M16#18

Question

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

* 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

Bunuel · Answer

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.

vicksikand · Answer

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.

Bunuel · Answer

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.

M16#18

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards