All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Scrutiny (Posted on 2023-05-01)
A puzzle by V. Dubrovsky, from Quantum, January-February 1992:

In a certain planetary system, no two planets are separated by the same distance. On each planet sits an astronomer who observes the planet closest to hers.

Prove that if the total number of planets is odd, there must be a planet that no one is observing.

 Submitted by Ady TZIDON Rating: 5.0000 (1 votes) Solution: (Hide) Of the n planets , assume that two watch each other. If any of the remaining n – 2 astronomers are looking at one of these planets, then there won’t be enough astronomers left to observe all the remaining n – 2 planets. On the other hand, if none of the remaining astronomers are looking at the two closest-set planets, then we can discard those two and ask our original question of the n – 2 planets that remain: Which two are closest together? Again, those two astronomers must be observing each other. And so on. Each time we discard a pair of planets we’ll have some odd number remaining. But that number can’t decrease forever (it can’t be negative). We must eventually arrive at 1, a planet that no astronomer is observing.

 Subject Author Date My version broll 2023-05-02 11:36:51 proof by contradiction xdog 2023-05-02 10:14:55 A start, but incomplete Larry 2023-05-01 23:42:39 re(2): soln Steven Lord 2023-05-01 19:24:27 re: soln Kenny M 2023-05-01 17:28:04 soln Steven Lord 2023-05-01 16:34:21

 Search: Search body:
Forums (0)