The greatest number by which the product of three consecutive multiples of 3 is always divisible is

## Concept & Logic When dealing with consecutive multiples of a specific number, factor out the multiplier from each term first, then apply the standard rule for consecutive integers ($n!$). ## Step-by-Step Solution * Let the three consecutive multiples of 3 be $3n$, $3(n+1)$, and $3(n+2)$, where $n$ is a natural number. * Multiply them together: $$ 3n \times 3(n+1) \times 3(n+2) = 27[n(n+1)(n+2)] $$ * The term $n(n+1)(n+2)$ is the product of three consecutive natural numbers. As established, this is always divisible by $3!$, which is 6. * Therefore, the entire expression is always divisible by $27 \times 6 = 162$. ## Exam Strategy & Shortcut Bypass the algebra completely by finding the product of the smallest possible sequence. The smallest consecutive multiples of 3 are 3, 6, and 9. Their product is $3 \times 6 \times 9 = 162$. Since the question asks for the greatest number that *always* divides the product, it cannot be larger than the smallest possible product itself. ## Common Pitfall A common error is factoring out 3 only once instead of three times, mistakenly believing the product is $3 \times [n(n+1)(n+2)] = 18$. Remember that each of the three terms contributes a factor of 3. ## Final Answer **Therefore, the correct answer is 162.**