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If x and y are integers, is 4x^2-y^2 an odd number?

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Math Revolution GMAT Instructor
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If x and y are integers, is 4x^2-y^2 an odd number?  [#permalink]

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New post Updated on: 09 Feb 2018, 23:00
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[GMAT math practice question]

If \(x\) and \(y\) are integers, is \(4x^2-y^2\) an odd number?

1) \(x\) is an odd number
2) \(y\) is an odd number

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Originally posted by MathRevolution on 07 Feb 2018, 00:46.
Last edited by amanvermagmat on 09 Feb 2018, 23:00, edited 2 times in total.
changed y2 to y^2
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Re: If x and y are integers, is 4x^2-y^2 an odd number?  [#permalink]

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New post 07 Feb 2018, 02:36
MathRevolution wrote:
[GMAT math practice question]

If \(x\) and \(y\) are integers, is \(4x^2-y2\) an odd number?

1) \(x\) is an odd number
2) \(y\) is an odd number


I believe it should read as '4x^2 - y^2'

Since x,y are integers:- 4x^2 is always going to be even, so whether 4x^2 - y^2 will be odd or not, depends on whether y is odd or not.

(1) x is an odd number, but nothing about y. Not sufficient.

(2) y is odd, so 4x^2 - y^2 = even - odd = odd. So sufficient.

Hence B answer
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Re: If x and y are integers, is 4x^2-y^2 an odd number?  [#permalink]

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New post 07 Feb 2018, 02:58
If x and y are integers, is \(4x^2-y2 an odd number?\)

4x2 would be even whether x is odd or even.

The answer depends on the value of y alone.

If y is odd, y2 is odd. therefore 4x2-y2 = even-odd = odd (sufficient).

So the correct choice is B
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Re: If x and y are integers, is 4x^2-y^2 an odd number?  [#permalink]

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New post 09 Feb 2018, 01:40
=>

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

Modifying the question:
In order for \(4x^2 – y^2\) to be odd, \(y^2\) must be odd since \(4x^2\) is always even. This is equivalent to y being odd. So, the question asks if y is odd. Thus, the answer is B.

Condition 1)
If \(y\) is an odd number, then both \(2x + y\) and \(2x – y\) are odd numbers and \((2x+y)(2x-y)\) is an odd number.
If \(y\) is an even number, both \(2x + y\) and \(2x – y\) are even numbers and \((2x+y)(2x-y)\) is an even number.
Since the question does not have a unique answer, condition 1) is not sufficient.


Therefore, the answer is B.

Answer: B
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Re: If x and y are integers, is 4x^2-y^2 an odd number?  [#permalink]

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New post 12 Feb 2018, 17:13
MathRevolution wrote:
[GMAT math practice question]

If \(x\) and \(y\) are integers, is \(4x^2-y^2\) an odd number?

1) \(x\) is an odd number
2) \(y\) is an odd number


Let’s recall some facts about odds and evens: (1) Even x even = even and even x odd = even. Thus, 4x^2 will always be an even number. (2) Odd x odd = odd. So if y is odd, then y^2 will be odd.

Thus, in order for 4x^2 - y^2 to be odd, y must be odd.

Statement One Alone:

x is an odd number

Since we know nothing regarding y, statement one alone is not sufficient to answer the question.

Statement Two Alone:

y is an odd number

Since y is odd, 4x^2 - y^2 must be odd since even - odd = odd. Statement two alone is sufficient to answer the question.

Answer: B
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Re: If x and y are integers, is 4x^2-y^2 an odd number? &nbs [#permalink] 12 Feb 2018, 17:13
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