If $n \ge 25$ and $n = 8r_1 + 5r_2$ (with $r_1, r_2$ being natural numbers), prove that either $r_1 \ge 2$ or $r_2 \ge 2$. So is uttered the question which I found in a book on introductory combinatorics. I believe it involves using the pigeonhole principle, and so the way I attempted at it is:
If we distribute $2(2 - 1) + 1 = 3$ units into $r_1, r_2$, then either $r_1$ has at least 2 ($\ge 2$) or $r_2$ has at least 2 units ($\ge 2$). So if we show that $r_1 + r_2 \ge 3$ then the conclusion can be reached. I believe this is supposed to be shown through the fact that $8r_1 + 5r_2 \ge 26$ but I don't know how. Thank you in advance.
A scale is loaded up with several starfish and octopuses. Each starfish weighs five ounces, each octopus weighs eight ounces. The total weight of the scale is $n$.
Each ounce is a pigeon, and there are two holes. The ounce goes in the "starfish" hole if it came from a starfish, and the "octopus" hole otherwise. Since there are at least $25$ pigeons, and two holes, by the generalized pigeonhole principle, there must exist a hole with at least $\lceil 25/2\rceil=13$ pigeons. That is, there are either $13$ ounces of starfish, or $13$ ounces of octopus. In the first case, there must be at least $\lceil 13/5\rceil =3$ starfish. In the second case, there must be $\lceil 13/8\rceil=2$ octopuses. Either way, some creature appears at least twice, completing the proof.