If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even

If x and y are positive integers, is \((x+2)^{y+2}+x^y (y−2)^{x+2} \) even?

(1) \(5x+8x^2+12x^3+9\) is odd.
(2) \(3y+(11+y)+35(3y+2) \) is even.

i) if \(5x+8x^2+12x^3+9\) = odd

9 = odd; 12*Term = even; 8*Term = even => O+E+E+unknown = odd => O+Unknown = odd => Unknown = even

Unknown = 5x = even => x = even

If x is even => \((x+2)^{y+2}+x^y (y−2)^{x+2} \) => \(Even^{anything}\) + (\(Even^{Anything}*(Anything)\)) = Even + Even = Even

Sufficient

ii) if \(3y+(11+y)+35(3y+2) \) = Even =>

Case i) y = odd => Odd*Odd + (Odd+Odd) + (Odd(Odd*Odd)+(Even)) => Odd+Even+(Odd+Even) => Odd+Even+Odd = Even

Case ii) y = even => Odd*Even+ (Odd+Even) + (Odd(Odd*Even)+(Even)) => Even+Odd+(Even+Even) => Even+Odd+Even = Odd

2 cases - 2 Answers - Insufficient

Answer - A

Hey Shameev,

Thanks for posting your analysis.

As far as Statement 1 is concerned, your analysis is perfect. However, you made a small mistake in Statement 2.

Statement 2 clearly states that \(3y+(11+y)+35(3y+2) \) is even. This means that we need to find the even-odd nature of "y", that will make the given expression in statement 2 even.

Now, your calculation is perfect (highlighted in your solution). However, you did not realize that when y = even, statement 2 is not being satisfied. That means we need to discard this case and infer that "y" must be an odd number for statement 2 to be true.

Now, if y is odd, we need to check if the statement given in the question stem \((x+2)^{y+2}+x^y (y−2)^{x+2} \), is even or not.

I hope this explanation helped you in identifying your mistake. You can also watch the video solution of this question to gain more clarity.


Regards,
GMATWhiz Team

GMATWhizTeam :- Thank you so much for pointing out my mistake and providing a detailed analysis. Really appreciate your time.

It's my pleasure Shameek. Happy to know that the explanation helped

I am having tough time in doing pre-analysis, your videos will surely help me. Surprisingly I marked the correct answer here and checked the video later, my approach was correct. Thank you so much

The problem is about EVEN vs ODD.
If x=0 and y>0, none of the expressions in the problem will yield a fraction.
Implication:
We can ignore the condition that x must be positive and test x=0 in each statement.

Statement 1:
Case 1: x=0, with result that \(5x+8x^2+12x^3+9 = 9\)
If y=2, then \((x+2)^{y+2}+x^y (y−2)^{x+2} = 16\)
Here, the answer to the question stem is YES.
If y=1, then \((x+2)^{y+2}+x^y (y−2)^{x+2} = 8\)
Here, the answer to the question stem is YES.

Case 2: x=1, with result that \(5x+8x^2+12x^3+9 = 34\)
Not viable, since the sum must be ODD.

Since only Case 1 is viable -- and the answer is YES in Case 1 whether y is even or odd -- SUFFICIENT.

Statement 2:
For the sole purpose of determining whether y can be even in Statement 2, we can test y=0.
If y=0, then \(3y+(11+y)+35(3y+2) = 81\) -- not viable, since the sum must be EVEN.
Implication:
In Statement 2, y must be ODD.

If y=1 and x=0, then the answer to the question stem is YES, as shown in Statement 1.
If y=1 and x=1, then \((x+2)^{y+2}+x^y (y−2)^{x+2} = 26\), so the answer to the question stem is YES.
Since the answer is YES whether x is even or odd, SUFFICIENT.

(x + 2)^(y +2) + x^y (y-2)^(x +2) will be even when each of them are either odd or even. Since x is a base here, if x is even, the value will be even regardless the value of y except y =0.

1) 5x +8x^2+12x^3 +9 is odd. 8x^2 +12x^3 is even as even * (odd/even) =even . So 5x + 9 will have to odd as Odd + odd =even and even + odd =odd. That means 5x is even and for this x has to be even. The powers of x in the question stem cannot be 0. Sufficient.
2) 3y+(11+y)+35(3y+2) is even. WE can rewrite it as : 3y +11+y+105y +70 . 109y +81 is even. As Odd +odd =even and odd * odd =odd, so y is odd. We need the value of x. Not sufficient.
A is the answer.

The assumption in red is incorrect.
If y=odd and x=odd, then \((x+2)^{y+2}+x^y (y−2)^{x+2} = \) ODD + ODD = EVEN
If y=odd and x=even, then \((x+2)^{y+2}+x^y (y−2)^{x+2} = \) EVEN + EVEN = EVEN
Regardless of the value of x, the answer to the question stem is YES.
Thus, statement 2 is SUFFICIENT.