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can someone show me how to answer this without having to find actual number

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

1. 18 is a factor of N^2 2. 27 is a factor of N^3

Algebraic approach:

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

(1) 18 is a factor of n^2 --> \(n^2=18p\), where \(p\) is a positive integer --> \(n=3\sqrt{2p}\) --> as \(n\) 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, ... So \(n\) may or may not be a multiple of 9. Not sufficient.

(2) 27 is a factor of n^3 --> \(n^3=27q\), where \(q\) is a positive integer --> \(n=3\sqrt[3]{q}\) --> as \(n\) 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, ... So \(n\) 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 --> \(n\) 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.

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

Posted from my mobile device

\(n=3\sqrt{2p}=integer\) --> \(\sqrt{2p}=integer\), 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.

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?

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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