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

When A and B are positive integers, is AB a multiple of 4?

1) The greatest common divisor of A and B is 6
2) The least common multiple of A and B is 30

Another beautiful problem, Max. Congrats!
$$A,B\,\,\, \geqslant 1\,\,\,{\text{ints}}$$
$$\frac{{A \cdot B}}{4}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int} $$
$$\left( 1 \right)\,\,\,GCD\left( {A,B} \right) = 6\,\,\,\, \Rightarrow \,\,\,\left\{ \matrix{
\,A = 6M\,,\,\,\,M \ge 1\,\,{\mathop{\rm int}} \hfill \cr
\,B = 6N\,,\,\,\,N \ge 1\,\,{\mathop{\rm int}} \hfill \cr} \right.\,\,\,\,\,\,\,\,\,{\rm{with}}\,\,\,\,M,N\,\,\,{\rm{relatively}}\,\,\,{\rm{prime}}$$
$$?\,\,\,\,:\,\,\,\,\frac{{A \cdot B}}{4}\,\,\, = \,\,\,\frac{{6M \cdot 6N}}{4}\,\,\, = \,\,\,9MN\,\,\, = \,\,\,\operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle $$
$$\left( 2 \right)\,\,\,LCM\left( {A,B} \right) = 30\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {A,B} \right) = \left( {1,30} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {A,B} \right) = \left( {2,30} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.$$

The correct answer is therefore (A).


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.

Asking if AB is a multiple of 4 is equivalent to asking if AB = 4k for some integer k.

Condition 1)
Since A = 6a = 2*3*a and B = 6b = 2*3*b for some integers a and b, AB = 2^2*3^2*ab = 4*3^2ab.
Thus, AB is a multiple of 4 and condition 1) is sufficient.

Condition 2)
If A = 6 and B = 5, then lcm(A,B) = 30, but AB = 30 is not a multiple of 4, and the answer is 'no'.
If A = 6 and B = 10, then lcm(A,B) = 30, and AB = 60 is a multiple of 4. The answer is 'yes'.
Since we don't have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A