If n is an integer, is m is an integer ?

Updated the question. Sorry for confusion folks!

(1) \(\frac{m\sqrt{3}}{\sqrt{12}}\) is an integer

\(\frac{m\sqrt3}{\sqrt 3 \sqrt 4}\)

\(\frac{m}{2}\) = Integer

\(m = 2 \)* integer

Hence \(m\) is an integer.

SUFF.

(2) \(3m = n^3 - n\)

\(3m =n(n^2-1)\)

\(3m=(n-1)n(n+1)\)

\(m= \frac{(n-1)n(n+1)}{3}\)

Out of \(3\) consecutive integers in the numerator, one will always be a multiple of \(3 \), hence the \(3\) in the denominator will cancel out and \(m\) will be an integer.

Hence m is an integer.

SUFF.

Ans D

Hope it's clear.

For statement 1, what if m = 6/3? then M is not an integer. (Yes, we can reduce it, however, my question is do we assume such a thing?)

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From (1) we get that m/2 = integer, which means that m = 2*integer = integer. Hence, (1) is sufficient to get a YES answer to the question.

To answer your question: note that 6/3 = 2, so it's an integer no matter how you write it.­

­I have seen a couple of wrong responses for statement 1, so I wanted to give it a go:

Statement 1 tells you that the result of (m*3^(1/2))/(12^(1/2)) is an integer. Therefore, it can be rewritten as: (m*3^(1/2))/(12^(1/2)) = int (int being the integer). 

When you then solve for it like so:

(m*3^(1/2))/(12^(1/2)) = int   | separate the (12^(1/2) into (3^(1/2)*4^(1/2)

=> (m*3^(1/2))/(3^(1/2)*4^(1/2)) = int   | cancel the 3^(1/2)

=> (m)/4^(1/2)) = int  | 4^(1/2) is equal to 2

Therefore the entire equation can be rewritten as: m/2 = int

If you then times both sides by 2, you end up with m= int*2. An integer times 2 is always an integer, thus the statement is sufficient. 

 Hope this helps. See other posts for the solution to Statement 2. Answer D is correct.