Number System Practice Test 1

Question : 30 of 40

Marks: +1, -0

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

Solution:

Let the three consecutive multiples of 3 are 3n, 3(n + 1), 3(n + 2)
Now, 3n × 3(n + 1) × 3(n + 2)

= 27n (n^2 + 3n + 2)

Putting, n = 1 then,

27 × 1 (1^2 + 3 × 1 + 2)

= 27 × 6 = 162
Hence, required number = 162

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