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

If \(n\) is a positive integer, is 9 a factor of \(n\) ?

Given: \(n\) is a positive integer. The question is: can \(n\) be expressed as \(n=9k\), where \(k\) is a positive integer?

(1) 18 is a factor of \(n^2\).

From the given information, \(n^2 = 18p\), where \(p\) is a positive integer. This means \(n = 3\sqrt{2p}\). Since \(n\) must be an integer, \(\sqrt{2p}\) must also be an integer, specifically one with even values. Therefore, \(n = 3\sqrt{2p}\) can take values that are multiples of \(3*2 = 6\), such as 6, 12, 18, etc. Based on this, \(n\) might or might not be a multiple of 9. Not sufficient.

(2) 27 is a factor of \(n^3\).

From the given information, \(n^3 = 27q\), where \(q\) is a positive integer. This means \(n = 3\sqrt[3]{q}\). Since \(n\) is an integer, \(\sqrt[3]{q}\) must also be an integer. Therefore, \(n = 3\sqrt[3]{q}\) can take values that are multiples of 3, such as 3, 6, 9, 12, etc. Thus, \(n\) might or might not be a multiple of 9. Not sufficient.

(1)+(2) From (1), we know that \(n\) is a positive multiple of 6. From (2), we know that \(n\) is a positive multiple of 3. Combining these pieces of information, we only know that \(n\) is a positive multiple of 6, and thus \(n\) could be 6, 12, 18, etc. This means that \(n\) might or might not be a multiple of 9. Not sufficient.


Answer: E