Mongolian Mathematical Olympiad

Let us denote the cell in row $i$ and column $j$ by $a_{i,j}$, where $1\le i,j\le 5$. Initially, $a_{i,j}=5(i-1)+j$.

Each operation increases all numbers in a row or a column by $1$. Suppose we perform $r_i$ operations on row $i$ and $c_j$ operations on column $j$. Then the final value in cell $(i,j)$ is:

$$a_{i,j}^{\prime }=a_{i,j}+r_i+c_j$$

We want all $a_{i,j}^{\prime }$ to be equal. That is, for all $i,j$:

$$a_{i,j}+r_i+c_j=K$$ for some constant $K$.

Rewriting: $$r_i+c_j=K-a_{i,j}$$

Consider two cells in the same row $i$, columns $j$ and $k$: $$r_i+c_j=K-a_{i,j}$$ $$r_i+c_k=K-a_{i,k}$$ Subtract: $$c_j-c_k=a_{i,k}-a_{i,j}$$ But $a_{i,k}-a_{i,j}=k-j$ (since $a_{i,j}=5(i-1)+j$). So $c_j-c_k=k-j$.

This means the sequence $c_1,c_2,\dots ,c_5$ must satisfy $c_j-c_k=j-k$ for all $j,k$. This is only possible if $c_j=j+d$ for some constant $d$.

Similarly, consider two cells in the same column $j$, rows $i$ and $k$: $$r_i+c_j=K-a_{i,j}$$ $$r_k+c_j=K-a_{k,j}$$ Subtract: $$r_i-r_k=a_{k,j}-a_{i,j}$$ But $a_{k,j}-a_{i,j}=5(k-i)$. So $r_i-r_k=5(k-i)$, so $r_i=5i+e$ for some constant $e$.

Now, plug $c_j=j+d$ and $r_i=5i+e$ into the equation: $$r_i+c_j=K-a_{i,j}$$ But $a_{i,j}=5(i-1)+j$, so $$r_i+c_j=K-(5(i-1)+j)$$ $$(5i+e)+(j+d)=K-5(i-1)-j$$ $$5i+e+j+d=K-5i+5-j$$ $$5i+j+e+d+5i+j=K+5$$ Wait, this is inconsistent. Let's simplify:

Left: $5i+e+j+d$ Right: $K-5(i-1)-j$

Bring all terms to one side: $$5i+e+j+d+5(i-1)+j-K=0$$ But this is not matching. Let's try plugging $r_i=5i+e$, $c_j=j+d$ into $r_i+c_j=K-a_{i,j}$: $$5i+e+j+d=K-(5(i-1)+j)$$ $$5i+e+j+d=K-5i+5-j$$ $$5i+e+j+d+5i+j-K-5=0$$ Wait, this is not matching. Let's try to solve for $K$:

Left: $5i+e+j+d$ Right: $K-5(i-1)-j$

So: $$5i+e+j+d=K-5(i-1)-j$$ $$5i+e+j+d+5(i-1)+j-K=0$$ $$5i+e+j+d+5i-5+j-K=0$$ $$10i+e+d+2j-5-K=0$$ $$K=10i+e+d+2j-5$$ But $K$ must be constant for all $i,j$, which is only possible if the coefficients of $i$ and $j$ are zero, i.e., $10i+2j$ must be constant, which is impossible as $i$ and $j$ vary.

Therefore, it is not possible to make all the numbers in the grid equal by performing such operations.