Help Analyze whether, at a given moment, the following cases are possible....

Question

Question

The numbers $4$, $8$, $9$ and $15$ are written on the board. Carla deletes three of them and then writes three more numbers following the rule: if the numbers $a$, $b$, $c$ are deleted, the replacement numbers are $\frac{a+b}{2}$, $\frac{a+c}{2}$, $\frac{c+b}{2}$ and then the operation is repeated.

Analyze whether, at a given moment, the following cases are possible:

a) In a set of four numbers, the number $16$ appears on the board.

b) The set of four numbers on the board is $6, 8, 9, 10$.

c) The set of four numbers on the board is $5, 5, 12, 14$.

My idea

I was able to solve the first request. For point a) we can use min and max which are between the arithmetic mean, this means that all the numbers will be between $4$ and $15$. $16$ is greater than $15$, which means it can't appear on the table.

I don't know what to do for the forward points. I thought of showing that numbers are always different, which means that $8,9$ can't appear again or showing that numbers shouldn't equal which means $2$ of number $5$ can't be on the same string.

Hope on of you can help me! Thank you!

Answer 1

Let us show that case (c) is impossible.

The first solution.

There are at most two stages to consider. Namely, let at the first stage the number $15$ remains undeleted and the second stage starts when the number $15$ is deleted. Similarly to you answer to case (a), we observe that at the first stage the second biggest number is at most $9$. Then when the second stage begins, the biggest number can be at most $\frac{15+9}2=12$, so again similarly to you answer to case (a) we see that at the second stage none of the numbers can be bigger than $12$.

The second solution.

For any real number $a$, $b$, and $c$ we have $$\left(\frac{a+b}{2}\right)^2+\left(\frac{a+c}{2}\right)^2+\left(\frac{b+c}{2}\right)^2=$$ $$\frac{a^2+b^2+c^2+ab+ac+bc}{2}\le a^2+b^2+c^2,$$ because $ab\le\frac{a^2+b^2}{2}$, $ac\le\frac{a^2+c^2}{2}$, and $bc\le\frac{b^2+c^2}{2}$.

Therefore at each step the sum of the squares of the numbers is nonincreasing. But for the numbers $4$, $8$, $9$, and $15$ the sum of the squares equals $16+64+81+225=386$, whereas for the numbers $5$, $5$, $12$, and $14$ the sum of the squares equals $25+25+144+196=390$.

Answer 2

To solve $(b)$, note that the total sum of the numbers on the board is an invariant - it remains the same after each operation.

To solve $(c)$, I would use that, as you yourself noted, at each step you can never produce a number smaller than the previous minimum. I would therefore argue that at each step, either you haven't used $4$ in an operation yet, or your minimum would be strictly greater than $5$. Thus $5$ can never appear on the board.

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By request, a more detailed answer of $(c)$.

Consider the numbers split in two groups: The first group is the single number $4$, and the three others in the second group. Now, if we apply the operation and pick only the numbers of the second group, then the minimal number of the second group does not decrease, i.e. it is at least $8$. Thus, no matter how many times you have applied the operation on the second group, the first time you pick $4$ as one of the numbers in the operation, the new smallest number is bounded below by $\frac{4+8}{2}=6$. Thus, at any point in the game, either you have a $4$, or your smallest number is at least $6$, and hence, $5,5,12,14$ will never occur.