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

n is an integer. Is n(n+2) a multiple of 8?

1) n is an even integer
2) n is a multiple of 4

$${{n\left( {n + 2} \right)} \over 8}\,\,\mathop = \limits^? \,\,{\mathop{\rm int}} \,\,\,\,\,\,\,\left( {n\,\,\,{\mathop{\rm int}} } \right)$$
$$\left( 1 \right)\,\,n\,\,{\rm{even}}\,\,\, \Rightarrow \,\,\,{\rm{n = 2M}}{\rm{,}}\,\,\,{\rm{M}}\,\,{\mathop{\rm int}} $$
$${{n\left( {n + 2} \right)} \over 8} = {{2M\left( {2M + 2} \right)} \over 8} = {{4\overbrace {M\left( {M + 1} \right)}^{even\,\, = \,\,2L}} \over 8} = L\,\,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$
$$\left( 2 \right)\,\,n = 4K\,\,,\,\,K\,\,{\mathop{\rm int}} $$
$${{n\left( {n + 2} \right)} \over 8} = {{4K\left( {4K + 2} \right)} \over 8} = {{8K\left( {2K + 1} \right)} \over 8} = K\left( {2K + 1} \right)\,\,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

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.

Note that the product of two consecutive even integers is a multiple of 8 since one of them is a multiple of 4 and the other is an even integer.

Thus, each of conditions is sufficient since each implies that n and n+2 are two consecutive even integers.

Therefore, D is the answer.
Answer: D