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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If x and y are integers, and x² + xy + 2y + x is even, which of the fo
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27 Sep 2021, 07:10
27 Sep 2021, 07:10
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If x and y are integers, and x² + xy + 2y + x is even, which of the following MUST be true?
I. xy is even
II. x is even
III. x + y is odd
A) I only
B) III only
C) I and II
D) I and III
E) II and III
I. xy is even
II. x is even
III. x + y is odd
A) I only
B) III only
C) I and II
D) I and III
E) II and III
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If x and y are integers, and x² + xy + 2y + x is even, which of the fo
[#permalink]
27 Sep 2021, 08:51
27 Sep 2021, 08:51
1
GreenlightTestPrep wrote:
If x and y are integers, and x² + xy + 2y + x is even, which of the following MUST be true?
I. xy is even
II. x is even
III. x + y is odd
A) I only
B) III only
C) I and II
D) I and III
E) II and III
I. xy is even
II. x is even
III. x + y is odd
A) I only
B) III only
C) I and II
D) I and III
E) II and III
With this question type, I find it useful to first test all 4 possible cases, by plugging in 0 for even numbers, and 1 for odd numbers.
When we do so, we get:
Since we're told x² + xy + y + x is EVEN, we can see that cases i, ii and iii are all possible.
Now let's examine the three statements...
I. xy is even
For all three possible cases (i.e., cases i, ii and iii), xy will be even.
So, statement I must be true.
II. x is even
x is even for cases i and ii, but x is ODD for case iii.
Therefore, statement II need not be true.
III. x + y is odd
x + y is odd for cases ii and iii, but x + y is EVEN for case i.
Therefore, statement III need not be true.
Answer: A
Re: If x and y are integers, and x² + xy + 2y + x is even, which of the fo
[#permalink]
27 Sep 2021, 07:39
27 Sep 2021, 07:39
\(x² + xy + 2y + x\) is even
\(( x^2 + xy + x ) + 2y\)
Now, \(2y\) will always be even. So let's consider the bracket.
\(x^2 + xy + x = x( x + y + 1)\) This part must be even.
if \(y = 2\) & \(x = 1\) __________________ (1)
\(x( x + y + 1) = 4\) Possible
If we take both as even \(x = y = 2\), the bracket will always be even & same in the case of \(x = 2\) & \(y = 1\)
I. xy is even - ALWAYS
II. x is even - Not necessary. See \((1)\)
III. x + y is odd - Not necessary as explained above.
Answer A
\(( x^2 + xy + x ) + 2y\)
Now, \(2y\) will always be even. So let's consider the bracket.
\(x^2 + xy + x = x( x + y + 1)\) This part must be even.
if \(y = 2\) & \(x = 1\) __________________ (1)
\(x( x + y + 1) = 4\) Possible
If we take both as even \(x = y = 2\), the bracket will always be even & same in the case of \(x = 2\) & \(y = 1\)
I. xy is even - ALWAYS
II. x is even - Not necessary. See \((1)\)
III. x + y is odd - Not necessary as explained above.
Answer A
