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

If m and n are integers, is m+m^2-n an even number?

1) m is an even number
2) n is an even number

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

The correct answer is therefore (B).

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.

m+m^2-n = m(m+1) - n. Now, m(m+1) is an even number since m(m+1) is the product of two consecutive integers. Thus, the parity of m+m^2-n depends on the parity of n only.
Thus, condition 2) is sufficient.

Therefore, B is the answer.
Answer: B