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M17-24

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M17-24 [#permalink]

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New post 16 Sep 2014, 01:01
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52% (01:11) correct 48% (01:12) wrong based on 141 sessions

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M17-24 [#permalink]

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New post 16 Sep 2014, 01:01
Official Solution:

If n is a positive integer, is 9 a factor of n ?

Statements (1) and (2) combined are insufficient. If \(N\) is 6, both statements hold but the answer is "no". If \(N = 18\), both statements hold and the answer is "yes".

Alternative Explanation

Algebraic approach:

Given: \(n=integer \gt 0\). Question: is \(n=9k\) (where \(k\) is a positive integer)?

(1) 18 is a factor of \(n^2\). Given, \(n^2=18p\), where \(p\) is a positive integer. So, \(n=3\sqrt{2p}\). Now, since \(n\) is an integer \(\sqrt{2p}\) must be an integer too, taking the even integer values. So, basically \(n=3\sqrt{2p}\) would be a multiple of \(3*2=6\), taking the following values: 6, 12, 18, ... Thus \(n\) may or may not be a multiple of 9. Not sufficient.

(2) 27 is a factor of \(n^3\). Given, \(n^3=27q\), where \(q\) is a positive integer. So, \(n=3\sqrt[3]{q}\). Now, since \(n\) is an integer \(\sqrt[3]{q}\) must be an integer too. So, basically \(n=3\sqrt[3]{q}\) would be just a multiple of 3: 3, 6, 9, 12, ... Thus \(n\) may or may not be a multiple of 9. Not sufficient.

(1)+(2) From (1) \(n\) is a positive multiple of 6 and from (2) \(n\) is a positive multiple of 3. Hence, we just know that \(n\) is a positive multiple of 6, so \(n\) could be 6, 12, 18, ... Thus \(n\) may or may not be a multiple of 9. Not sufficient.


Answer: E
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Re: M17-24 [#permalink]

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New post 06 Dec 2014, 17:09
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I used the method described in books from MGMAT (prime box)
for N to be divisible by 9, it must contain in its prime box two of 3's

1) 18 is a factor of N^2
prime factorization of 18 = 2 * 3^2
n^2 = n * n
which means that one n, in its prime box, will take a two, and a three, since we do not know whether n might contain another 3, statement 1 is insufficient.
2) same method used, we have that n has only one 3, it might have more than one, but we do not know, hence the statement is insufficient.

1+2 -> we know for sure that n has one three in its prime box, but it might have more than one, therefore 1+2 insufficient.

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Re: M17-24 [#permalink]

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New post 15 Apr 2017, 03:10
Please could you explain the concept of Prime boxes

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Re: M17-24 [#permalink]

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New post 12 May 2017, 07:42
The answer must be option E.

From statement 1: n^2=18p ; n=3*sqrt(2p). Not sufficient as we can't be sure if the second part has a factor of 3 or not.
From statement 2: n^3=27q ; n=3*((q)^(1/3)). Not sufficient as we can't be sure if the second part has a factor of 3 or not.

Statement 1 & 2: n^5=(3^5)*(r) ---> n=3*((3)*(r^1/5)). Not sufficient as we can't be sure if the second part has a factor of 3 or not.

Option E is the correct answer
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Re: M17-24 [#permalink]

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New post 10 Oct 2017, 08:03
Hello wwambugu14,

As a concept, the 'Prime Box' is very well explained in MGMAT strategy guide. It is explained in their 'Number properties' book. Give it a quick read and you would be able to handle problems such as this.

Also, there is a Thursdays with Ron video on this topic. You may want to have a look there as well. The link for it is -

wwambugu14 wrote:
Please could you explain the concept of Prime boxes

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Re: M17-24   [#permalink] 10 Oct 2017, 08:03
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