Sum of cells in chessboard

Question

Can you help me solve this problem?

A chessboard cells are numbered 1 to 64 sequentially from cell(1,1) to cell(8,8). You have to place 32 pawns in the cells such that every row and column contains 4 pawns. Prove that in any such configuration, the sum of values of cells the pawn is places is 1040.

EDIT: This question was put on hold for not adding my effort. So adding my thoughts. Once we have a configuration ready we can create further answers fairly easy. One method is a @Anatoly suggested. I also thought of a method of rearranging rows and columns as a whole. Lets say we put first row to bottom and renumber. This will also be a valid configuration. In the new configuration, rows (2-8) becomes(1-7) and row 1 becomes 8. So rows (2-8) will have a reduction of values (2(no. of pawns)*8(increase of one row)*1(no of each rows increased)*7(total rows) and row 1 will get an increase by (2(no.of pawns)*8(increase of one row)*7(no of each rows increased)*1(total rows).

Similarly we can show column shuffle also results in same sum.

I think what is pending is a formal proof for showing every configuration is reachable from any initial configuration. This is intuitively doable from a column, row shuffle

Answer 1

Label the squares with ordered pairs $(a,b)$ where $0\le a\le7$ and $0 \le b\le 7$, with $a$ corresponding to the row and $b$ corresponding to the column, so the ordered pairs range from $(0,0)$ to $(7,7)$. If we define $f(a,b) = 8a+b$, then $f(a,b) + 1$ is the number in square $(a,b)$. Notice that $f(a,b) + f(c,d) = f(a+c, b+d)$

Now suppose the 32 pawns are in squares $(a_1,b_1), (a_2, b_2), (a_3,b_3), \dots (a_{32},b_{32})$. According to the restrictions on pawn placement, the sequence $a_1, a_2, a_3, \dots, a_{32}$ consists of four each of the numbers $0$ through $7$, so $$a_1 + a_2 + a_3 \dots + a_{32} = 4 \cdot (0 + 1 + 2 + \dots +7) = 112$$ and the same is true for the sequence $b_1, b_2, b_3, \dots ,b_{32}$. So the sum of the numbers in these squares is $$\begin{align} \sum_{i=1}^{32} \left( f(a_i,b_i) + 1 \right) &= f \left( \sum_{i=1}^{32} a_i, \sum_{i=1}^{32} b_i \right) + 32 \\ &= f(112, 112) + 32 \\ &= 1008 + 32 \\ &= 1040 \end{align}$$

Answer 2

As correctly stated in the comments, probably the problem requires to demonstrate that the sum is $1040$. If this is the case, assuming that cells are numbered in sequential order (e.g., increasing from $1$ to $8$ in the first row, from $9$ to $16$ in the second row, and so on), we can identify an arbitrary initial configuration that satisfies the condition of the OP. For example we could place, in half of the rows, a pawn in the $1^\text{st}$, $4^\text{th}$, $5^\text{th}$, and $8^\text{th}$ column. Then we could place, in the remaining half of the rows, a pawn in the $2^\text{nd}$, $3^\text{rd}$, $6^\text{th}$, and $7^\text{th}$ column. In this configuration, we have $4$ pawns per row and $4$ pawns per columns. Also, in each row the sum of the occupied cells clearly equals that of the free cells, so that the sum of all occupied cells in the chessboard is equal to half of the total sum of all cells. The total sum of all cells is $64 \cdot 65/2=2080\,\,\,\,$, so that the sum of the occupied cells is $4160/2=1040\,\,\,$.

Now note that we can perform, from this initial valid configuration, a transformation by moving one pawn within a row from the $i^\text{th}$ to the $j^\text{th}$ column, and another pawn in another row from the $j^\text{th}$ to the $i^\text{th}$ column. This transformation is always possible because, among the $4$ possible rows that contain a pawn in the $j^\text{th}$ column, at least one has a free cell in the $i^\text{th}$ column (otherwise this would imply that, before the initial movement, the $i^\text{th}$ column contains more than $4$ pawns). Also, this transformation does not alter the condition that there are $4$ pawns per row and $4$ pawns per column, and does not change the total sum of the occupied cells, since the two movements decrease and increase the sum by an identical quantity.

Performing a sequence of these transformations, it is not diffucult to show that we can reach any other valid configuration that satisfies the conditions of the OP. Therefore, the final risult is that, in any such configuration, the sum of the values of the occupied cells is $1040\,$.