Push Z150 to the Quiet Western Front

Let's first see how many numbers we can pick before a pair is inevitably found. If we choose $1, 2, 3, ..., 19$ , we cannot choose $20, 21, ..., 38$ because $20-1=19, 21-2=19, ...38-19=19$ . So we must choose alternating sets of 19 numbers that fit within 150. Here is one possible group of such sets: $1-19$ $39-57$ $77-95$ $115-133$ We obtain another such group if we start at 2 instead of 1 and so on. So how many numbers are in this set? Well each set has 19 numbers and 4 sets can fit in 150 so we can pick 4*19=76 numbers such that no pair exists. If we pick 77 numbers, then we are guaranteed to have 1 pair so the answer is 1.

Bullet Proof (3 bullets)

• First, consider congruence classes modulo $19$ . $\boxed{0,1,2,3,4,...,18}$ . For sure, $19$ classes .

• Now, using PHP , we find, $77=19\times 4+1$ yields there are at least $5$ numbers $n_1<n_2<n_3<n_4<n_5$ of our sample are of the same congruence class.

• If, somehow, $n_{i+1}-n_i \ne 19$ , it would be as clear as Plexiglas (Polymethylmethacrylate) that, $n_{i+1}-n_i \ge 38$ , and so, $n_5-n_1 \ge 154$ .

Well, friends, that is impossible. No two numbers of our set are more than $149$ miles apart.

Thus, the answer is $e^{2i \pi }$ .