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If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y

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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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New post 29 Jan 2017, 06:26
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A
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Question Stats:

51% (01:50) correct 49% (01:37) wrong based on 158 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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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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New post 29 Jan 2017, 07:09
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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New post 29 Jan 2017, 10:55
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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New post 29 Jan 2017, 12:01
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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If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y  [#permalink]

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New post 29 Jan 2017, 21:21
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.
Therefore, answer A.
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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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New post 19 Mar 2017, 10:02
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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New post 10 Aug 2018, 18:47
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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New post 11 Aug 2018, 07:00
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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Re: If x any y are positive integers,is x even? (1) x^2+y^2=98 (2) x=y &nbs [#permalink] 11 Aug 2018, 07:00
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