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# M16#18

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Senior Manager
Joined: 29 Sep 2009
Posts: 396
GMAT 1: 690 Q47 V38
Followers: 2

Kudos [?]: 35 [0], given: 5

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13 Nov 2010, 10:12
Is the product $$2*x*5*y$$ an even integer?

1. $$2 + x + 5 + y$$ is an even integer
2. $$x - y$$ is an odd integer

(C) 2008 GMAT Club - m16#18

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

I did get the right answer, but I overshot on time. I have a fundamental question that I want to ask you guys:
if x + y = Odd and
x - y = Odd , then can we conclude that x and y will always be integers? In the question solution we did prove it algebraically and thus I am inclined to "remember" this fact rather than solve it, if encountered again in a problem.
Extrapolating from the official explanation:
x+y=n ....1
x-y=m.....2
1: x=n-y , subst in 2
2 can be re-written as 2y=(n-m)
LHS is always Even - this implies y is an Integer
Examine RHS : n-m , now difference of 2 Integers is Even when both are odd or both are even. Thus we can conclude the following:
if x+y = Odd and x-y = Odd then both x, and y are integers.
if x+y = Even and x-y = Even then both x, and y are integers.

FYI - O.A. is C
Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7091

Kudos [?]: 93344 [0], given: 10557

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13 Nov 2010, 10:30
vicksikand wrote:
Is the product $$2*x*5*y$$ an even integer?

1. $$2 + x + 5 + y$$ is an even integer
2. $$x - y$$ is an odd integer

(C) 2008 GMAT Club - m16#18

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

I did get the right answer, but I overshot on time. I have a fundamental question that I want to ask you guys:
if x + y = Odd and
x - y = Odd , then can we conclude that x and y will always be integers? In the question solution we did prove it algebraically and thus I am inclined to "remember" this fact rather than solve it, if encountered again in a problem.

O.A. is C

Question: is $$2*x*5*y=even$$? As there is 2 as a multiple, then this expression will be even if $$5xy=integer$$. Basically we are asked is $$5xy=integer$$ true?

Note that $$x$$ and $$y$$ may not be integers for $$2*x*5*y$$ to be even (example $$x=\frac{7}{9}$$ and $$y=\frac{9}{7}$$) BUT if they are integers then $$2*x*5*y$$ is even.

(1) $$2+x+5+y=even$$ --> $$7+x+y=even$$ --> $$x+y=odd$$. Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=1.7 answer NO)

(2) $$x-y=odd$$. Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=0.3 answer NO)

(1)+(2) Sum (1) and (2) $$(x+y)+(x-y)=odd_1+odd_2$$ --> $$2x=even$$ --> $$x=integer$$ --> $$y=integer$$ --> Both $$x$$ and $$y$$ are integers. Hence sufficient.

As for your question: if $$x+y=odd_1$$ and $$x-y=odd_2$$ then $$x$$ and $$y$$ must be integers (see proof above).

Hope it helps.
_________________
Senior Manager
Joined: 29 Sep 2009
Posts: 396
GMAT 1: 690 Q47 V38
Followers: 2

Kudos [?]: 35 [0], given: 5

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13 Nov 2010, 10:48
Bunuel
I didn't have a problem understanding the explanation ,however what i am saying is that shouldn't the following always hold true:
if x+y=odd and
x-y=odd
then both x and y are always integers
similarly
if x+y=even and
x-y=even then both x and y are always integers.
This result may come in handy to save time solving some complex number prop problems.
Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7091

Kudos [?]: 93344 [0], given: 10557

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13 Nov 2010, 10:54
vicksikand wrote:
Bunuel
I didn't have a problem understanding the explanation ,however what i am saying is that shouldn't the following always hold true:
if x+y=odd and
x-y=odd
then both x and y are always integers
similarly
if x+y=even and
x-y=even then both x and y are always integers.
This result may come in handy to save time solving some complex number prop problems.

I thought I answered this question.

If $$x+y=odd_1$$ and $$x-y=odd_2$$ then $$x$$ and $$y$$ must be integers: add them up $$(x+y)+(x-y)=odd_1+odd_2$$ --> $$2x=even$$ --> $$x=integer$$ --> $$y=integer$$ --> Both $$x$$ and $$y$$ are integers.

If $$x+y=even_1$$ and $$x-y=even_2$$ then $$x$$ and $$y$$ must be integers: add them up $$(x+y)+(x-y)=even_1+even_2$$ --> $$2x=even$$ --> $$x=integer$$ --> $$y=integer$$ --> Both $$x$$ and $$y$$ are integers.

Hope it's clear.
_________________
Re: M16#18   [#permalink] 13 Nov 2010, 10:54
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# M16#18

Moderator: Bunuel

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