The greatest number by which the product of three consecutive multiples of 3 is always divisible is :
Options:
A . 54
B . 81
C . 162
D . 243
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
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