Four sages stand around a non-transparent baobab. Each of the sages wears red, blue, or green hat. A sage sees only his two neighbors. Each of them at the same time must make a guess about the color of his hat. If at least one sage guesses correctly, the sages win. They could consult before the game started. How should they act to win?
Problem
Source: Tuymaada 2015, Day 1, Problem 8, Junior League
Tags: Tuymaada, logic, combinatorics
RagvaloD
10.07.2017 16:07
Can you set source of this problem as "Tuymaada 2015, Day 1, Problem 8, Junior League" ,please ?
BogdanB
11.07.2017 12:14
Anyone ?
jt314
09.09.2017 09:35
Let 0 denote red hat, 1 blue hat and 2 green hat If the people (in clockwise order) have hats $a, b, c, d$ respectively then: First person says: $b+d$ Second person: $-(a+c)$ Third person: $b-d$ Fourth person: $c-a$ (mod 3) Suppose this doesn't work, then $s=b+d-a$ and $t=b-d-c$ are either 1 or 2 mod 3 otherwise the first person or third person has gotten it right If $3|t-s=(a-c)-2d \equiv d-(c-a) (mod 3)$ then the fourth person wins, else $t, s$ are 1 and 2 mod 3 in some order and $3|t+s=2b-(a+c) \equiv (-a-c)-b (mod 3)$ so the second person wins.
richrow12
05.04.2020 19:15
jt314 wrote: Let 0 denote red hat, 1 blue hat and 2 green hat If the people (in clockwise order) have hats $a, b, c, d$ respectively then: First person says: $b+d$ Second person: $-(a+c)$ Third person: $b-d$ Fourth person: $c-a$ (mod 3) Suppose this doesn't work, then $s=b+d-a$ and $t=b-d-c$ are either 1 or 2 mod 3 otherwise the first person or third person has gotten it right If $3|t-s=(a-c)-2d \equiv d-(c-a) (mod 3)$ then the fourth person wins, else $t, s$ are 1 and 2 mod 3 in some order and $3|t+s=2b-(a+c) \equiv (-a-c)-b (mod 3)$ so the second person wins. You can prove it easier: just note that $$ (b+d-a)^2+(-a-c-b)^2+(b-d-c)^2+(c-a-d)^2=3(a^2+b^2+c^2+d^2)\equiv 0\pmod 3, $$so at least one person will be right.