1. The problem is to solve multiplication problems from the multiplication table. 2. The multiplication operation is defined as repeated addition. For example, $a \cdot b$ means adding $a$ to itself $b$ times. 3. We solve each multiplication problem by multiplying the two numbers. 4. Examples from the first column: - $9 \cdot 9 = 81$ - $6 \cdot 6 = 36$ - $7 \cdot 1 = 7$ - $9 \cdot 2 = 18$ - $8 \cdot 4 = 32$ 5. For the second column, some answers are missing, so we calculate: - $4 \cdot 3 = 12$ - $7 \cdot 4 = 28$ - $5 \cdot 5 = 25$ - $5 \cdot 7 = 35$ - $3 \cdot 9 = 27$ - $1 \cdot 8 = 8$ - $0 \cdot 9 = 0$ - $4 \cdot 1 = 4$ - $2 \cdot 2 = 4$ - $0 \cdot 3 = 0$ - $10 \cdot 2 = 20$ - $6 \cdot 3 = 18$ - $3 \cdot 4 = 12$ - $5 \cdot 6 = 30$ 6. Similarly, for the third column: - $0 \cdot 7 = 0$ - $7 \cdot 7 = 49$ - $0 \cdot 8 = 0$ - $9 \cdot 7 = 63$ - $5 \cdot 1 = 5$ - $4 \cdot 2 = 8$ - $8 \cdot 3 = 24$ - $4 \cdot 4 = 16$ - $0 \cdot 5 = 0$ - $6 \cdot 9 = 54$ - $6 \cdot 8 = 48$ - $2 \cdot 7 = 14$ - $3 \cdot 6 = 18$ - $0 \cdot 6 = 0$ - $4 \cdot 1 = 4$ - $6 \cdot 4 = 24$ - $2 \cdot 5 = 10$ - $10 \cdot 5 = 50$ - $4 \cdot 9 = 36$ - $6 \cdot 7 = 42$ - $2 \cdot 6 = 12$ - $7 \cdot 2 = 14$ - $1 \cdot 1 = 1$ - $9 \cdot 3 = 27$ - $2 \cdot 8 = 16$ 7. For the fourth column: - $7 \cdot 6 = 42$ - $3 \cdot 7 = 21$ - $10 \cdot 9 = 90$ - $10 \cdot 8 = 80$ - $5 \cdot 2 = 10$ - $1 \cdot 4 = 4$ - $1 \cdot 3 = 3$ - $0 \cdot 4 = 0$ - $7 \cdot 3 = 21$ - $9 \cdot 4 = 36$ - $10 \cdot 3 = 30$ - $8 \cdot 5 = 40$ - $4 \cdot 7 = 28$ - $3 \cdot 5 = 15$ - $5 \cdot 3 = 15$ - $3 \cdot 1 = 3$ - $10 \cdot 1 = 10$ - $8 \cdot 2 = 16$ - $6 \cdot 5 = 30$ - $0 \cdot 2 = 0$ - $7 \cdot 8 = 56$ - $8 \cdot 8 = 64$ - $8 \cdot 9 = 72$ - $9 \cdot 8 = 72$ - $0 \cdot 1 = 0$ 8. These problems reinforce multiplication skills and understanding of the multiplication table. Final answers are as calculated above for each problem.