Has anyone a solution for this problem?

Anyone has solution to this one?

There is my solution : a) in this configuration the answer is no we cant obtain what we want, proove: first, let say that the cell$(1;1)$ is $a_1$ and $(1;2)$ is $a_2$ ,.... and the cell $(3;3)$ is $c_3$ then we can say that in this configuration the cell $b_3$ for exemple has the value $6$. Fine, now let take $R$ and $T$ such as $R= a_1 + a_3 + c_1 + c_3$ and $ T= a_2 + b_1 + b_3 + c_2$ we notice that if we replace $3$ numbers of a column or of a row we will have (WLOG we take the case of the row) $ R_1 = a_1 + x + a_3 - x + c_1 + c_3 = R_0$ so $R$ is constant but we also notice that for $T$ we find that $T_{i+1} = T_i - x$ or $T_{i+1}=T_i$ if we change the numbers of the second row or the second column, so $T$ decreases or stay the same. In this configuration we have $ T_0 = 20$ and $R_0=20$ and we want to have $T = R = 4m$ but we have to change the first or last row or column, so we will have $T<R$ and then we cannot have all numbers of the board equals. b) Now we need to find the maximum value of $m$ : Suppose that $m > 4$ So we cannot have numbers $1$, $2$, $3$ and $4$ in the cell $b_2$ because its value can never increase Let take $S$ the sum of all the cells of the board, so we can say that in every operation, $S$ decrease all the time, we need to have after all these operations $S_k=9m$ (with $S_k$ the sum of the numbers of the final configuration of the board), we also have $S_0 = 45$, we notice that if we do an operation and change three numbers $a$, $b$ and $c$ by $a+x$, $b-x$, $c-x$ or $a-x$,... we will find that $S_{i+1}=S_i - x \Rightarrow$ we need to make as little operations as possible and with the smallest $x$ , but we're obliged to do some operations in the the rows and columns which contain these four numbers $(1 ; 2 ; 3 ; 4)$ so they can reach $m$ which is bigger than $4$. So we will have at some point $S_k \leq S_0 - (m-1) - (m-2) - (m-3) - (m-4) = 55 - 4m$ $ \Rightarrow 9m \leq 55 - 4m \Rightarrow m\leq \frac{55}{13}$ So $R = 4m \leq \frac{220}{13}$ and we know that $R$ is a positive integer because $R$ is constant and is an integer in the first configuration We also know that $16 < \frac{220}{13} < 17 \Rightarrow S\leq 16$ because it s an integer so $m\leq 4$ So our first supposition of qst $b$ is wrong and we can find a right configuration where $m=4$ : $(a_1= 1, a_2= 8, a_3= 9,b_1= 6, b_2= 5,b_3= 7,c_1=4 ,c_2= 3, c_3= 2 )$ We can easily reach the configuration where they are all equal to $4$ except the cells $b_1=6$ and $b_3=5$ but you can simply do two opposite operations on the first column where $x=1$ and two other operations on the 3rd column where $x=\frac{1}{2}$

(a) the answer is no note that when changing $(a,b,c) \Rightarrow (a+x,b-x,c-x)$ or $ (a-x,b-x,c+x)$ final sum: $(a+x)+(c-x)=a+c$(final sum), that is, it remains invariant In addition, after the changes, we note that the sum of the corners remains invariant. Suppose it is possible to put all the numbers equal then the numbers of the $4$ corners are equal. But then the corners should be $5$, then the central cell does not receive operations, so the operations are carried out only on the edges (see attached file), now the number $4$ cannot reach $5$, contradiction. therefore it is not possible.