1. Problem: Construct a multiplication table modulo 9. 2. The multiplication modulo 9 means we multiply two numbers and then take the remainder when divided by 9. 3. The table entries are calculated as $$a \times b \mod 9$$ for $$a,b=0,1,2,...,8$$. 4. Constructing the table: | \times | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |-------|---|---|---|---|---|---|---|---|---| | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | | 1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | | 2 | 0 | 2 | 4 | 6 | 8 | 1 | 3 | 5 | 7 | | 3 | 0 | 3 | 6 | 0 | 3 | 6 | 0 | 3 | 6 | | 4 | 0 | 4 | 8 | 3 | 7 | 2 | 6 | 1 | 5 | | 5 | 0 | 5 | 1 | 6 | 2 | 7 | 3 | 8 | 4 | | 6 | 0 | 6 | 3 | 0 | 6 | 3 | 0 | 6 | 3 | | 7 | 0 | 7 | 5 | 3 | 1 | 8 | 6 | 4 | 2 | | 8 | 0 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 5. Explanation: Each cell is the product of the row and column headers modulo 9. 6. Next, consider the equation $$-14 = 3(-3) - 5$$. 7. The Division Algorithm states for integers $$a$$ and $$b>0$$, there exist unique integers $$q$$ and $$r$$ such that $$a = bq + r$$ with $$0 \leq r < b$$. 8. Here, $$a = -14$$, $$b = 3$$, $$q = -3$$, and $$r = -5$$. 9. Check if $$r$$ satisfies $$0 \leq r < b$$: $$-5 \not\geq 0$$, so $$r$$ is not a valid remainder. 10. Therefore, 3 is the divisor, -14 is the dividend, but the remainder -5 does not satisfy the Division Algorithm conditions. 11. The correct remainder should be non-negative and less than 3. 12. Hence, 3 is the divisor in this equation, but the remainder is invalid under the Division Algorithm.