Operations with Integers | A

Question 2

Try again, and choose different numbers this time. What product did you get? Was it different from the first time? Try a few more times with different numbers!

Play the same game with the grid below. What answer do you get?

What is so special about these grids? Is the magic in the numbers or the way they are arranged or both? Can you make more such grids?

Solution

Understand the Question

In this grid game, we select four numbers such that no two numbers belong to the same row or column (exactly one number is picked from each row and each column).
Multiplying the chosen numbers always results in the same constant product: $-30240$ .
This happens because each cell in the grid is generated by multiplying a row factor $R_i$ and a column factor $C_j$ ( $M_{ij} = R_i \times C_j$ ). Selecting one entry from each row and column multiplies all row factors and column factors together: $(R_1 R_2 R_3 R_4) \times (C_1 C_2 C_3 C_4)$ , which is constant regardless of the selection.

(i) Play the game with different selections of numbers from the grid. What product do you get? Is it different from the first time?

Step 1 · Calculate Product for First Selection (Main Diagonal)

Select one number from each row and column along the main diagonal: $8$ , $14$ , $18$ , and $-15$ . Diagram 1

\begin{aligned} 8 \times 14 \times 18 \times (-15) &= 112 \times 18 \times (-15) \\ &= 2016 \times (-15) \\ &= -30240 \end{aligned}

Step 2 · Calculate Product for Second Selection (Anti-Diagonal)

Select another set of numbers along the anti-diagonal: $-6$ , $-42$ , $-6$ , and $20$ . Diagram 2

\begin{aligned} (-6) \times (-42) \times (-6) \times 20 &= 252 \times (-6) \times 20 \\ &= -1512 \times 20 \\ &= -30240 \end{aligned}

Answer

(i) $-30240$

(ii) What is so special about these grids? Is the magic in the numbers or the way they are arranged or both?

Step 1 · Find Row and Column Factors

Let each cell $M_{ij} = R_i \times C_j$ , where $R_i$ is the row factor and $C_j$ is the column factor. Setting $R_1 = 1$ , we find the column factors from the first row:

\begin{aligned} M_{11} &= R_1 \times C_1 = 8 \implies 1 \times C_1 = 8 \implies C_1 = 8 \\ M_{12} &= R_1 \times C_2 = -4 \implies 1 \times C_2 = -4 \implies C_2 = -4 \\ M_{13} &= R_1 \times C_3 = 12 \implies 1 \times C_3 = 12 \implies C_3 = 12 \\ M_{14} &= R_1 \times C_4 = -6 \implies 1 \times C_4 = -6 \implies C_4 = -6 \end{aligned}

Using $C_1 = 8$ , we find the remaining row factors from the first column:

\begin{aligned} M_{11} &= R_1 \times 8 = 8 \implies R_1 = 1 \\ M_{21} &= R_2 \times 8 = -28 \implies R_2 = -3.5 \\ M_{31} &= R_3 \times 8 = 12 \implies R_3 = 1.5 \\ M_{41} &= R_4 \times 8 = 20 \implies R_4 = 2.5 \end{aligned}

Verifying for $M_{22}$ : $R_2 \times C_2 = (-3.5) \times (-4) = 14$

For any selection picking one number from each row and column:

\begin{aligned} \text{Product} &= (R_1 C_{j_1}) \times (R_2 C_{j_2}) \times (R_3 C_{j_3}) \times (R_4 C_{j_4}) \\ &= (R_1 R_2 R_3 R_4) \times (C_1 C_2 C_3 C_4) \end{aligned}

Calculating the constant value:

\begin{aligned} 1 \times (-3.5) \times 1.5 \times 2.5 &= -3.5 \times 3.75 = -13.125 \\ 8 \times (-4) \times 12 \times (-6) &= -32 \times (-72) = 2304 \end{aligned}

$\text{Total Product} = -13.125 \times 2304 = -30240$

Answer

(ii) Each grid cell is the product of a row factor and a column factor ( $M_{ij} = R_i \times C_j$ ). Choosing one number from each row and column always yields the product of all row and column factors, $(R_1 R_2 R_3 R_4) \times (C_1 C_2 C_3 C_4) = -30240$ . The magic is in both the numbers and their arrangement.

(iii) Can you make more such grids?

Answer

(iii) Yes, we can create more grids by choosing any set of row factors and column factors and multiplying them to fill each cell of the grid.

Common Mistakes

Picking Duplicate Rows/Columns: Selecting two numbers from the same row or column breaks the rule and leads to an incorrect product.
Sign Errors: Mishandling the negative signs during sequential multiplication; an odd number of negative factors always results in a negative final product.

Question 2

More questions in A