Is 3 a factor of x?

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Is 3 a factor of x?

by Max@Math Revolution » Thu Jan 31, 2019 12:57 am

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[GMAT math practice question]

Is 3 a factor of x?

1) x-3 is divisible by 6
2) x+3 is divisible by 6

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Is 3 a factor of x?

by fskilnik@GMATH » Thu Jan 31, 2019 4:14 am

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

Is 3 a factor of x?

1) x-3 is divisible by 6
2) x+3 is divisible by 6
$${x \over 3}\,\,\mathop = \limits^? \,\,{\mathop{\rm int}} \,\,\,\,\left( {{\rm{and}}\,\,x\mathop = \limits^? {\mathop{\rm int}} } \right)$$

$$\left( 1 \right)\,\,{{x - 3} \over 6} = {\mathop{\rm int}} \,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,2} \,\,\,\,{{x - 3} \over 3} = {\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,{x \over 3} = {\mathop{\rm int}} + 1 = {\mathop{\rm int}} \,\,\,\,\left( {{\rm{and}}\,\,x = 3 \cdot {\mathop{\rm int}} + 3 = {\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$
$$\left( 2 \right)\,\,{{x + 3} \over 6} = {\mathop{\rm int}} \,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,2} \,\,\,\,{{x + 3} \over 3} = {\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,{x \over 3} = {\mathop{\rm int}} - 1 = {\mathop{\rm int}} \,\,\,\,\left( {{\rm{and}}\,\,x = 3 \cdot {\mathop{\rm int}} - 3 = {\mathop{\rm int}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

The correct answer is therefore (D).


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Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Max@Math Revolution » Mon Feb 04, 2019 4:10 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 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

To solve remainder questions, plugging in numbers is recommended.

Condition 1)
If we plug in x = 9, then x - 3 = 6 is divisible by 6 and x is a multiple of 3. Condition 1) is sufficient.

Condition 2)
If we plug in x = 9, then x + 3 = 12 is divisible by 6 and x is a multiple of 3. Condition 2) is sufficient.

Therefore, the answer is D.
Answer: D