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

If x, y are integers, is x^2+x+y an odd integer?

1) x is an odd integer
2) y is an odd integer

$$x,y\,\,{\rm{ints}}\,\,\,\left( * \right)$$
$${{\text{x}}^{\text{2}}} + x + y\,\, = \,\,\underbrace {x\left( {x + 1} \right)}_{\left( * \right)\,\,{\text{even}}} + y\,\,\mathop = \limits^? \,\,\,{\text{odd}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,\,y\,\,\mathop = \limits^? \,\,\,{\text{odd}}\,\,}$$
$$\left( 1 \right)\,\,x\,\,{\rm{odd}}\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,y\,\,{\rm{odd}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

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

Regards,
Fabio.

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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 parity of x^2+x+y = x(x+1) + y is same as the parity of y, since x^2+x = x(x+1) is the product of two consecutive integers and so it is always an even integer.
Thus, asking whether x^2+x+y = x(x+1) + y is odd is equivalent to asking whether y is odd.

Therefore, B is the answer.
Answer: B