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.

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.

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