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If a positive even number n is not divisible by 3 or 4, then the produ

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If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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New post 14 Sep 2015, 05:19
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If a positive even number n is not divisible by 3 or 4, then the product (n + 6)(n + 8)(n + 10) must be divisible by which of the following?

I. 24
II. 32
III. 96

A. None
B. I only
C. II only
D. I and II only
E. I, II, and III

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If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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New post Updated on: 25 Aug 2016, 21:58
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1
Bunuel wrote:
If a positive even number n is not divisible by 3 or 4, then the product (n + 6)(n + 8)(n + 10) must be divisible by which of the following?

I. 24
II. 32
III. 96

A. None
B. I only
C. II only
D. I and II only
E. I, II, and III


Given : n is a multiple of 2 but not a multiple of 3 or 4

Possible values of n = 2, 10, 14, 22.....

@n= 2 : (n + 6)(n + 8)(n + 10) = 8*10*12 i.e. Divisible by 24, 32 and 96 all
@n=10 : (n + 6)(n + 8)(n + 10) = 16*18*20 i.e. Divisible by 24, 32 and 96 all
@n=14 : (n + 6)(n + 8)(n + 10) = 20*22*24 i.e. Divisible by 24, 32 and 96 all
@n=22 : (n + 6)(n + 8)(n + 10) = 28*30*32 i.e. Divisible by 24, 32 and 96 all

Answer: Option E
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Originally posted by GMATinsight on 14 Sep 2015, 05:54.
Last edited by GMATinsight on 25 Aug 2016, 21:58, edited 1 time in total.
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If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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New post 14 Sep 2015, 19:55
GMATinsight wrote:
Bunuel wrote:
If a positive even number n is not divisible by 3 or 4, then the product (n + 6)(n + 8)(n + 10) must be divisible by which of the following?

I. 24
II. 32
III. 96

A. None
B. I only
C. II only
D. I and II only
E. I, II, and III


Given : n is a multiple of 2 but not a multiple of 3 or 4

Possible values of n = 10, 14, 22.....

@n=10 : (n + 6)(n + 8)(n + 10) = 16*18*20 i.e. Divisible by 24, 32 and 96 all
@n=14 : (n + 6)(n + 8)(n + 10) = 20*22*24 i.e. Divisible by 24, 32 and 96 all
@n=22 : (n + 6)(n + 8)(n + 10) = 28*30*32 i.e. Divisible by 24, 32 and 96 all

Answer: Option E


Why does not your possible values include 2?
since 2 is also a multiple of 2.
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Re: If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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New post 14 Sep 2015, 20:14
1
If a positive even number n is not divisible by 3 or 4, then the product (n + 6)(n + 8)(n + 10) must be divisible by which of the following?

I. 24
II. 32
III. 96

A. None
B. I only
C. II only
D. I and II only
E. I, II, and III

Ans is E.

even multiple of 3 --- 6,12,18,24 and so on
multiple of 4 ---- 4,8,12,16 and so on

There is n not a multiple of 3(even) and 4 ----hence n possible value is ---2,10,14,22

Taking the n=2 and putting the value in equation (n + 6)(n + 8)(n + 10)
we get = 8*10*12
check option 1,2,3 all can divide 8*10*12

Hence ans is E
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Re: If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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New post 14 Sep 2015, 21:43
1
Aves wrote:
GMATinsight wrote:
Bunuel wrote:
If a positive even number n is not divisible by 3 or 4, then the product (n + 6)(n + 8)(n + 10) must be divisible by which of the following?

I. 24
II. 32
III. 96

A. None
B. I only
C. II only
D. I and II only
E. I, II, and III


Given : n is a multiple of 2 but not a multiple of 3 or 4

Possible values of n = 10, 14, 22.....

@n=10 : (n + 6)(n + 8)(n + 10) = 16*18*20 i.e. Divisible by 24, 32 and 96 all
@n=14 : (n + 6)(n + 8)(n + 10) = 20*22*24 i.e. Divisible by 24, 32 and 96 all
@n=22 : (n + 6)(n + 8)(n + 10) = 28*30*32 i.e. Divisible by 24, 32 and 96 all

Answer: Option E


Why does not your possible values include 2?
since 2 is also a multiple of 2.


2 also is a possible value and with value 2, the expression become 8*10*12 which is again divisible by all (I.e.24, 32 and 96) hence answer remains same.

However I wish to thank you here and want to admit that not mentioning 2 was a mistake done on my part which made me use bigger values to check the divisibility expression.

Smaller the values you check, easier it is for you to solve the question. So never miss smallest possible values and never check values randomly.

I hope this helps!!!
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Re: If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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New post 20 Sep 2015, 09:13
1
2
Bunuel wrote:
If a positive even number n is not divisible by 3 or 4, then the product (n + 6)(n + 8)(n + 10) must be divisible by which of the following?

I. 24
II. 32
III. 96

A. None
B. I only
C. II only
D. I and II only
E. I, II, and III


VERITAS PREP OFFICIAL SOLUTION:

This question offers several constraints for the possible values of n, and missing any one of those key traits will likely cost you the question. N must be:

* Positive
* Even
* Not divisible by 3
* Not divisible by 4

As this question was discussed by aspiring MBA students in a forum thread earlier this week, the most common reason for someone to miss it was that they missed that small word “even”; they’d plug in a number like 1 or 5 for n – numbers that ARE NOT possible given the constraints – and incorrectly believe that the answer is E. In actuality, however, these constraints taken together mean that the only possible values for n are numbers like:

2, 10, 14, 22, 34…

And for these numbers, the product of (n+6)(n+8)(n+10) is divisible by all of the above. We are assured that, if n is even but not divisible by 4, then n + 6 and n + 10 WILL BE divisible by 4 (try it: 2, 6, 10, 14…all of these numbers are even but not divisible by 4; add 6 and you get 8, 12, 16, 20 —> all divisible by 4). So the product will give us:

(n + 6)(n + 8)(n + 10)

which is equivalent to:

Divisible by 4 * Even * Divisible by 4

Providing prime factors of (at minimum):

(2*2)(2)(2*2) –> the product must be divisible by 2^5.

And because we have three consecutive even integers, exactly one will be divisible by 3 (try it: 2, 4, 6, 8, 10, 12, 14, 16, 18… – every third value is divisible by 3).

So we know that the product is divisible by 3*2^5, and to be divisible by I, II, and III we need:

24: 3*2^3 (check!)

32: 2^5 (check!)

96: 3*2^5 (check!)
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Re: If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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New post 28 Oct 2016, 06:05
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2
Let’s focus on one restriction. N is not divisible by 4. That means N has only one factor of 2. Now let’s represent N as 2x (where x is any prime factor except 3 and 2). We have (2x+6)(2x+8)(2x+10). Now take common factor 2. We’ll have:
2^3(x+3)(x+4)(x+5)=8(x+3)(x+4)(x+5)
Now we know that this product is divisible by 8.
Let’s check the divisibility of the remaining part (x+3)(x+4)(x+5)
Also notice that because we have factored out 2s all remaining Xs are now odd. In this case odd+3 and odd+5 are both even, and we know that this product is also divisible by 4.
Now, because we have 3 consecutive numbers, one of these numbers in any case will be a multiple of 3 (you can check this by yourself by plugging in any values). So this product is also divisible by 3.
Now let’s check available options.
24=8*3. After cancelling 8 the remaining part is divisible by 3.
32=8*4. Remaining part is divisible by 4.
96=8*4*3. Same logic. Divisible by 8, 3 and 4.
Answer E
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Re: If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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New post 01 Mar 2017, 02:59
We cross check few of the numbers like 0, 2, 10, 14….. We can see that the outcome for the product is divisible by 24,32,96. Hence Option E is the answer
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Re: If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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New post 13 Apr 2017, 18:05
Let n = 2k, n is not divisible by 3 and 4 --> k is not divisible by 3 and 2 --> k is odd

(n+6)(n+8)(n+10) = (2k+6) (2k+8)(2k+10) = 2^3(k+3)(k+4)(k+5)

We have (k+3)(k+4)(k+5) is the product of 3 consecutive integers --> (k+3)(k+4)(k+5) is divisible by 3. Futhermore, k is odd --> (k+3) and (k+5) are even --> (k+3)(k+4)(k+5) is divisible by 2^2.

In sum, (n+6)(n+8)(n+10) is divisible by 2^5 and 3 --> (n+6)(n+8)(n+10) is divisible by 24, 32 and 96

Hence, E
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Re: If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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New post 13 Apr 2017, 20:34
When you first look at this question, you must identify the numbers which are positive, even and not multiple of 4 or 3 which are 2, 10, 14, and 22 and so on. Looking at these numbers you will recognize that each number can have exactly one 2 when being factorized. For example 2-->2, 10-->2*5, 14-->2*7 and so on.
There is a number property that whenever a number which exactly have one 2 adds into any other number which also exactly have one 2, their sum have at least two 2s. So based on this property look at
(n+6) -->implies it will have at least two 2 (prime factorize n you will have one 2 and prime factorize 6 you will have one 2 so property applies over it that sum of n and 6 will have at least two 2s when factorized)
(n+10) -->implies it will have at least two 2
(n+8)-->Property does not apply over this as 8 have more than one 2 when factorized. However you can imply that (n+8) at lease have one 2 as this is an even number anyway.
(n+6)(n+8)(n+10) -->this combine expression indicates three consecutive even numbers. So this means this will have at least one 3.
Now you have everything just put into it (n+6)(n+8)(n+10) -->(2^2) 2 (2^2) 3 so it implies you can have 3*2^5 at maximum in answer choices. So when factories you will find no number in denominator exceeds than this.
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Re: If a positive even number n is not divisible by 3 or 4, then the produ  [#permalink]

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