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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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 Updated on: 07 Apr 2020, 23:11
Originally posted by GMATWhizTeam on 05 Apr 2020, 07:34.
Last edited by GMATWhizTeam on 07 Apr 2020, 23:11, edited 2 times in total.
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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.


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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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 Updated on: 05 Apr 2020, 21:23
Originally posted by GMATWhizTeam on 05 Apr 2020, 07:37.
Last edited by GMATWhizTeam on 05 Apr 2020, 21:23, edited 2 times in total.
I hope you were able to internalize the first skill that we discussed in the Prepathon. Here's the detailed solution to the first question. Even if you got the question right make sure that you check out the method in the solution to ensure that you have solved it using the right method. All the best!


Feel free to post any doubts that you may have in the forum below. We will be happy to help you with your queries :)

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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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05 Apr 2020, 07:56
Statement 1: 5x+8x^2+12x^3= even
So 5x= even
So x=even
Now (x+2)^(y+2)+x^y(y-2)^(x+2)
= (even)^(y+2)+even^y(y-2)^(even+2)
= even
Sufficient
Statement 2: 3y+11+y+35(3y+2)=even
So 3y+y+35(3y+2)= odd
Case 1: y= odd
Odd+odd+odd= odd
Case 2: y=even
Even + even + even = odd
This case is not valid
Therefore y=odd
Back to question stem
(X+2)^odd+ x^odd(odd-2)^(x+2)
Now case 1: x=odd
odd^odd+ odd*odd= even
Case 2: x=even
Even+even=even
Both cases of x gives the same answer to the target question
Sufficient
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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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05 Apr 2020, 16:14
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
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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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06 Apr 2020, 04:10
shameekv1989
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
Show more


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

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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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06 Apr 2020, 04:18
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shameekv1989
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
Show more

GMATWhizTeam :- Thank you so much for pointing out my mistake and providing a detailed analysis. Really appreciate your time.
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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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06 Apr 2020, 05:47
shameekv1989
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shameekv1989
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.
Show more

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

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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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09 Apr 2020, 02:08
GMATWhizTeam
I hope you were able to internalize the first skill that we discussed in the Prepathon. Here's the detailed solution to the first question. Even if you got the question right make sure that you check out the method in the solution to ensure that you have solved it using the right method. All the best!


Feel free to post any doubts that you may have in the forum below. We will be happy to help you with your queries :)
Show more

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
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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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09 Apr 2020, 04:43
GMATWhizTeam
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.
Show more

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.

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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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09 Apr 2020, 10:34
(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.
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If x and y are positive integers, is (x+2)^(y+2)+x^y(y−2)^(x+2) even
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09 Apr 2020, 10:52
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2)so y is odd. We need the value of x. Not sufficient.
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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.

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