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# If x is an integer, is (x^2+1) (x+5) an even number? 1. x is

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Manager
Joined: 11 Apr 2009
Posts: 160

Kudos [?]: 114 [0], given: 5

If x is an integer, is (x^2+1) (x+5) an even number? 1. x is [#permalink]

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07 Jul 2009, 18:44
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If x is an integer, is (x^2+1) (x+5) an even number?

1. x is an odd number.

2. Each prime factor of x^2 is greater than 7.

Please discuss St. 2 in detail. Thanks

Kudos [?]: 114 [0], given: 5

Senior Manager
Joined: 04 Jun 2008
Posts: 286

Kudos [?]: 155 [0], given: 15

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07 Jul 2009, 18:58
If a number is even, it will definitely have 2 as a prime factor. If it doesn't have 2 as prime factor, it will always be an odd number.

Kudos [?]: 155 [0], given: 15

Manager
Joined: 28 Jan 2004
Posts: 202

Kudos [?]: 28 [0], given: 4

Location: India

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09 Jul 2009, 00:58
E

Stmt 1 -
Insuff.
For X= 0 ........Odd
For X > 0........Even
For x < 0 .......Can not say.

Stmt 2 -
If X^2 is even then 2 need be a prime factor.
According to stmt.2 , 2 is not a prime factor of x^2. This means that X^2 is odd or X is odd.
This is the same info. as Stmt. 1.

So it is not possible to determine.

Kudos [?]: 28 [0], given: 4

Senior Manager
Joined: 04 Jun 2008
Posts: 286

Kudos [?]: 155 [0], given: 15

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09 Jul 2009, 06:40
No, the answer would be D. Each sttmt is enough

Odd + odd = even

Square of an odd number will always be odd and vice versa

If a number is even, it will definitely have 2 as a prime factor. If it doesn't have 2 as prime factor, it will always be an odd number.

Combining the above 3 rules, you will get answer as D

Stmt 1 says x is odd, so x^2 is odd, 1 and 5 both are odd, so when 1 OR 5 is added to x or x^2, both will give even results. So the answer will be even.

Stmt 2 says x doesn't have 2 as prime factor, ie, same as above, x is odd.

each stmt is giving results as even, so each is sufficient.

Kudos [?]: 155 [0], given: 15

Senior Manager
Joined: 23 Jun 2009
Posts: 360

Kudos [?]: 134 [0], given: 80

Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago

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09 Jul 2009, 07:20
D
No need to explain 1
In 2; If all prime factors are greater than 7; that means, all of factors are odd. So their multiplication is odd. So 2 is suff.

Kudos [?]: 134 [0], given: 80

Manager
Joined: 28 Jan 2004
Posts: 202

Kudos [?]: 28 [0], given: 4

Location: India

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09 Jul 2009, 11:34
I completely messed up with this question.
agree with D.

Kudos [?]: 28 [0], given: 4

Senior Manager
Joined: 01 Mar 2009
Posts: 367

Kudos [?]: 96 [0], given: 24

Location: PDX

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09 Jul 2009, 11:49
So 1 is pretty obviously correct .

Regarding 2 - Since x^2 has all prime factors > 7 , x^2 is odd => x is odd so if x is odd - same reasoning as 1 and hence D.
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Kudos [?]: 96 [0], given: 24

Re: Inequality   [#permalink] 09 Jul 2009, 11:49
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