is x/140 an integer?

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by deloitte247 » Sun Sep 30, 2018 11:04 am

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Question = is x/140 an integer?
$$\frac{x}{140}=\ \frac{x}{4\cdot5\cdot7}=\ \frac{x}{2^2\cdot5\cdot7}$$
For x to be an integer it must have 2 power of 2 and 1 power each of 5 and 7
Statement 1 = LCM of x and y = 360
= 5* 8 * 9
$$=5\ \cdot\ 2^3\ \cdot\ 3^2$$
$$=2^3\ \cdot\ 3^2\ \cdot\ 5$$
This means there is no powers of 7 in x and y and we are able to answer with all certainty that,
$$\frac{x}{140}\ is\ not\ an\ integer,\ hence\ statement\ 1\ is\ SUFFICIENT$$

Statement 2 = GCF of x and y = 40
GCF is same thing as HCF
Therefore, HCF of x and y = 40
=5 * 8
$$=5\ \cdot\ 2^3$$
$$=2^3\ \cdot\ 5$$

The HCF just provides us with the common powers and it doesn't provide enough information about the powers of other prime numbers.
Hence, Statement 2 is INSUFFICIENT.

Option A is CORRECT.

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by fskilnik@GMATH » Sun Sep 30, 2018 1:01 pm

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AbhishekRyu wrote:is x/140 an integer?
1) LCM of x & y is 360
2) GCF of x & y is 40
$${x \over {{2^2} \cdot 5 \cdot 7}}\,\,\,\mathop = \limits^? \,\,\,{\mathop{\rm int}} $$
$$\left( 1 \right)\,\,\,LCM\left( {x,y} \right) = 360\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,x\,\,{\mathop{\rm int}} \,\,\,\,({\rm{implicitly}}) \hfill \cr
\,x\,\,{\rm{is}}\,{\rm{a}}\,{\rm{factor}}\,{\rm{of}}\,\,360\,\,\,\,\, \Rightarrow \,\,\,\,\,{{{2^3} \cdot {3^2} \cdot 5} \over x} = {\mathop{\rm int}} \,\,\,\left( * \right) \hfill \cr} \right.$$
$$\left( * \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,7\,\,{\rm{is}}\,\,{\rm{not}}\,\,{\rm{a}}\,\,{\rm{factor}}\,\,{\rm{of}}\,\,x\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{x \over {{2^2} \cdot 5 \cdot 7}}\,\,\, \ne \,\,\,{\mathop{\rm int}} \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,$$

$$\left( 2 \right)\,\,\,GCF\left( {x,y} \right) = 40 = {2^3} \cdot 5\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {{2^3} \cdot 5 \cdot 7,{2^3} \cdot 5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {{2^3} \cdot 5,{2^3} \cdot 5 \cdot 7} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$

The correct answer is therefore (A).


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

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Jay@ManhattanReview » Sun Sep 30, 2018 8:52 pm

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AbhishekRyu wrote:is x/140 an integer?

1) LCM of x & y is 360
2) GCF of x & y is 40
We have to determine whether x/140 is an integer.

Let's take each statement one by one.

1) LCM of x & y is 360.

For x/140 to be an integer, x must be a multiple of 140. Upon factorizing 140, we found that 7 is one of the prime factors of 140; however, there is no prime factor 7 in the 360, the LCM; thus. we can conclude that x/140 is not an integer. Sufficient.

2) GCF of x & y is 40.

Case 1: Say x = 280 and y = 40, we have GCF of x & y = 40, and x/140 is an integer. The answer is Yes.
Case 2: Say x = 40 and y = 280, we have GCF of x & y = 40; however x/140 is NOT an integer. The answer is No.

No unique answer. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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