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If n is a positive integer, is

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by Max@Math Revolution » Tue Dec 25, 2018 12:55 am

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[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
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Source: — Data Sufficiency |
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If n is a positive integer, is $$\sqrt{n+1}$$ an even integ

by fskilnik@GMATH » Wed Dec 26, 2018 6:05 am
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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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edit

by Max@Math Revolution » Thu Dec 27, 2018 1:00 am
=>

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
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edited

by deloitte247 » Fri Dec 28, 2018 10:28 pm
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.

$$answer\ is\ Option\ D$$
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