Is (x + y)^3 an even integer?

Question

(1) x and y are integers 
(2) xy = 9

VenoMfTw · Answer

IMO : C

St 1 : x and y integers 
x = even y = even then (x + y)^3 = even
x = even y = odd then (x + y)^3 = odd
x = odd y = odd then (x + y)^3 = even
Nt suff

St 2 : xy = 9
x, y can be integers or fractions 
nt suff

Combined : 
xy = 9
(x,y) can be (9,1), (1,9), (3,3)
both are odd. Thus case(3) from st 1 x = odd y = odd then (x + y)^3 = even.
Suff

GMATinsight · Answer

Question : Is (x + y)^3 an even integer?

Statement 1 : x and y are integers  
But for x even and y even (x + y)^3 will be Even and 
But for x even and y Odd (x + y)^3 will be Odd
 NOT SUFFICIENT

Statement 2 : xy = 9  
for x=3 and y = 3, (x + y)^3 will be Even and 
for x=3/4 and y = 12, (x + y)^3 will be NOT be Even 
 NOT SUFFICIENT

Combining the two statements  
x and y are integers and xy=9
i.e. only Six cases are possible
x=1 and y = 9
or x=3 and y = 3
or x = 9 and y = 1
or x=-1 and y = -9
or x=-3 and y = -3
or x = -9 and y = -1
and in all cases (x + y)^3 will be Even
 SUFFICIENT

Answer: option C

ENGRTOMBA2018 · Answer

(x+y)^3 = even ---> x+y = even?

Statement 1, x,y = integer ---> lets say x = 1 , y =2 , then x+y = 3 , odd , "no" for even but if x=2, y =4 , then x+y=6 , "yes" for even. Thus not sufficient.

Statement 2, xy =9 ---> could be x =9, y =1/9 , "no" for x+y=even but if x=3=y , then"yes" for even . Thus not sufficient.

Combining, xy =9 and x,y = integer.

Thus, x =y=3 or x=1 , y =9. Either case, you get even for "x+y". Thus C is the correct answer.

ashokk138 · Answer

if (x+y)^3 is even, then x+y is even
qn is if x+y even?

x and y will be even when both x and y are both even or both odd.

St 1: x and y are integers
x and y could be both even or both odd then => x+y is even
or, x could be even and y could be odd or x could be odd and y could be even => x+y is odd

Hence not sufficient.

St 2: xy = 9
both x and y could be integers => 3 * 3 => 3+3 => 6 (Even)
x and y could be fraction => 4.5 * 2 => 4.5 + 2 => 6.5 (Neither odd or even)
Hence not sufficient.

Combining 1 and 2,

the combinations are 
3*3=> 3+3 = Even
1*9=> 1+9 = Even
-3*-3 => -3-3 = Even
-1*-9 => -1-9 = Even

Hence Sufficient. Option C

stonecold · Answer

Great Question 
Here we need to get the even/odd nature of (x+y)^3
remember we aren't told if they are integer or not
Statement 1
they are both integers 
Hmm
x=even
y=even => yes, (x+y)^3 will be even
x=even 
y=odd
(x+y)^3 will be odd
hence insufficient 
Statement 2
Here xy=odd
x=3
y=3
(x+y)^3 will be odd
but 
if x=1/9
y=9
(x+y)^3 will not be an integer 
hence insufficient 
Combining the two statements we can say that x and y are integers and are both odd as there product is odd
hence x+y will be even
so (x+y)^3 will be even too
hence C

Abhishek009 · Answer

FROM STATEMENT - I   ( INSUFFICIENT )

We do not know whether x and y odd/even...

FROM STATEMENT - Ii   ( INSUFFICIENT )

There can be 2 possibilities -

1. x = y = 3
2. x = 1or 9 = y

FROM STATEMENT I & II   ( SUFFICIENT )

From statement I it is clear x and y are 2 distinct integers...

So, ( x + y ) = ( 1 + 9 ) = 10 = Even

Hence,  Even^3  = Even...

Thus, BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked, answer will be (C)...

auradediligodo · Answer

S1 does not tell us much. Insufficient

S2 is interesting but if X=1/9 and Y=81 S2 becomes insufficient

Taken together possible pairs of values are 1*9, 3*3. All these pairs yield an even integer in the initial formula. 
 Hence C

tirthshah2013 · Answer

Answer - C
Reason-
Statement 1 just tells us that x and y are integers. Nothing else. Statement 1 is not sufficient. A and D were eliminated.

Statement 2 tells us that xy = 9.
Case 1- 
when x and y both are integers. In this case, both x and y have to be odd, to get an odd product. Therefore x+y will be even. And thus (x + y)^3 will be an even integer.
Case 2- 
When one of the numbers is not an integer. In this case, lets assume x =1/3, y = 27. Therefore, x + y = 1/3+27 = 82/3 which is a fraction, and thus (x + y)^3 will also be a fraction. Thus 2 is not sufficient. Thus, B was eliminated

Combining both, we get a perfect answer to the question (case 1). Hence C is correct answer.

Is (x + y)^3 an even integer?

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Is (x + y)^3 an even integer?

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