Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If n is a positive integer, is $$\sqrt{n+1}$$ an even integer?

1) n is the product of 2 consecutive odd numbers
2) n is an odd number

Beautiful problem, Max. Congrats!

$$n \geqslant 1\,\,\,\operatorname{int} $$
$$\sqrt {n + 1} \,\,\,\mathop = \limits^? \,\,{\text{even}}\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\boxed{\,\,n + 1\,\,\,\mathop = \limits^? \,\,\,{{\left( {{\text{even}}} \right)}^2}\,\,}$$

$$\left( 1 \right)\,\,\,n = \left( {2M - 1} \right)\left( {2M + 1} \right) = {\left( {2M} \right)^2} - {\left( 1 \right)^2}\,\,\,\,\,\left[ {M\,\,\operatorname{int} \,} \right]\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,n + 1 = {\left( {2M} \right)^2}\,\,\,,\,\,\,\,M\,\,\operatorname{int} \,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,$$
$$\left( 2 \right)\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,n = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,n = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The question is equivalent to asking if $$\sqrt{n+1}=2k$$ for some positive integer k.
$$\sqrt{n+1}=2k$$
=> n+1 = 4k^2
=> n = 4k^2-1
=> n = (2k-1)(2k+1)
n is a product of two consecutive odd integers.
Thus, condition 1) is sufficient.

Condition 2)
If n = 3, then $$\sqrt{3+1}=\sqrt{4}=2$$ and the answer is 'yes'.
If n = 1, then $$\sqrt{1+1}=\sqrt{2}$$ is not an integer and the answer is 'no'.
Condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

Even integers = 2k , where k is an integer
$$For\ \sqrt{n+1}=\ integer\ $$
It must be evaluated to be a rational number, hence
$$\sqrt{n+1}=perfect\ square$$

Statement 1
n is the product of 2 consecutive odd numbers
Odd number = 2k +1 $$2\ con\sec utive\ odd\ numbers=\left(2k+1\right),\left(2k+1\right)+2$$ $$\Pr oduct\ of\ \left(2k+1\right)\ \left(2k+1\right)+2=Odd\ number\ $$ $$n=Odd\ number\ $$
Statement 1 is INSUFFICIENT.

Statement 2
n is an odd number
$$hence\ \sqrt{n+1}=Even\ integer\ $$
statement 2 is INSUFFICIENT.

Both statement alone are SUFFICIENT.

$$answer\ is\ Option\ D$$