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
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\(n=3\sqrt{2p}=integer\) --> \(\sqrt{2p}=integer\), which basically means that 2p is a even perfect square: 4, 16, 36, 64 --> so \(\sqrt{2p}\) can be \(\sqrt{4}=2\), \(\sqrt{16}=4\), \(\sqrt{36}=6\), \(\sqrt{64}=8\), ... even numbers --> \(n=3*even\), so n is multiple of 6 for sure.

Hope it's clear.
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Lol.. Try to understand the emotions

N2 and N3 means N^2 and N^3.
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I will post in the right place from now on. Also in part 1 why only take the even integers to realize is was a multiple of 6

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I feel stupid for asking, but why are prime numbers only positive? Can't negative numbers be prime numbers, i.e. -2 or -3? Is this something I just have to know and not worry too much about?

By definition a prime number is a positive integer that has no positive integer divisors other than 1 and itself.

So just remember it and don't worry about it.

(The main reason we have this restrictions in definition is that if we remove them there will be some problems with fundamental theorem of arithmetic (unique prime factorization theorem), so removing them would require many other adjustment).
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what on earth are N2 and N3

Regards
Stonecold
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Then E is the answer i guess
Logic used => N and N^p has same prime factor.
A better way => Take N =6 and Smash that E
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I really don't understand why it is writen as N2 instead of N^2

Condition of N is not clear so E will be ans

Sent from my Redmi 4A using GMAT Club Forum mobile app

(1) 18 is a factor of N^2

Let N = 6.............. 18 factor of (6)^2................Answer is No

Let N =18.............. 18 factor of (18)^2..............Answer is Yes

Insufficient


(2) 27 is a factor of N^3

Let N = 6.............. 27 factor of (6)^3................Answer is No

Let N =18.............. 27 factor of (18)^3..............Answer is Yes

Insufficient

Combine 1 & 2

Use same examples above.....No clear answer

Answer: E

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