If x and y are positive integers, is x2 + x + y + 1 an even number?

Question

If x and y are positive integers, is x^2 + x + y + 1 an even number?

1. x is equal to half the sum of the HCF of 12, 26, and 48 and LCM of 25, 35, and 50. 
2. y^4/16 is an even number.

sandeep4488 · Answer

1) is in the form of x(x+1) + y +1. No matter what the value of x is, we will still need to know whether y is even or odd to judge whether the whole thing is even or odd. 
Also x(x+1) is always even
2) No. is divisible by 16, hence it is even. Therefore according to the above rule, the whole expression in odd

Hence option B

Lucky2783 · Answer

X2 will be odd when x is odd , will be even when X is even .

4 possible scenarios for F(XY)=x^2 + x + y + 1

X=Odd , Y= Even , F(XY) = ODD
X=Even , Y= Even , F(XY) = ODD

X=Odd, Y=Odd, F(XY) = EVEN
X=Even, Y=Odd F(XY)=EVEN

1. x is equal to half the sum of the HCF of 12, 26, and 48 and LCM of 25, 35, and 50. 
 not sufficient as we do not know about Y (it may be odd or even) . 
2. y^4/16 is an even number. 
 clearly Y is even , let look at above scenarios. 
 we see that whenever Y is EVEN F(XY) is ODD.
 so this statement is sufficient.

not that hard question IMO.

Harley1980 · Answer

At first we should analyze question: exponents doesn't matter in the tasks about parity, so we have x + x = 2x and this part will be even, so for solving this task we should only know about parity of  y

1) in this statement we don't have information about  y  
 insufficient 
2) exponents doesn't matter so we have y / 16 = a so y = 16a and we know that  y  is even 
 sufficient

and answer is B

AbdurRakib · Answer

According to question ,  x^2 +x=even number (no matter what x could be even or odd integer,)

So basically the Scope of the answer lies on y (if y is odd, x^2 +x+y+1=even,But if y is even  x^2 +x+y+1=odd)

Statement 1. didn't say anything about y, Insufficient

Statement 2.  \frac{y^4}{16} =even,So y=even number. Sufficient

Correct Answer B

stonecold · Answer

This is a tremendous Question Testing our even/odd understanding 
We are asked if x^2+x+y+1 is even
here let V=x^2+x+y+1
V=x(x+1)+y+1
V=Even+y+odd
V=Odd+y
If y is odd => V is even
If y is even => V is odd
So the Question is asking "IS Y ODD"
Statement 1
x is some value
Dont stress on this statement as no clue of y is mentioned
hence insufficient 
Statement 2
y^4/16 => even
hence y^4 must be even too
so y*y*y*y is even => This can happen only if y is even
So as y is even => x^2+x+y+1=> odd
Hence sufficient

Hence B

rever08 · Answer

how is the whole expression odd?

testcracker · Answer

hi here you can see, x(x+1) is a product of 2 consecutive numbers, and hence the product will always be an even number. So the odd- even nature of the whole expression will rest on the odd-even nature of y only... statement 1 says nothing about y, not sufficient statement 2 says y is even, sufficient because, even + even + odd = odd cheers through the kudos button, if this helps thanks

chetan2u · Answer

x^2 + x + y + 1 
 x^2+x=x(x+1) , that is product of consecutive integers. Hence, we are left with only y+1. 
Thus our answer will be opposite of what type of number y is.

1) x is equal to half the sum of the HCF of 12, 26, and 48 and LCM of 25, 35, and 50. 
Nothing about y.

2)  \frac{y^4}{16}  is an even number.
Thus, y is even. 
y+1 and thus the entire expression is odd. 
Sufficient

B

If x and y are positive integers, is x2 + x + y + 1 an even number?

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GMAT Data Sufficiency (DS) Questions

If x and y are positive integers, is x2 + x + y + 1 an even number?

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