's not prime, that doesn't mean it will be a sum of a composite number and a multiple of 42. And why does the answer have to be composite? It seems arbitrary.
(I've completely rewritten this answer, to make it closer to that of the given solution.)
Let us consider such positive integers separately, by their remainder modulo $42$.
So fix an $r$, where $1 \le r \le 42$. (We're picking $1 \le r \le 42$ instead of the conventional $0 \le r < 42$, because we only care about positive numbers and this makes some arguments simpler.) Let us consider numbers $n \equiv r \mod 42$.
That is, we're looking for an $n$ in the sequence $$r, r + 42, r + 2(42), r+3(42), r+4(42), \dots$$ such that $n$ is not the sum of a positive composite integer and a positive multiple of $42$. This is equivalent to saying that $n$, wherever it occurs in this sequence, has no composite number earlier in the sequence than it.
Therefore, the largest "good" $n$ in the sequence is the largest $n$ such that all earlier elements of the sequence are non-composite, which means that $n$ is the first composite number in the sequence.
So for each $r$, where $1 \le r \le 42$, we simply need to write down the sequence $r, r + 42, r + 2(42), \dots$, and record the first composite number that occurs in it. The largest of these is the answer. As a shortcut (as e.g. considering $r=2$ shows that the the answer happens to be at least $44$), we can choose to start with only prime $r$.