Jinglander wrote:

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.

Answer: E.

Also:

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