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Is integer x prime?

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Senior Manager
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Is integer x prime? [#permalink]

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07 Mar 2010, 10:23
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15% (low)

Question Stats:

80% (02:09) correct 20% (01:14) wrong based on 70 sessions

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Is integer x prime?

(1) $$x^2-3$$ is an even number

(2) $$x+2$$ is an odd number
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 Jan 2014, 07:13, edited 3 times in total.
Edited the question and added the OA.
Senior Manager
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Re: Is integer x prime? [#permalink]

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07 Mar 2010, 11:02
kp1811 wrote:
jeeteshsingh wrote:
Is Integer x prime?
1. $$x^2-3$$ is an even number
2. $$x+2$$ is an odd number

E...

Made the question more clear for you...!
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JT...........
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Re: Is integer x prime? [#permalink]

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07 Mar 2010, 11:08
jeeteshsingh wrote:
kp1811 wrote:
jeeteshsingh wrote:
Is Integer x prime?
1. $$x^2-3$$ is an even number
2. $$x+2$$ is an odd number

E...

Made the question more clear for you...!

will try to make it more clearer.....

stmnt1) x^2 - 3 is even

let x = 3 (prime) then 3^2 - 3 = 6 even
let x =9 (non prime) then 9^2 - 3 = 78 even
hence insuff

stmnt2) x+2 is odd
let x = 3 (prime) then 3 + 2 = 5 odd
let x =9 (non prime) then 9 + 2 = 11 odd
hence insuff

even together they don't suffice. Hence E
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Re: Is integer x prime? [#permalink]

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07 Mar 2010, 11:26
kp1811 wrote:
jeeteshsingh wrote:

will try to make it more clearer.....

stmnt1) x^2 - 3 is even

let x = 3 (prime) then 3^2 - 3 = 6 even
let x =9 (non prime) then 9^2 - 3 = 78 even
hence insuff

stmnt2) x+2 is odd
let x = 3 (prime) then 3 + 2 = 5 odd
let x =9 (non prime) then 9 + 2 = 11 odd
hence insuff

even together they don't suffice. Hence E

Some how I dont get this approach as you use the ques to prove the statement below. This question is from PR 1012 and I see the same solution there which isnt convincing for me.

My approach is as follows:

Given x is an integer.
Ques is x prime?

S1: x^2 - 3 = even
x^2 - 3 = 2m where m is an integer
x = sqrt(2m + 3) where m is >= 0 as you cannot have - ve sqrt.

This gives x = $$\sqrt{3},\sqrt{5},\sqrt{7},3,\sqrt{11},\sqrt{13},\sqrt{15},\sqrt{17},\sqrt{19},\sqrt{21},\sqrt{23},5,\sqrt{27},\sqrt{29},.....$$
Since it is given that x is an integer we get only 3, 5, 7..... which are all prime. Hence SUFF.

S2: x + 2 is odd
x + 2 = 2k + 1 where k is an integer
x = 2k - 1
Therefore x = ....-7,-5,-3,-1,1,3,5,7,9,11... which means all odd numbers and hence not necessarily be prime. Therefore NOT SUFF.

Hence answer is A....!

Can someone highlight what is wrong in my approach!
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
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~~Better Burn Out... Than Fade Away~~

Senior Manager
Joined: 30 Aug 2009
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Location: India
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Re: Is integer x prime? [#permalink]

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07 Mar 2010, 11:34
2
KUDOS
jeeteshsingh wrote:
kp1811 wrote:
jeeteshsingh wrote:

will try to make it more clearer.....

stmnt1) x^2 - 3 is even

let x = 3 (prime) then 3^2 - 3 = 6 even
let x =9 (non prime) then 9^2 - 3 = 78 even
hence insuff

stmnt2) x+2 is odd
let x = 3 (prime) then 3 + 2 = 5 odd
let x =9 (non prime) then 9 + 2 = 11 odd
hence insuff

even together they don't suffice. Hence E

Some how I dont get this approach as you use the ques to prove the statement below. This question is from PR 1012 and I see the same solution there which isnt convincing for me.

My approach is as follows:

Given x is an integer.
Ques is x prime?

S1: x^2 - 3 = even
x^2 - 3 = 2m where m is an integer
x = sqrt(2m + 3) where m is >= 0 as you cannot have - ve sqrt.

This gives x = $$\sqrt{3},\sqrt{5},\sqrt{7},3,\sqrt{11},\sqrt{13},\sqrt{15},\sqrt{17},\sqrt{19},\sqrt{21},\sqrt{23},5,\sqrt{27},\sqrt{29},.....$$
Since it is given that x is an integer we get only 3, 5, 7..... which are all prime. Hence SUFF.

S2: x + 2 is odd
x + 2 = 2k + 1 where k is an integer
x = 2k - 1
Therefore x = ....-7,-5,-3,-1,1,3,5,7,9,11... which means all odd numbers and hence not necessarily be prime. Therefore NOT SUFF.

Hence answer is A....!

Can someone highlight what is wrong in my approach!

in highlighted part after 3,5, 7 .... 9 will come [ for m = 39] which is non prime
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Re: Is integer x prime? [#permalink]

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07 Mar 2010, 11:39
kp1811 wrote:
jeeteshsingh wrote:

Some how I dont get this approach as you use the ques to prove the statement below. This question is from PR 1012 and I see the same solution there which isnt convincing for me.

My approach is as follows:

Given x is an integer.
Ques is x prime?

S1: x^2 - 3 = even
x^2 - 3 = 2m where m is an integer
x = sqrt(2m + 3) where m is >= 0 as you cannot have - ve sqrt.

This gives x = $$\sqrt{3},\sqrt{5},\sqrt{7},3,\sqrt{11},\sqrt{13},\sqrt{15},\sqrt{17},\sqrt{19},\sqrt{21},\sqrt{23},5,\sqrt{27},\sqrt{29},.....$$
Since it is given that x is an integer we get only 3, 5, 7..... which are all prime. Hence SUFF.

S2: x + 2 is odd
x + 2 = 2k + 1 where k is an integer
x = 2k - 1
Therefore x = ....-7,-5,-3,-1,1,3,5,7,9,11... which means all odd numbers and hence not necessarily be prime. Therefore NOT SUFF.

Hence answer is A....!

Can someone highlight what is wrong in my approach!

in highlighted part after 3,5, 7 .... 9 will come [ for m = 39] which is non prime

Thanks mate! Got it Now! Kudos +1
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
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Re: Is Integer x prime? 1. x^2-3 is an even number 2. x+2 is an [#permalink]

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07 Jan 2014, 07:04
jeeteshsingh wrote:
Is Integer x prime?
1. $$x^2-3$$ is an even number
2. $$x+2$$ is an odd number

Yeah answer is clearly A cause in B, 1 is obviously not a prime number

Now question, when they say that something is an odd number

Do they mean odd integer? Cause I'm pretty sure the definitions of number and integer are different

Now, can a decimal be odd? Say like 1.5 is odd and 1.2 is even I guess so right?

Just wanted to clarify on those

Cheers!
J
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Re: Is Integer x prime? 1. x^2-3 is an even number 2. x+2 is an [#permalink]

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07 Jan 2014, 07:08
Expert's post
jlgdr wrote:
jeeteshsingh wrote:
Is Integer x prime?
1. $$x^2-3$$ is an even number
2. $$x+2$$ is an odd number

Yeah answer is clearly A cause in B, 1 is obviously not a prime number

Now question, when they say that something is an odd number

Do they mean odd integer? Cause I'm pretty sure the definitions of number and integer are different

Now, can a decimal be odd? Say like 1.5 is odd and 1.2 is even I guess so right?

Just wanted to clarify on those

Cheers!
J

Only integers can be odd or even. So, odd number and odd integer are the same thing.
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Re: Is Integer x prime? 1. x^2-3 is an even number 2. x+2 is an [#permalink]

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07 Jan 2014, 07:12
1
KUDOS
Expert's post
jlgdr wrote:
jeeteshsingh wrote:
Is Integer x prime?
1. $$x^2-3$$ is an even number
2. $$x+2$$ is an odd number

Yeah answer is clearly A cause in B, 1 is obviously not a prime number

Now question, when they say that something is an odd number

Do they mean odd integer? Cause I'm pretty sure the definitions of number and integer are different

Now, can a decimal be odd? Say like 1.5 is odd and 1.2 is even I guess so right?

Just wanted to clarify on those

Cheers!
J

The correct answer is not A, it's E. Consider x=1 and x=3.

Hope it helps.
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Re: Is integer x prime? [#permalink]

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08 Jan 2014, 00:16
jeeteshsingh wrote:
kp1811 wrote:
jeeteshsingh wrote:

will try to make it more clearer.....

stmnt1) x^2 - 3 is even

let x = 3 (prime) then 3^2 - 3 = 6 even
let x =9 (non prime) then 9^2 - 3 = 78 even
hence insuff

stmnt2) x+2 is odd
let x = 3 (prime) then 3 + 2 = 5 odd
let x =9 (non prime) then 9 + 2 = 11 odd
hence insuff

even together they don't suffice. Hence E

Some how I dont get this approach as you use the ques to prove the statement below. This question is from PR 1012 and I see the same solution there which isnt convincing for me.

My approach is as follows:

Given x is an integer.
Ques is x prime?

S1: x^2 - 3 = even
x^2 - 3 = 2m where m is an integer
x = sqrt(2m + 3) where m is >= 0 as you cannot have - ve sqrt.

This gives x = $$\sqrt{3},\sqrt{5},\sqrt{7},3,\sqrt{11},\sqrt{13},\sqrt{15},\sqrt{17},\sqrt{19},\sqrt{21},\sqrt{23},5,\sqrt{27},\sqrt{29},.....$$
Since it is given that x is an integer we get only 3, 5, 7..... which are all prime. Hence SUFF.

S2: x + 2 is odd
x + 2 = 2k + 1 where k is an integer
x = 2k - 1
Therefore x = ....-7,-5,-3,-1,1,3,5,7,9,11... which means all odd numbers and hence not necessarily be prime. Therefore NOT SUFF.

Hence answer is A....!

Can someone highlight what is wrong in my approach!

Hello,
Just a small thing
You say "x = sqrt(2m + 3) where m is >= 0 as you cannot have - ve sqrt."
but your m can be -1, still square root would be +ve i.e 1;
now you have to add x=1 to your list :1,3,5,7 etc.
As 1 is not prime.
A becomes insufficient.

Cheers!!!
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Re: Is integer x prime? [#permalink]

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08 Jan 2014, 00:30
jeeteshsingh wrote:
Is integer x prime?

(1) $$x^2-3$$ is an even number

(2) $$x+2$$ is an odd number

X is an integer.

Statement I is insufficient:

x = 5 (Prime) x = 9 (Not Prime)

Statement II is insufficient:

5 + 2 is odd 9 + 2 is odd

No need to combine since we are using the same set of numbers to disprove each statement individually

Hence answer is E
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Re: Is integer x prime? [#permalink]

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02 Feb 2014, 03:16
Stmt 1: $$x^2-3$$ is even simply means x is odd integer. Now x being odd is not sufficient for it to be prime. INSUFF
Stm2: x+2 is odd means x is odd integer. Therefore stmt 1 & 2 are same. INSUFF

Both statements together give no new information. hence E.
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Re: Is integer x prime?   [#permalink] 02 Feb 2014, 03:16
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