brand.scso.com.
Names of solvers will be posted on this page. Notify if you don't want
your name to be mentioned.
|
This month's riddle comes from Dominic van der Zypen who says that this is
a generalization of his design, based on a riddle he was presented by
Daniel Steiner.
Let ℕ be the naturals: 1, 2, 3,… and let Πℕ be the set of permutations over ℕ (that is, the set of bijections from ℕ to itself). Let us now define a property called k-resistance. We say that ℕ is k-resistant, for some k∈ℕ, if for each π∈Πℕ there exist a,d∈ℕ, such that
This month's riddle: for which values of k are the naturals k-resistant? |
List of solvers:Joseph DeVincentis (1 April 13:11)Lorenzo Gianferrari Pini and Radu-Alexandru Todor (2 April 03:31) Lorenz Reichel (3 April 18:41) Anurag Anshu (5 April 19:56) Sen Gu (5 April 23:54) Øyvind Grotmol (6 April 08:37) Suhail Sherif (6 April 15:01) Jan Fricke (7 April 18:17) Thomas Mack (8 April 14:43) Yuping Luo (9 April 11:32) Luke Pebody (26 April 10:32) Rohan Joshi (26 April 14:26) Dharmadeep Muppalla (30 April 00:01) |
Elegant and original solutions can be submitted to the puzzlemaster at riddlesbrand.scso.com.
Names of solvers will be posted on this page. Notify if you don't want
your name to be mentioned.
The solution will be published at the end of the month.
Enjoy!