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Is integer x prime?
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Updated on: 07 Jan 2014, 06:13
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85% (01:23) correct 15% (01:46) wrong based on 106 sessions
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Is integer x prime? (1) \(x^23\) is an even number (2) \(x+2\) is an odd number
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Originally posted by jeeteshsingh on 07 Mar 2010, 09:23.
Last edited by Bunuel on 07 Jan 2014, 06:13, edited 3 times in total.
Edited the question and added the OA.



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Re: Is integer x prime?
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07 Mar 2010, 10:08
jeeteshsingh wrote: kp1811 wrote: jeeteshsingh wrote: Is Integer x prime? 1. \(x^23\) 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?
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07 Mar 2010, 10:26
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?
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07 Mar 2010, 10:34
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?
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07 Mar 2010, 10:39
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^23 is an even number 2. x+2 is an
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07 Jan 2014, 06:04
jeeteshsingh wrote: Is Integer x prime? 1. \(x^23\) 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^23 is an even number 2. x+2 is an
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07 Jan 2014, 06:08



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Re: Is Integer x prime? 1. x^23 is an even number 2. x+2 is an
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07 Jan 2014, 06:12



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Re: Is integer x prime?
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07 Jan 2014, 23:16
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?
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07 Jan 2014, 23:30
jeeteshsingh wrote: Is integer x prime?
(1) \(x^23\) 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?
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02 Feb 2014, 02:16
Stmt 1: \(x^23\) 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?
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24 Jan 2018, 03:34
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Re: Is integer x prime? &nbs
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