If x and y are positive integers, is √15xy an integer?

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[Math Revolution GMAT math practice question]

If x and y are positive integers, is $$\sqrt{15xy}$$ an integer?

1) xy is a multiple of 15
2) x and y are prime numbers

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by fskilnik@GMATH » Mon Nov 26, 2018 4:56 am

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

If x and y are positive integers, is $$\sqrt{15xy}$$ an integer?

1) xy is a multiple of 15
2) x and y are prime numbers
$$x,y\,\, \ge 1\,\,{\rm{ints}}\,\,\,\left( * \right)$$
$$\sqrt {15xy} \,\,\mathop = \limits^? \,\,\operatorname{int} \,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\boxed{\,xy\,\,\mathop = \limits^? \,\,15 \cdot {M^2}\,\,\,,\,\,\,M \geqslant 1\,\,\operatorname{int} \,}$$
$$\left( 1 \right)\,\,{{xy} \over {15}} = {\mathop{\rm int}} \,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,15} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,15} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,x,y\,\,{\rm{primes}}\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {3,5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,\,\left( {x,y} \right) = \left( {3,5} \right)\,\,\,{\rm{or}}\,\,\,\left( {x,y} \right) = \left( {5,3} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,$$


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

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Max@Math Revolution » Wed Nov 28, 2018 12:17 am

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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.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
When we consider both conditions together, there are two sets of possible values of x and y: x = 3, y = 5 and x = 5, y = 3. In both cases, xy = 15, so
√15xy = √15*3*5 = √225 = 15 is an integer.
Thus, both conditions together are sufficient.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If x = 3 and y = 5, then √15xy = √15*3*5 = √225 = 15 is an integer.
If x = 6 and y = 5, then √15xy = √15*6*5 = √450 = 15√2 is not an integer.
Since we don't have a unique answer, condition 1) is not sufficient by CMT (Common Mistake Type) 2.

Condition 2)
If x = 3 and y = 5, then √15xy = √15*3*5 = √225 = 15 is an integer.
If x = 2 and y = 5, then √15xy = √15*2*5 = √150 = 5√6 is not an integer.
Since we don't have a unique answer, condition 2) is not sufficient by CMT (Common Mistake Type) 2.

Therefore, C is the answer.
Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.