1. **State the problem:** We are given a number trick with the following steps: - Choose a number $n$ - Double it: $2n$ - Add 6: $2n + 6$ - Double it again: $2(2n + 6) = 4n + 12$ - Subtract 4: $(4n + 12) - 4 = 4n + 8$ - Divide by 4: $\frac{4n + 8}{4} = n + 2$ - Subtract 2: $(n + 2) - 2 = n$ 2. **Make a conjecture:** After performing all the steps, the final result is the original number $n$. This means the trick returns the starting number regardless of what $n$ is. 3. **Prove the conjecture using algebra:** - Start with $n$ - Double it: $2n$ - Add 6: $2n + 6$ - Double it again: $2(2n + 6) = 4n + 12$ - Subtract 4: $4n + 12 - 4 = 4n + 8$ - Divide by 4: $\frac{4n + 8}{4} = n + 2$ - Subtract 2: $n + 2 - 2 = n$ Since the final expression simplifies exactly to $n$, the original number, the conjecture is proven. 4. **Explanation:** This trick always returns the original number because the operations simplify algebraically to $n$. The additions and subtractions cancel out after doubling and dividing, leaving the starting number unchanged. **Final answer:** The number trick returns the original number $n$ after all the steps.