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# Is X divisible by 3^k?

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Manager
Joined: 22 Apr 2015
Posts: 63
Is X divisible by 3^k?  [#permalink]

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Updated on: 07 May 2015, 03:16
2
9
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Difficulty:

55% (hard)

Question Stats:

64% (02:05) correct 36% (02:08) wrong based on 136 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, 03:13.
Last edited by Bunuel on 07 May 2015, 03:16, edited 1 time in total.
Edited the question.
Manager
Joined: 03 Sep 2014
Posts: 74
Concentration: Marketing, Healthcare
Re: Is X divisible by 3^k?  [#permalink]

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07 May 2015, 07:54
3
1
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

Manager
Joined: 06 Mar 2014
Posts: 95
Is X divisible by 3^k?  [#permalink]

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31 May 2015, 00:37
Bunuel :- Can you please suggest or solve this by some easier method.
I didnt get what king did.

Manager
Joined: 22 Apr 2015
Posts: 63
Re: Is X divisible by 3^k?  [#permalink]

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02 Jun 2015, 00:08
1
Shree9975 wrote:
Bunuel :- Can you please suggest or solve this by some easier method.
I didnt get what king did.

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.

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

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02 Jun 2015, 01:45
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

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Senior Manager
Joined: 15 Jan 2017
Posts: 356
Re: Is X divisible by 3^k?  [#permalink]

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10 Oct 2017, 12:09
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, 12:09
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