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Is X divisible by 3^k?

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Is X divisible by 3^k?  [#permalink]

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New post Updated on: 07 May 2015, 04:16
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Question Stats:

66% (01:29) correct 34% (01:55) wrong based on 134 sessions

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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

Originally posted by PrepTap on 07 May 2015, 04:13.
Last edited by Bunuel on 07 May 2015, 04:16, edited 1 time in total.
Edited the question.
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Re: Is X divisible by 3^k?  [#permalink]

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New post 07 May 2015, 08:54
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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. -- Insufficient

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

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 ---- Sufficient


Hence Answer is B
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Is X divisible by 3^k?  [#permalink]

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New post 31 May 2015, 01:37
Bunuel :- Can you please suggest or solve this by some easier method.
I didnt get what king did.

Thanks in advance
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Re: Is X divisible by 3^k?  [#permalink]

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New post 02 Jun 2015, 01:08
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Shree9975 wrote:
Bunuel :- Can you please suggest or solve this by some easier method.
I didnt get what king did.

Thanks in advance


Hi Shree9975

Please let us know where exactly you are facing problem?

This question is about simplifying the algebraic equation.

Statement 1: As we do not know the value of k, we cannot say if 30! is divisible by 3^k or not.

Statement 2: X = [(3n+6)(3n+9)(n+1)]^(k/3) = [3*(n+2)*3*(n + 3)*(n+1)]^(k/3) = [3*3*(n+1)(n+2)(n+3)]^(k/3)

(n+1)(n+2)(n+3) is a product of 3 consecutive integers and as every third number is divisible by 3, a product of 3 consecutive integers is divisible by 3.

Hence, (n+1)(n+2)(n+3) = 3*p where p is an integer.

Thus X = [3*3*(n+1)(n+2)(n+3)]^(k/3) = [3*3*3*p]^(k/3) = [3^3 * p]^(k/3) = 3^k * p^(k/3)

So we can conclude that X is divisible by 3^k.

Answer: Option (B)
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Re: Is X divisible by 3^k?  [#permalink]

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New post 02 Jun 2015, 02:45
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


Statement 1: X= 30!

The statement doesn't provide any information about k therefore this is insufficient to answer the Question

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

i.e. X = [3(n+2)x 3(n+3)x (n+1)]^(k/3), where n and k/3 are positive integers [Just took number 3 common from each of the first two brackets]

X = [3^(2k/3)][(n+2)x(n+3)x(n+1)]^(k/3)

but (n+2)x(n+3)x(n+1) is a product of three consecutive Integers and every three consecutive integer include one number as multiple of 3 therefore (n+2)x(n+3)x(n+1) will be a multiple of 3

Let (n+2)x(n+3)x(n+1) = 3a

X = [3^(2k/3)][(n+2)x(n+3)x(n+1)]^(k/3) = [3^(2k/3)](3a)^(k/3)

i.e. X = [3^(k)](a)^(k/3)

i.e. X is divisible by 3^k, Hence Sufficient

Answer: Option
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Re: Is X divisible by 3^k?  [#permalink]

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New post 10 Oct 2017, 13:09
Went through a couple of answers, and I felt that I needed to further simplify the factorization.

St 2: [(3n+6) (3n +9) (n + 1)]^k/3 --> [3(n+2) X 3(n +3) X (n+1)]^k/3 => 9^k/3 [(n+3)(n+2)(n+1)]/3^k

=> 3^2/3k / 3^k * [(n+3)(n+2)(n+1)]
=>3^k - 2/3k * [(n+3)(n+2)(n+1)] {BECAUSE larger power minus smaller power when base is the same}
=> 3^k[(n+3)(n+2)(n+1)] --> thus this expression is divisible by 3^k as not only the [] is a multiple of 3, it is multiplied by the given denominator 3^k
Re: Is X divisible by 3^k? &nbs [#permalink] 10 Oct 2017, 13:09
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