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If x and y are integers, is x + y an even integer?

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If x and y are integers, is x + y an even integer? [#permalink]

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[GMAT math practice question]

If x and y are integers, is \(x + y\) an even integer?

(1) \(x\) is an odd integer.
(2) \(x^2 + y^2\) has a remainder of 2 when it is divided by 4.
[Reveal] Spoiler: OA

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Re: If x and y are integers, is x + y an even integer? [#permalink]

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New post 16 Nov 2017, 07:00
IF x and y are both integers, then x+y is even when either both x/y are odd, or both x/y are even.

(1) x is odd. But we dont know anything about y. Insufficient.

(2) x^2 + y^2 gives remainder 2 when divided by 4. Which means x^2 + y^2 is even (since its of the form 4K + 2, where k is a non negative integer). x^2 + y^2 will be even when either x/y are both even or x/y are both odd. So now we know that x+y is also going to be even (an even number stays even and an odd number stays odd when raised to any integer power).
Sufficient.

Hence B answer
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Re: If x and y are integers, is x + y an even integer? [#permalink]

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New post 16 Nov 2017, 08:34
MathRevolution wrote:
[GMAT math practice question]

If x and y are integers, is \(x + y\) an even integer?

(1) \(x\) is an odd integer.
(2) \(x^2 + y^2\) has a remainder of 2 when it is divided by 4.


Statement 1: No information about \(y\). Insufficient

Statement 2: \(x^2 + y^2=4q+2=Even\) this is only possible if both \(x\) & \(y\) are even or both \(x\) & \(y\) are odd. In either case \(x+y=Even\). Sufficient

Another approach-
\((x+y)^2=x^2 + y^2+2xy=Even+Even\)

\(=>x+y=\sqrt{Even}\) and as \(x\) & \(y\) are integer so \(x+y=Even\). Sufficient

Option B
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Re: If x and y are integers, is x + y an even integer? [#permalink]

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New post 16 Nov 2017, 09:05
MathRevolution wrote:
[GMAT math practice question]

If x and y are integers, is \(x + y\) an even integer?


(1) \(x\) is an odd integer.


x = odd
y = odd
x + y = even

x = odd
y = even
x + y = odd

Insufficient. AD are OUT.

Quote:
(2) \(x^2 + y^2\) has a remainder of 2 when it is divided by 4.


Multiples of 4 -> 4,8,12,.....
=> Multiples of 4 are always EVEN.

We are given
=> \(\frac{(x^2+y^2)}{4}\) = 4Q + 2
=> The resulting integer will also be an even number.

Sufficient. B is the answer.
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Re: If x and y are integers, is x + y an even integer? [#permalink]

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New post 19 Nov 2017, 17:15
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since the question includes 2 variables (x and y) and no equation, C is most likely to be the answer. Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Conditions 1) & 2)
Since \(x^2 + y^2\) has a remainder of 2 when it is divided by 4, \(x^2 + y^2\) must be even. Since x is odd, \(x^2\) is odd and so \(y^2\) must also be odd. Therefore, y is odd, and \(x + y\) is even. The answer is ‘yes’.

Condition 1)
Since we don’t know whether y is even or odd, this is not sufficient.

Condition 2)
The condition tells us that \(x^2+y^2=4k+2=2(2k+1)\) is even. Since\(x^2+y^2=(x+y)^2-2xy\), and \(2xy\) is even, this implies that \((x+y)^2\) is also even. But this can only happen if x+y is even. So, the answer is ‘yes’.
This condition is sufficient.

Therefore, the answer is B.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E).

Answer: B
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Re: If x and y are integers, is x + y an even integer? [#permalink]

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New post 19 Nov 2017, 21:04
(1) x is an odd integer => absolutely insufficient since we don't know anything about y.
(2) x2+y2 has a remainder of 2 when it is divided by 4 => x2 + y2 is even => we have 2 cases: x2 and y2 are both odd or both even.
- x2, y2 are odd => x, y are both odd => x+y = even
- x2, y2 are odd => x, y are both even => x + y = even

Hence, the answer is B
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Re: If x and y are integers, is x + y an even integer?   [#permalink] 19 Nov 2017, 21:04
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