parmenides51 wrote: Let $M$ be a set containing positive integers with the following three properties: (1) $2018 \in M$. (2) If $m \in M$, then all positive divisors of m are also elements of $M$. (3) For all elements $k, m \in M$ with $1 < k < m$, the number $km + 1$ is also an element of $M$. Prove that $M = Z_{\ge 1}$. (Proposed by Walther Janous) Since $2018 \in M$, we have that $1,2,4,1009 \in M$. So, $2,4 \in M \Rightarrow 9 \in M$, and $3 \mid 9$, so $3 \in M$. Now, $2,3 \in M \Rightarrow 7 \in M$, and $2,7 \in M \Rightarrow 15 \in M$ and since $5 \mid 15$, $5 \in M$. By choosing $m=2<n$, we obtain that $2n+1 \in M$, for $n \geqslant 3$, that is, all odd integers greater than $7$ belong to $M$. But, $1,3,5 \in M$, so all odd integers belong to $M$. Now, for $t>1$, by choosing $k=2t-1,m=2t+1$, $4t^2$ belongs to $M$ and since $2t \mid 4t^2$, $2t \in M$ for each $t \geqslant 2$. It suffices to show that $2 \in M$, but that's obvious. The proof is complete.

I liked this problem a lot. Although it is easy, that's rare for me to say since it's Combinatorics. Central claim. If $1, 2,\dots, n\in M$, then $n+1\in M$ for any $n\geq 5$. Proof. We split the proof into two cases. Case 1. If $n$ is even, then $1<2<\frac{n}{2}<n$ and (since $\frac{n}{2}<n$ then $\frac{n}{2}\in M$) we obtain that $2\cdot\frac{n}{2}+1=n+1\in M$ by the third condition, as needed. Case 2. If $n$ is odd then $1<2<\frac{n+1}{2}<n$ and (since $\frac{n+1}{2}<n$ clearly $\frac{n+1}{2}\in M$) we obtain that $2\cdot\frac{n+1}{2}+1=n+2\in M$. Now since $1<n<n+2$ and $n, n+2\in M$, by the third condition we have $n(n+2)+1=n^2+2n+1=(n+1)^2\in M$ and thus, by the second condition, $n+1\in M$, as needed. Possibly employing an inductive argument, we now only need to prove that $i\in M$ for each $i=\overline{1, 5}$. Since $2018\in M$, the second condition gives that $1, 2, 1009\in M$. Since $2, 1009\in M$, by the third condition $2019\in M$ and thus $3\in M$. Now since $2, 3\in M$ the third condition gives that $7\in M$ and since $2, 7\in M$ we have that $15\in M$ and thus $5\in M$. Now since $3, 5\in M$ we finally obtain that $16\in M$ and thus $4\in M$, which ends our proof.