The greatest number by which the product of three consecutive multiples of 3…

Step-by-Step Solution

To find the greatest number that always divides the product of three consecutive multiples of 3, we can represent these multiples algebraically and simplify their product.

1. Define the consecutive multiples: Let the three consecutive multiples of 3 be 3m, 3(m + 1), and 3(m + 2), where m is any positive integer.

2. Calculate the product: Product = (3m) × (3(m + 1)) × (3(m + 2)) Product = 27 × m × (m + 1) × (m + 2)

3. Analyze the product: The term m × (m + 1) × (m + 2) represents the product of three consecutive integers. A known property in mathematics is that the product of any three consecutive integers is always divisible by 6 (or 3! = 3 × 2 × 1).

4. Find the greatest divisor: Since the product is 27 × (m × (m + 1) × (m + 2)), and (m × (m + 1) × (m + 2)) is always divisible by 6: Product = 27 × (6k) = 162k (where k is some integer). Therefore, the product is always divisible by 162.