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

Answer: Option C

Three consecutive multiples of 3 are 3m, 3(m + 1) and 3(m + 2)
Their product = 3m × 3(m + 1) × 3(m + 2)
                      = 27 × m × (m + 1) × (m + 2)
Putting m = 1, this product is (27 × 1 × 2 × 3) = 162
So, this product is always divisible by 162