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Last edited by Bunuel on 07 Jan 2014, 06:13, edited 3 times in total.

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...!
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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| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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

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| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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.

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, 02:16

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