1. **Problem (A):** If $n$ is an integer such that $3n + 2$ is odd, then $n$ is odd. 2. **Choosing the proof method for (A):** The hint suggests assuming the opposite, i.e., assume $n$ is even and show a contradiction. This is a proof by contrapositive or contradiction. 3. **Proof for (A):** - Assume $n$ is even, so $n = 2k$ for some integer $k$. - Substitute into $3n + 2$: $$3(2k) + 2 = 6k + 2 = 2(3k + 1)$$ - This expression is even, contradicting the assumption that $3n + 2$ is odd. - Therefore, $n$ must be odd. 4. **Problem (B):** For every integer $n \geq 1$, $n^3 + 2n$ is divisible by 3. 5. **Choosing the proof method for (B):** Since the statement is for all integers $n \geq 1$, mathematical induction is the natural choice. 6. **Proof for (B) by induction:** - **Base case:** For $n=1$, $1^3 + 2(1) = 1 + 2 = 3$, divisible by 3. - **Inductive hypothesis:** Assume for some $k \geq 1$, $k^3 + 2k$ is divisible by 3. - **Inductive step:** Consider $(k+1)^3 + 2(k+1)$: $$ (k+1)^3 + 2(k+1) = k^3 + 3k^2 + 3k + 1 + 2k + 2 = (k^3 + 2k) + 3k^2 + 3k + 3 $$ - By the inductive hypothesis, $k^3 + 2k$ is divisible by 3. - The remaining terms $3k^2 + 3k + 3$ are clearly divisible by 3. - Therefore, $(k+1)^3 + 2(k+1)$ is divisible by 3. - By induction, the statement holds for all $n \geq 1$. **Final answers:** - (A) Use proof by contradiction assuming $n$ is even. - (B) Use mathematical induction.