PrepTap wrote:

Is X divisible by \(3^k\)?

(1) X = 30!

(2) X = [(3n+6)(3n+9)(n+1)]^k/3, where n and k/3 are positive integers

A) \(30!/3^k\), since k can assume any value, it will vary the result. --

InsufficientB) \((3n+6)(3n+9)(n+1)]^\frac{k}{3}\)

This can be written as \([3(n+3)*3(n+2)*(n+1)]^\frac{k}{3}\) => \(9^\frac{k}{3}*[(n+3)*(n+2)*(n+1)]^\frac{k}{3}\)

Now \(X/3^k\) = \(9^\frac{k}{3}*[(n+3)*(n+2)*(n+1)]^\frac{k}{3}/3^k\)

=> \(3^\frac{2k}{3}*[(n+3)*(n+2)*(n+1)]^\frac{k}{3}/3^k\)

=>\([(n+3)*(n+2)*(n+1)]^\frac{k}{3}/3^\frac{k}{3}\)

Since, (n+3)*(n+2)*(n+1) are consecutive integers, at least on of them is divisible by 3; and since numerator and denominator has same exponent, it is divisible ----

SufficientHence Answer is B