Question 2

This grid has a special multiplicative property.

Step 1 — Calculate the first product

Let us choose numbers from the main diagonal. We pick one number from each row. We make sure no two numbers are from the same column. The numbers are 8, 14, 18, and -15. Let us multiply these numbers.

$8 \times 14 \times 18 \times (-15)$

$= 112 \times 18 \times (-15)$

$= 2016 \times (-15)$

$\boxed{\mathbf{-30240}}$

Step 2 — Calculate a second product

Let us choose a different set of numbers. We pick numbers from the anti-diagonal. The numbers are -6, -42, -6, and 20. Let us multiply these numbers.

$(-6) \times (-42) \times (-6) \times 20$

$= 252 \times (-6) \times 20$

$= -1512 \times 20$

$\boxed{\mathbf{-30240}}$

Step 3 — Understand the special property

We see both products are the same. Let us find the factors for each row and column. Let $R_i$ be the factor for row $i$. Let $C_j$ be the factor for column $j$. Each number in the grid, $M_{ij}$, is $R_i \times C_j$. Let us set the first row factor $R_1$ to 1. From the first row, we find column factors. $M_{11} = R_1 \times C_1 = 8$, so $1 \times C_1 = 8$. $M_{12} = R_1 \times C_2 = -4$, so $1 \times C_2 = -4$. $M_{13} = R_1 \times C_3 = 12$, so $1 \times C_3 = 12$. $M_{14} = R_1 \times C_4 = -6$, so $1 \times C_4 = -6$. So, the column factors are $C_1 = \mathbf{8}$, $C_2 = \mathbf{-4}$, $C_3 = \mathbf{12}$, $C_4 = \mathbf{-6}$.

Now, let us find the row factors. From the first column, we use $C_1 = 8$. $M_{11} = R_1 \times C_1 = 8$, so $R_1 \times 8 = 8$. This means $R_1 = \mathbf{1}$. $M_{21} = R_2 \times C_1 = -28$, so $R_2 \times 8 = -28$. This means $R_2 = \mathbf{-3.5}$. $M_{31} = R_3 \times C_1 = 12$, so $R_3 \times 8 = 12$. This means $R_3 = \mathbf{1.5}$. $M_{41} = R_4 \times C_1 = 20$, so $R_4 \times 8 = 20$. This means $R_4 = \mathbf{2.5}$. So, the row factors are $R_1 = \mathbf{1}$, $R_2 = \mathbf{-3.5}$, $R_3 = \mathbf{1.5}$, $R_4 = \mathbf{2.5}$.

Let us check if these factors work for $M_{22}$. $M_{22}$ is 14. $R_2 \times C_2 = (-3.5) \times (-4) = \mathbf{14}$. It works! This property is true for all numbers in the grid.

When we choose one number from each row and column, say $M_{1j_1}, M_{2j_2}, M_{3j_3}, M_{4j_4}$. Their product is $(R_1 C_{j_1}) \times (R_2 C_{j_2}) \times (R_3 C_{j_3}) \times (R_4 C_{j_4})$. We can rearrange the terms. Product $= (R_1 R_2 R_3 R_4) \times (C_{j_1} C_{j_2} C_{j_3} C_{j_4})$. The column indices $j_1, j_2, j_3, j_4$ are just $1, 2, 3, 4$ in some order. So, $(C_{j_1} C_{j_2} C_{j_3} C_{j_4})$ is the same as $(C_1 C_2 C_3 C_4)$. The product is always $(R_1 R_2 R_3 R_4) \times (C_1 C_2 C_3 C_4)$. This value is constant.

Let us calculate this constant product. Product of row factors: $1 \times (-3.5) \times 1.5 \times 2.5$. $1 \times (-3.5) \times 1.5 \times 2.5$ $= -3.5 \times 3.75$ $= -13.125$

Product of column factors: $8 \times (-4) \times 12 \times (-6)$. $8 \times (-4) \times 12 \times (-6)$ $= -32 \times (-72)$ $= 2304$

The total product is the product of these two results. $-13.125 \times 2304$ $= -30240$

$\boxed{\mathbf{-30240}}$

This confirms the constant product.

Answer

(i) When I chose different numbers, the product I got was -30240. This was not different from the first time. I tried a few more times, and the product was always -30240. This is the answer for playing the game with the given grid. (ii) The special thing about these grids is that each number is the product of a row factor and a column factor. This means that if you choose one number from each row and each column, their product will always be the same. The magic is in both the numbers and their arrangement. (iii) Yes, we can make more such grids. We choose a set of row factors. We choose a set of column factors. We multiply them to fill the grid.