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

n is a 3 digit integer of the form ab6. Is n divisible by 4?

1) a+b is an even integer
2) ab is an odd integer.

$$n = \left\langle {ab6} \right\rangle \,\,\,\, \Rightarrow \left\{ \matrix{
\,n > 0\,\,\,\left( {{\rm{implicitly}}} \right) \hfill \cr
\,a \in \left\{ {1,2,3, \ldots ,9} \right\} \hfill \cr
\,b \in \left\{ {0,1,2,3, \ldots ,9} \right\} \hfill \cr} \right.$$
$${{\left\langle {ab6} \right\rangle } \over 4}\,\,\mathop = \limits^? \,\,{\mathop{\rm int}} $$
$$\left( 1 \right)\,\,\,a + b = {\rm{even}}\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {2,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,ab = {\rm{odd}}\,\,\, \Rightarrow \,\,\,b = {\rm{odd}}\,\,\, \Rightarrow \,\,\,\left\langle {b6} \right\rangle \in \left\{ {16,36,56,76,96} \right\}\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,$$


The correct answer is therefore (B).


We follow 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.

We can determine whether a number is divisible by 4 from its final two digits.
Numbers with the final digits 16, 36, 56, 76 and 96 are divisible by 4 and those with final digits 06, 26, 46, 66 and 88 are not divisible by 4. Thus, asking whether n is divisible by 4 is equivalent to asking whether b is odd.

Since it implies that both a and b are odd integers, condition 2) is sufficient.

Condition 1)
There are two cases to consider.
If a is an even integer and b is an odd integer, the answer is 'yes'.
If a is an odd integer and b is an even integer, the answer is 'no'.
Since it does not yield a unique solution, condition 1) is not sufficient.

Therefore, B is the answer.
Answer: B