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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Please could you explain the concept of Prime boxes

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 -


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