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Examveda

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

A. 54

B. 81

C. 162

D. 243

Answer: Option C

Solution(By Examveda Team)

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

This Question Belongs to Arithmetic Ability >> Number System

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