Subject (4). Consider the sets:

$A=\{5 p+7 q \mid p, q \in \mathbb{N}\}, B=\{5 p+7 q \mid p, q \in \mathbb{Z}\}, C=\{35 p+14 q \mid p, q \in \mathbb{Z}\}$.

a) Determine $\operatorname{card}(\mathbb{N} \backslash A)|\mathbb{N} \backslash A|$.

b) Determine $\mathbb{Z} \backslash B$.

c) Determine the set $C$.
Solution.

a) The official solution takes advantage of the relatively small values 5 and 7 to arrive, through the calculation of relatively few cases (but without details, only a kind of handwaving), at the incorrect result $133^{3}$. In fact, it is a well-known result by Sylvester (known in Anglo-Saxon countries under the infamous name of Chicken McNugget Theorem). This result states that for $m, n \in \mathbb{N}^{*}$ coprime, all natural numbers greater than or equal to $(m-1)(n-1)$ can be represented in the form $m p+n q$, for some $p, q \in \mathbb{N}$, and among the $(m-1)(n-1)$ smaller natural numbers, exactly half are thus representable. Therefore, $|\mathbb{N} \backslash A|=\frac{(5-1)(7-1)}{2}=12$. By the way, why $\operatorname{card}(\mathbb{N} \backslash A)$, when $|\mathbb{N} \backslash A|$ is the standard notation?[^1]b) Having $\gcd(5,7)=1$, it follows that $B=\mathbb{Z}$ directly from Bézout's relation (a result from the textbook, but omitted in the official solution).

c) Having $\gcd(35,14)=7$, it follows that $C=7 \mathbb{Z}$ from Bézout's relation. The official solution contains the statement $7 \mathbb{Z} \subseteq C$ twice, incorrectly; it should have been written $C \subseteq 7 \mathbb{Z}$.

I am very curious how a solution invoking Sylvester's result, extremely well-known in literature and in competition practice, would be graded. Most likely with zero points ... isn't that so?