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Is integer x prime?

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Is integer x prime?  [#permalink]

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New post Updated on: 07 Jan 2014, 07:13
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A
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E

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Is integer x prime?

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

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

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Originally posted by jeeteshsingh on 07 Mar 2010, 10:23.
Last edited by Bunuel on 07 Jan 2014, 07:13, edited 3 times in total.
Edited the question and added the OA.
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Re: Is integer x prime?  [#permalink]

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New post 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

Please explain


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

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New post 07 Mar 2010, 11:34
2
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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New post 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
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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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New post 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

Please explain


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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New post 07 Jan 2014, 07:08
jlgdr wrote:
jeeteshsingh wrote:
Is Integer x prime?
1. \(x^2-3\) is an even number
2. \(x+2\) is an odd number

Please explain


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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New post 07 Jan 2014, 07:12
1
jlgdr wrote:
jeeteshsingh wrote:
Is Integer x prime?
1. \(x^2-3\) is an even number
2. \(x+2\) is an odd number

Please explain


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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New post 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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New post 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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New post 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]

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Re: Is integer x prime?   [#permalink] 24 Jan 2018, 04:34
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