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13 Nov 2010, 11:12
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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 - [t]m16#18[/t]
* 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
1. \(2 + x + 5 + y\) is an even integer
2. \(x - y\) is an odd integer
(C) 2008 GMAT Club - [t]m16#18[/t]
* 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
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13 Nov 2010, 11:30
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vicksikand
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.
Answer: C.
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.
13 Nov 2010, 11:48
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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 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.
