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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y

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Senior SC Moderator V
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If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y  [#permalink]

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5 00:00

Difficulty:   65% (hard)

Question Stats: 54% (01:48) correct 46% (01:37) wrong based on 169 sessions

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If $$x$$ and $$y$$ are positive integers, is $$x$$ even ?

(1) $$x^{2} + y^{2} = 98$$
(2) $$x = y$$

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Manager  D
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Re: If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y  [#permalink]

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Try every positive integer starting from 1. You'll understand option B is only to trap you.
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Re: If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y  [#permalink]

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ziyuenlau wrote:
If $$x$$ and $$y$$ are positive integers, is $$x$$ even ?

(1) $$x^{2} + y^{2} = 98$$
(2) $$x = y$$

1. we have only possible combination x=y=7..Suff
2.clearly in suff

So A.
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Re: If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y  [#permalink]

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1
Given that x and y are positive integer.
Hence x>0 and y>0

Statement 1
$$x^2+y^2=98$$
Start by checking x=1, $$y^2=97$$ => But since y is an integer, x cannot be 1.
By checking each integer from x=1 to x=9, we see that only for x=7, y has an integer value which is also 7.
For all other values of x, y is not an integer.

So $$x^2+y^2=98$$ is possible only at x=7 and y=7. Hence we know x is not even.
Statement 1 is sufficient to answer the question with a definite 'No'

Statement 2
x=y
From the above statement, nothing can be deduced about the nature of x and y
for x=1, y=1 in which case, x is odd
for x=2, y=2 in which case x is even.
Statement 2 is not sufficient to answer the question with a definite 'yes' or 'no'.

Hence A
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GMAT 1: 730 Q49 V41 If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y  [#permalink]

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Both X and Y are +ve integers.

now x^2 + y^2 = 98
obviously x & y < 10.

We can check the values to get two perfect squares:

X^2 + Y^2
--------------------
9^2 + 17
8^2 + 34
7^2 + 7^2
6^2 + 62
5^2 + 73
4^2 + 82 ( 82>9^2)

Hence only possible values of X and Y are 7 & 7.
So (1) is sufficient.

(2) Does not provide any values. Hence Insufficient.
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If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y  [#permalink]

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ziyuen wrote:
If $$x$$ and $$y$$ are positive integers, is $$x$$ even ?

(1) $$x^{2} + y^{2} = 98$$
(2) $$x = y$$

Hi,
Here's a quick way to solve this one...
considering statement (1)

(1) $$x^{2} + y^{2} = 98$$

The sum of two numbers is even..this is only possible when both are odd or both are even.
We're gonna play the units digit card here. The units digit of squares can be only any of the following..

0,1,4,5,6,9

Now, if you notice..there is no possible way in which the units digit of the sum of two odd unit digits can come out to be 8. Thus, both the numbers have to be Even numbers. No need to try numbers. Sufficient

Statement (2)

$$x = y$$

This by itself tells us nothing. Not sufficient.

So our answer is (A) _________________
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Re: If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y  [#permalink]

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This may not be true.....
7x7= 49 and
49 +49 =98

Posted from my mobile device
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Re: If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y  [#permalink]

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ShashankDave wrote:
ziyuen wrote:
If $$x$$ and $$y$$ are positive integers, is $$x$$ even ?

(1) $$x^{2} + y^{2} = 98$$
(2) $$x = y$$

Hi,
Here's a quick way to solve this one...
considering statement (1)

(1) $$x^{2} + y^{2} = 98$$

The sum of two numbers is even..this is only possible when both are odd or both are even.
We're gonna play the units digit card here. The units digit of squares can be only any of the following..

0,1,4,5,6,9

Now, if you notice..there is no possible way in which the units digit of the sum of two odd unit digits can come out to be 8. Thus, both the numbers have to be Even numbers. No need to try numbers. Sufficient

Statement (2)

$$x = y$$

This by itself tells us nothing. Not sufficient.

So our answer is (A) Your statement is not correct.

The units digit of the sum of two odd unit digits can come out to be 8 (for example $$...7^2 + ...7^2 = ...9 + ...9 = ...8$$)
The units digit of the sum of two even unit digits can also come out to be 8 (for example $$...2^2 + ...2^2 = ...4 + ...4 = ...8$$)
This units digit rule doesn't work in this question I think.

I can't find better ways than number plugging to solve this question.
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