Positive integers 1 to 9 are written in each square of a $ 3 \times 3 $ table. Let us define an operation as follows: Take an arbitrary row or column and replace these numbers $ a, b, c$ with either non-negative numbers $ a-x, b-x, c+x $ or $ a+x, b-x, c-x$, where $ x $ is a positive number and can vary in each operation. (1) Does there exist a series of operations such that all 9 numbers turn out to be equal from the following initial arrangement a)? b)? \[ a) \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} )\] \[ b) \begin{array}{ccc} 2 & 8 & 5 \\ 9 & 3 & 4 \\ 6 & 7 & 1 \end{array} )\] (2) Determine the maximum value which all 9 numbers turn out to be equal to after some steps.
Problem
Source: North Korea Team Selection Test 2013 #4
Tags: combinatorics proposed, combinatorics
mszew
22.05.2014 22:07
(1) Notice that the sum of the four vertices is an invariant. After each operation the sum of all the numbers is always decreasing by $x$. After each operation the central vertex is equal or decreasing by $x$ So it is not possible for the (a) and (b) case.
ahhha
17.05.2020 18:37
maybe... the first question was the part of seoul science high school's entrance exam
CrazyInMath
11.02.2023 03:57
First we notice that every time we do an operation, the sum decrease by $x$. Second we notice that the four corners would add up to a non-changing constant, so the answer must be $\frac{k}{4}$, where $k$ is an integer. We would prove that $k=17$ fails first. To reach $4.25$ for every square, you need to bring up the $1,2,3,4$ to $4.25$, so you will need to increase them by a total of $0.25+1.25+2.25+3.25=7$, but $4.25\times9=38.25>45-7$, so $k=17$ will fail. $k=16$ works by taking the first row to be $1,9,7$, second row to be $6,8,2$, third row to be $3,4,5$. Change the first and third column all to $4$ first. Then you make the $6$ in the top spot decrease to $5$ by taking $x=0.5$ twice on the first row. now fix the top and the bottom square, Finally, fix the middle one.