1. **Problem Statement:** Prove that the product $n(n+1)(n+2)$ is divisible by 3 for all integers $n > 1$ using the principle of mathematical induction. 2. **Induction Principle:** To prove a statement $P(n)$ for all $n \geq k$, we do: - Base Case: Verify $P(k)$ is true. - Inductive Step: Assume $P(m)$ is true for some $m \geq k$, then prove $P(m+1)$ is true. 3. **Base Case:** For $n=2$, calculate $2 \times 3 \times 4 = 24$. Since $24$ is divisible by 3, the base case holds. 4. **Inductive Hypothesis:** Assume for some $n = m \geq 2$, the product $m(m+1)(m+2)$ is divisible by 3. That is, there exists an integer $k$ such that: $$m(m+1)(m+2) = 3k$$ 5. **Inductive Step:** We need to prove that $(m+1)(m+2)(m+3)$ is divisible by 3. 6. **Proof of Inductive Step:** Consider the three consecutive integers $m+1$, $m+2$, and $m+3$. Among any three consecutive integers, exactly one is divisible by 3 because integers modulo 3 cycle every 3 numbers. 7. Since $m(m+1)(m+2)$ is divisible by 3 by the inductive hypothesis, and the set $(m+1)(m+2)(m+3)$ is just shifted by 1, it also contains exactly one multiple of 3. 8. Therefore, $(m+1)(m+2)(m+3)$ is divisible by 3. 9. **Conclusion:** By the principle of mathematical induction, $n(n+1)(n+2)$ is divisible by 3 for all integers $n > 1$. **Final answer:** The product of any three consecutive integers is divisible by 3 for all $n > 1$.