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Is x an even integer?

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Is x an even integer?  [#permalink]

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New post 30 May 2020, 03:52
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Is \(x\) an even integer?

(1) \(\sqrt{x}\) is a multiple of 2.

(2) \(x^2\) is a multiple of 4.

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Re: Is x an even integer?  [#permalink]

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New post 30 May 2020, 04:25
SajjadAhmad wrote:
Is \(x\) an even integer?

(1) \(\sqrt{x}\) is a multiple of 2.

(2) \(x^2\) is a multiple of 4.


Analyze statement (1) alone
(1) \(\sqrt{x}\) is a multiple of 2.
\(\sqrt{x}\)=2k, k is an integer
[(\(\sqrt{x}\)\()]^2\)= \((2k)^2\)
x=4\(k^2\), \(k^2\) is an integer
=> x is an even integer
Sufficient

Analyze statement (2) alone
(2)\(x^2\) is a multiple of 4.
\(x^2\)=4=> x=2 is an even integer
\(x^2\)=8=> x=\(\sqrt{8}\) is not an even integer
Not Sufficient

Answer: A

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Re: Is x an even integer?  [#permalink]

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New post 30 May 2020, 04:29
IMO D.

Statement 1: √x is a multiple of 2 i.e. √x can be written as 2^a * P^b. where P can be any other prime number. Surely x ( {√x} ^ 2) will be having 2 as one of its prime factor and we know EVEN * Odd = Even . Hence x is even. Sufficient

Statement 2: x^2 is a multiple of 4 i.e. x^2 has prime factors of the form: 2^2 * P^2b where P can be any other prime number. Surely x will be having 2 as a prime factor and we know EVEN * Odd = Even . Hence x is even. Sufficient
Ex: 4, 16, 36
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Re: Is x an even integer?  [#permalink]

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New post 30 May 2020, 04:31
Mannisha wrote:
SajjadAhmad wrote:
Is \(x\) an even integer?

(1) \(\sqrt{x}\) is a multiple of 2.

(2) \(x^2\) is a multiple of 4.


Analyze statement (1) alone
(1) \(\sqrt{x}\) is a multiple of 2.
\(\sqrt{x}\)=2k, k is an integer
[(\(\sqrt{x}\)\()]^2\)= \((2k)^2\)
x=4\(k^2\), \(k^2\) is an integer
=> x is an even integer
Sufficient

Analyze statement (2) alone
(2)\(x^2\) is a multiple of 4.
\(x^2\)=4=> x=2 is an even integer
\(x^2\)=8=> x=\(\sqrt{8}\) is not an even integer
Not Sufficient

Answer: A


Hello Mannisha,

8 cannot be written as x^2.

x^2 as a multiple of 4 will be : 4,16, 36, 64....
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Re: Is x an even integer?  [#permalink]

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New post 30 May 2020, 04:42
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NitishJain wrote:
IMO D.

Statement 1: √x is a multiple of 2 i.e. √x can be written as 2^a * P^b. where P can be any other prime number. Surely x ( {√x} ^ 2) will be having 2 as one of its prime factor and we know EVEN * Odd = Even . Hence x is even. Sufficient

Statement 2: x^2 is a multiple of 4 i.e. x^2 has prime factors of the form: 2^2 * P^2b where P can be any other prime number. Surely x will be having 2 as a prime factor and we know EVEN * Odd = Even . Hence x is even. Sufficient
Ex: 4, 16, 36



You do not know if x is integer..
x^2=4y
say y=2, so \(x^2=4*2...x=2\sqrt{2}\),...Not an even integer.
when y=4...x=4..Yes
So Statement II is insuff
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Re: Is x an even integer?  [#permalink]

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New post 30 May 2020, 04:47
chetan2u wrote:
NitishJain wrote:
IMO D.

Statement 1: √x is a multiple of 2 i.e. √x can be written as 2^a * P^b. where P can be any other prime number. Surely x ( {√x} ^ 2) will be having 2 as one of its prime factor and we know EVEN * Odd = Even . Hence x is even. Sufficient

Statement 2: x^2 is a multiple of 4 i.e. x^2 has prime factors of the form: 2^2 * P^2b where P can be any other prime number. Surely x will be having 2 as a prime factor and we know EVEN * Odd = Even . Hence x is even. Sufficient
Ex: 4, 16, 36



You do not know if x is integer..
x^2=4y
say y=2, so \(x^2=4*2...x=2\sqrt{2}\),...Not an even integer.
when y=4...x=4..Yes
So Statement II is insuff



ohh ok thanks for pointing my mistake.

Mannisha: Please ignore my reply on your quote, understood my mistake.
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Re: Is x an even integer?  [#permalink]

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New post 30 May 2020, 04:50
NitishJain wrote:
chetan2u wrote:
NitishJain wrote:
IMO D.

Statement 1: √x is a multiple of 2 i.e. √x can be written as 2^a * P^b. where P can be any other prime number. Surely x ( {√x} ^ 2) will be having 2 as one of its prime factor and we know EVEN * Odd = Even . Hence x is even. Sufficient

Statement 2: x^2 is a multiple of 4 i.e. x^2 has prime factors of the form: 2^2 * P^2b where P can be any other prime number. Surely x will be having 2 as a prime factor and we know EVEN * Odd = Even . Hence x is even. Sufficient
Ex: 4, 16, 36



You do not know if x is integer..
x^2=4y
say y=2, so \(x^2=4*2...x=2\sqrt{2}\),...Not an even integer.
when y=4...x=4..Yes
So Statement II is insuff



ohh ok thanks for pointing my mistake.

Mannisha: Please ignore my reply on your quote, understood my mistake.


No worries it happens:)
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Re: Is x an even integer?  [#permalink]

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New post 30 May 2020, 04:50
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Is \(x\) an even integer?

(1) \(\sqrt{x}\) is a multiple of 2.
\(\sqrt{x}=2y\), where y is an integer.
So \(x=4y^2\), and so x is even.
Suff

(2) \(x^2\) is a multiple of 4.
\(x^2=4y\) where y is an integer.
so x will be even integer only when y is a perfect square.....\(y=4.....x^2=4*4=16...x=4\)
If y= any other number, x is not an integer....\(y=3.....x^2=4*3=12...x=2\sqrt{3}\)
Insuff

A
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Re: Is x an even integer?   [#permalink] 30 May 2020, 04:50

Is x an even integer?

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