There are five smart kids sitting around a round table. Their teacher says: "I gave a few apples to some of you, and none of you have the same amount of apple. Also each of you will know the amount of apple that the person to your left and the person to your right has." The teacher tells the total amount of apples, then asks the kids to guess the difference of the amount of apple that the two kids in front of them have. $a)$ If the total amount of apples is less than $16$, prove that at least one of the kids will guess the difference correctly. $b)$ Prove that the teacher can give the total of $16$ apples such that no one can guess the difference correctly.
Problem
Source: 2017 Iran MO 2nd round P5
Tags: combinatorics, Iran
mszew
21.04.2017 22:08
b) This distribution is an example with 16 apples: $1,4,3,2,6$ Kid 1: two options with: $1,4,(0,5),6$ Kid 4: two options with: $1,4,3,(0,8)$ Kid 3: two options with: $0),4,3,2,(7$ Kid 4: two options with: $(0,5),3,2,6$ Kid 5: two options with: $1,(0,7),2,6$ For the a) point consider the case for the highest posible value of the lowest two consecutive kids, and notice that if it is $4$ or less there are only two posible options such as $0,4$ or $1,3$ however the other lowest sum of the other three numbers is $2,5,6$ is $13$ contradiction.
mszew
21.04.2017 22:47
I ll explain in details the first point a) WLOG consider that $l=k_1+k_2$ is the lowest from two consecutives kids $k_1+k_2,...,k_4+k_5,k_5+k_1$ Assume that there is a distribution with $s<16$ apples and no kid can guess. $l=0$ then the only difference for $k_1$ and $k_2$ is $0$ thus $k_4$ can guess. contradiction. $l=1$ then the only difference for $k_1$ and $k_2$ is $1$ thus $k_4$ can guess. contradiction. $l=2$ then the only difference for $k_1$ and $k_2$ is $2$ thus $k_4$ can guess. contradiction. $l=3$ two possible cases: $(0,3)$ and $(1,2)$ then the mínimum sum for the other three numbers is $4,5,6$ contradiction for the number of apples. $l=4$ two possible cases: $(0,4)$ and $(1,3)$ then the mínimum sum for the other three numbers is $2,5,6$ contradiction for the number of apples. Notice that for $l=5$ there are three possible cases: $(0,5)$, $(1,4)$ and $(2,3)$ giving the example for point b
enthusiast101
19.06.2019 14:41
Let $m$ be the maximum number of apples held by any of the students. Note that $max(a)=5$ since the number of apples have to be different for each student. So $max(a)=15-(1+2+3+4)=5$. Since $1,2,3,4$ result in the least sum.
Hence the number of possible differences of apples is $5-1=4$. But there are five students, so by PHP, one must be able to guess correctly.
AYKW1
19.06.2019 15:33
Is that a fake solve, since someone can have 0 apples?
enthusiast101
19.06.2019 16:09
AYKW1 wrote: Is that a fake solve, since someone can have 0 apples? Ahh. Sorry about that. I should read the question carefully next time.