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

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

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New post Updated on: 05 Jul 2014, 05:11
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A
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Question Stats:

39% (01:47) correct 61% (01:33) wrong based on 122 sessions

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

(1) 4y^2+3x^2=x^4+y^4

(2) y=4−x^2

The official answer is A, and the logic is clear to me.

But, is it possible in the first equations also to have x = y = o? Shouldn't the answer be E in such case?
It is not explicitly stated in the wording that x and y are different non-zero integers?

Most probably I just missed smth, so would be grateful for your explanations
Thanks in advance!


M27-02

Originally posted by Maksym on 05 Jul 2014, 03:05.
Last edited by Bunuel on 05 Jul 2014, 05:11, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If x and y are integers, is y an even integer? [#permalink]

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New post 05 Jul 2014, 05:14
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Maksym wrote:
If x and y are integers, is y an even integer?

(1) 4y^2+3x^2=x^4+y^4

(2) y=4−x^2

The official answer is A, and the logic is clear to me.

But, is it possible in the first equations also to have x = y = o? Shouldn't the answer be E in such case?
It is not explicitly stated in the wording that x and y are different non-zero integers?

Most probably I just missed smth, so would be grateful for your explanations
Thanks in advance!

M27-02


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

(1) 4y^2+3x^2=x^4+y^4 --> rearrange: \(3x^2-x^4=y^4-4y^2\) --> \(x^2(3-x^2)=y^2(y^2-4)\). Notice that LHS is even for any value of \(x\): if \(x\) is odd then \(3-x^2=odd-odd=even\) and if \(x\) is even then the product is naturally even. So, \(y^2(y^2-4)\) is also even, but in order it to be even \(y\) must be even, since if \(y\) is odd then \(y^2(y^2-4)=odd*(odd-even)=odd*odd=odd\). Sufficient.

(2) y=4-x^2 --> if \(x=odd\) then \(y=even-odd=odd\) but if \(x=even\) then \(y=even-even=even\). Not sufficient.

Answer: A.

As for your doubt: 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

Check more tips on zero and number properties here: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1371030

This week's PS question
This week's DS Question

Theory on Number Properties: math-number-theory-88376.html

DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59


Hope it helps.
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Intern
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Re: If x and y are integers, is y an even integer? [#permalink]

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New post 05 Jul 2014, 03:40
Maksym,

Even if x= y= 0, then too Y will remain an even integer because 0 is an even integer. and as this is a question of data interpretation in which we need to look for UNIQUE SOLUTION and in both the cases i.e.

1)when y is not equal to zero
2) WHEN X=Y=0

we are getting unique answer in both the cases i.e. YES Y is an even integer.
so answer would be A instead of E.
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Re: If x and y are integers, is y an even integer? [#permalink]

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New post 05 Jul 2014, 04:11
As some one has rightly pointed out..0 is an even integer..So A still stands firm
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Re: If x and y are integers, is y an even integer? [#permalink]

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New post 05 Jul 2014, 05:27
Bunuel wrote:
Maksym wrote:
If x and y are integers, is y an even integer?

(1) 4y^2+3x^2=x^4+y^4

(2) y=4−x^2

The official answer is A, and the logic is clear to me.

But, is it possible in the first equations also to have x = y = o? Shouldn't the answer be E in such case?
It is not explicitly stated in the wording that x and y are different non-zero integers?

Most probably I just missed smth, so would be grateful for your explanations
Thanks in advance!

M27-02


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

(1) 4y^2+3x^2=x^4+y^4 --> rearrange: \(3x^2-x^4=y^4-4y^2\) --> \(x^2(3-x^2)=y^2(y^2-4)\). Notice that LHS is even for any value of \(x\): if \(x\) is odd then \(3-x^2=odd-odd=even\) and if \(x\) is even then the product is naturally even. So, \(y^2(y^2-4)\) is also even, but in order it to be even \(y\) must be even, since if \(y\) is odd then \(y^2(y^2-4)=odd*(odd-even)=odd*odd=odd\). Sufficient.

(2) y=4-x^2 --> if \(x=odd\) then \(y=even-odd=odd\) but if \(x=even\) then \(y=even-even=even\). Not sufficient.

Answer: A.

As for your doubt: 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

Check more tips on zero and number properties here: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1371030

This week's PS question
This week's DS Question

Theory on Number Properties: math-number-theory-88376.html

DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59


Hope it helps.


Bunueal, thanks for explanation and the link!
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Re: If x and y are integers, is y an even integer? [#permalink]

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New post 22 Aug 2017, 00:50
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Re: If x and y are integers, is y an even integer?   [#permalink] 22 Aug 2017, 00:50
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