1. **State the problem:** We have two 3x3 grids where each row, column, and diagonal must sum to the same total. Each of the given numbers can be used exactly 3 times in its respective grid. We need to fill the blank boxes and find the product of the numbers in the red boxes. 2. **Left grid given numbers:** 8 at (1,3), 6 at (2,2), 4 at (3,1), red box at (3,3). 3. **Right grid given numbers:** 3 at (1,1), 7 at (1,2), 5 at (2,2), red box at (3,1). --- ### Left grid solution: 4. Let the magic sum be $S$. The grid uses numbers 8, 6, 4 each exactly 3 times. 5. Since 6 is at center, and 8 and 4 are corners, the sum $S$ must satisfy rows, columns, diagonals. 6. The sum of all numbers in the grid is $3 \times (8 + 6 + 4) = 3 \times 18 = 54$. 7. Since there are 3 rows each summing to $S$, total sum is $3S = 54 \Rightarrow S = 18$. 8. Using $S=18$, fill the grid so each row, column, diagonal sums to 18. 9. The red box is at (3,3). By balancing sums, the red box number is 8. --- ### Right grid solution: 10. Numbers 3, 7, 5 each used 3 times, total sum $= 3 \times (3 + 7 + 5) = 3 \times 15 = 45$. 11. Magic sum $S$ satisfies $3S = 45 \Rightarrow S = 15$. 12. Fill the grid so each row, column, diagonal sums to 15. 13. The red box is at (3,1). By balancing sums, the red box number is 7. --- ### Final step: 14. Product of numbers in red boxes $= 8 \times 7 = 56$. **Answer:** $56$