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M05 #16

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M05 #16 [#permalink] New post 08 Oct 2008, 15:27
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\frac{(40 * x^2 * y^2)}{z} is divisible by prime number Q. Is Q an even prime number?

1. Product of x^2 * y^2 is even
2. z * Q = 6

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Statement 1 does not help by itself. It tells us that x or y or both are even.

Statement 2 states that either z or Q is equal to 2.

Combining the two statements does not provide the answer either, we can figure out that either x or y has to be divisible by 3.

To make sure, we can pick numbers 4 for x and 3 for y. When we divide by 6, there is still no way to tell if z or Q is equal to 2.


I don't understand why z or Q has to be equal to 2; can't Q be 11 and z equal 6/11?
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Re: M05 #16 [#permalink] New post 09 Oct 2008, 02:18
SomeWheer must mentioned in the question x,y & z are integers or whole numbers
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Re: M05 #16 [#permalink] New post 14 Dec 2010, 06:50
make sure that the posted question is correct in wording.
i think that there has to some information about the x,y &z.
if we say that x,y & z are integers,
then,
statement 1: Product of x^2 * y^2 is even
doesn't make out any solution.
because it doesnot tell anything about z.
whether it ll be divisible by Q or not.

Statement 2: z * Q = 6

this statement doesn't say anything about the x & y.


so,
using both statement alone or together we cant answer the question.

correct my explanation if wrong.
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Re: M05 #16 [#permalink] New post 14 Dec 2010, 08:37
E.

Rephrasing the question: Is Q = 2?

S1: for x^2 *y^2 to be even any of x or y should be even or both can be even. This does not help nail the value of Q as z could be odd or even and based on that Q could be even or odd
S2 : this shows that either of z and Q can be 2 or 3

Together too they don't answer the question as z could be odd or even and based on that Q could be even or odd.
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Re: M05 #16 [#permalink] New post 14 Dec 2010, 08:42
snowy2009 wrote:

I don't understand why z or Q has to be equal to 2; can't Q be 11 and z equal 6/11?


even if we assume that there is no mention of the fact that x, y and z are integers and your solution is correct the answer still remains E. However the explanation provided by GMAT club can be questioned in that case.
In all probability it should have been mentioned that all variables in this question should be assumed to be integers.
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Re: M05 #16 [#permalink] New post 14 Dec 2010, 08:56
I got it down to an unsure C or E and guessed E. :/
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Re: M05 #16 [#permalink] New post 14 Dec 2010, 19:08
Using both stmt Q can be either 3 or 2 .

Now --> 40 = 8 * 5

So, it depends upon the values of X and Y, it is only provided that either of X an Y is even, we cant confirm the value of Q .

So, E
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Re: M05 #16 [#permalink] New post 14 Dec 2010, 23:17
That 40*x2*y2 is even is evident even without looking at stmt 1. Stmt 1 is insuff as it doesn't say anything abt z or q
z*q = 6 means either z or q is 2 or 3. If z is 2 then q is 3, an odd prime. But if z is 3 then q is 2, an even prime. But stmt 2 doesn't say anything conclusively. So stmt 2 insuff

Together, they don't indicate the value of z or q either

Hence, the ans is E
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Re: M05 #16 [#permalink] New post 15 Dec 2010, 01:46
here we use the formula

dividend = divisor*quotient+remainder and we shud find if Q=2??

here remainder = 0

since 40*x^2*y^2 = even = z*Q

(1) is not sufficient.

(2) says z*Q = 6 which leaves the option of Q to be 2 or 3.

(1) and (2) together does not make any difference.

So E
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Re: M05 #16 [#permalink] New post 22 Jan 2011, 03:16
sonnco wrote:
I got it down to an unsure C or E and guessed E. :/


Me too.. haha..

Must say it was a nervous few seconds before I clicked "Reveal".
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Re: M05 #16 [#permalink] New post 16 Dec 2011, 11:46
The question is asking if q is 2.

In order to know that, we must figure out what z is (or q, but they dont usually give that directly.
S1: Says nothing about the denominator.
S2: gives us Z and Q together, but we don't know which is 3 and which is 2.

Together, no help. Answer is E.
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Re: M05 #16 [#permalink] New post 02 Jun 2012, 11:24
MODS:

The link to this thread seems outdated..

I came here to M5 #16 looking for:

If x and y are positive integers and xy is divisible by prime number p . Is p an even number?

1 x^2 * y^2 is an even number

2 xp = 6
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Re: M05 #16 [#permalink] New post 02 Jun 2012, 11:55
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macjas wrote:
MODS:

The link to this thread seems outdated..

I came here to M5 #16 looking for:

If x and y are positive integers and xy is divisible by prime number p . Is p an even number?

1 x^2 * y^2 is an even number

2 xp = 6


Here is a solution for that problem.

If x and y are positive integers and xy is divisible by prime number p. Is p an even number?

Notice that as given that p is a prime number and the only even prime is 2, then the question basically asks whether p=2.

(1) x^2 * y^2 is an even number --> xy=even (this means that at least one of the unknowns is even). We have that some even number is divisible by prime number p, not sufficient to say whether p=2, for example if xy=6 then p can be ether 2 or 3.

(2) xp = 6 --> since x is a positive integer and p is a prime number then either x=2 and p=3 (answer NO) or x=3 and p=2 (answer YES). Not sufficient.

(1)+(2) If y=6 then xy=even, so the first statement is satisfied irrespective of the value of x and thus we have no constraints on its value. So from (2) x can take any of the two values 2 or 3, which means that p can also take any of the two values 2 or 3, respectively. Not sufficient.

Answer: E.
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Re: M05 #16 [#permalink] New post 02 Jun 2012, 20:45
Thanks Bunuel,

I kinda forgot that E/E and E/O can yield an integer if the numerator is a multiple of the denominator. The GMAT Club questions are great memory joggers!
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Re: M05 #16 [#permalink] New post 16 Jun 2012, 09:38
Basically asking whether Q=2
Statement 1 not very useful since it is mum on what Q could be. Not sufficient.
Statement 2 implies that Q is 2 or 3. Not sufficient.

Combining leaves us still unsure since x^2y^=even could mean that x^2y^2 could be a multiple of 2 or 3, thus masking the identity of Q, which could still be a 2 0r 3.
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Re: M05 #16 [#permalink] New post 19 Dec 2012, 15:21
Why OA E IMO E



x^2.y^2=even mean the multiple of these two is an integer (can be a factor of 3) so 40*even integer = even INTEGER

Q*z=6 if Q is an prime number it is also an integer thus z must be an integer so when we take these facts together Q can be (1,2,3, NOT 6 THAT IS WHY Z IS AN INTEGER)

1,2,3 Q can be so both statements together insufficient
Re: M05 #16   [#permalink] 19 Dec 2012, 15:21
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