Question 7

We will find the prime numbers for each empty box in both grids.

Step 1 — Find prime factors for Grid 1

Let us list the prime factors for each product. This helps us find the numbers for the boxes.

$63 = 3 \times 3 \times 7$

$27 = 3 \times 3 \times 3$

$190 = 2 \times 5 \times 19$

$45 = 3 \times 3 \times 5$

$42 = 2 \times 3 \times 7$

$171 = 3 \times 3 \times 19$

Step 2 — Fill Grid 1 cells

Look at the cell in the third row, third column. Its number must be a factor of 190 and 171. The common prime factor is 19. So, the number in cell (3,3) is 19.

Look at the cell in the first row, third column. Its number must be a factor of 63 and 171. The common prime factors are 3 and 3. We can use one 3 for cell (1,3).

Look at the cell in the second row, third column. Its number must be a factor of 27 and 171. The remaining common prime factor is 3. So, the number in cell (2,3) is 3.

Now the third column is 3, 3, 19. For row 1, the product is 63. We have 3 in cell (1,3). The other two numbers multiply to $63 \div 3 = 21$. The prime factors of 21 are 3 and 7. So, cells (1,1) and (1,2) are 3 and 7.

For row 2, the product is 27. We have 3 in cell (2,3). The other two numbers multiply to $27 \div 3 = 9$. The prime factors of 9 are 3 and 3. So, cells (2,1) and (2,2) are 3 and 3.

For row 3, the product is 190. We have 19 in cell (3,3). The other two numbers multiply to $190 \div 19 = 10$. The prime factors of 10 are 2 and 5. So, cells (3,1) and (3,2) are 2 and 5.

Let us use the column products to place numbers. Column 1 product is 45. The numbers are from (1,1), (2,1), (3,1). Let us try cell (1,1) as 3. Then cell (1,2) must be 7. For column 1, we have 3 from (1,1). The remaining product for (2,1) and (3,1) is $45 \div 3 = 15$. Cell (2,1) must be 3 (from row 2 factors). Then cell (2,2) must be 3. If cell (2,1) is 3, then cell (3,1) must be $15 \div 3 = 5$. So, cell (3,1) is 5. Then cell (3,2) must be 2 (from row 3 factors).

Let us check column 2. The numbers are 7 (from (1,2)), 3 (from (2,2)), 2 (from (3,2)). Their product is $7 \times 3 \times 2 = 42$. This matches the given column 2 product. All numbers are prime.

The filled grid 1 is:

$\boxed{\begin{array}{|c|c|c|} \hline 3 & 7 & 3 \\ \hline 3 & 3 & 3 \\ \hline 5 & 2 & 19 \\ \hline \end{array}}$

Step 3 — Find prime factors for Grid 2

Let us list the prime factors for each product. This helps us find the numbers for the boxes.

$343 = 7 \times 7 \times 7$

$66 = 2 \times 3 \times 11$

$44 = 2 \times 2 \times 11$

$28 = 2 \times 2 \times 7$

$154 = 2 \times 7 \times 11$

$231 = 3 \times 7 \times 11$

Step 4 — Fill Grid 2 cells

Look at the cell in the first row, third column. Its number must be a factor of 343 and 231. The common prime factor is 7. So, the number in cell (1,3) is 7.

Look at the cell in the third row, third column. Its number must be a factor of 44 and 231. The common prime factor is 11. So, the number in cell (3,3) is 11.

Look at the cell in the second row, third column. Its number must be a factor of 66 and 231. The third column product is 231. We have 7 and 11 in this column. The remaining number is $231 \div (7 \times 11)$. $231 \div 77 = 3$ So, the number in cell (2,3) is 3.

Now the third column is 7, 3, 11. For row 1, the product is 343. We have 7 in cell (1,3). The other two numbers multiply to $343 \div 7 = 49$. The prime factors of 49 are 7 and 7. So, cells (1,1) and (1,2) are 7 and 7.

For row 2, the product is 66. We have 3 in cell (2,3). The other two numbers multiply to $66 \div 3 = 22$. The prime factors of 22 are 2 and 11. So, cells (2,1) and (2,2) are 2 and 11.

For row 3, the product is 44. We have 11 in cell (3,3). The other two numbers multiply to $44 \div 11 = 4$. The prime factors of 4 are 2 and 2. So, cells (3,1) and (3,2) are 2 and 2.

Let us use the column products to place numbers. Column 1 product is 28. The numbers are from (1,1), (2,1), (3,1). Let us try cell (1,1) as 7. Then cell (1,2) must be 7. For column 1, we have 7 from (1,1). The remaining product for (2,1) and (3,1) is $28 \div 7 = 4$. Cell (3,1) must be 2 (from row 3 factors). Then cell (3,2) must be 2. If cell (3,1) is 2, then cell (2,1) must be $4 \div 2 = 2$. So, cell (2,1) is 2. Then cell (2,2) must be 11 (from row 2 factors).

Let us check column 2. The numbers are 7 (from (1,2)), 11 (from (2,2)), 2 (from (3,2)). Their product is $7 \times 11 \times 2 = 154$. This matches the given column 2 product. All numbers are prime.

The filled grid 2 is:

$\boxed{\begin{array}{|c|c|c|} \hline 7 & 7 & 7 \\ \hline 2 & 11 & 3 \\ \hline 2 & 2 & 11 \\ \hline \end{array}}$

Answer

(i) The filled grid 1 is as shown in Step 2. (ii) The filled grid 2 is as shown in Step 4.