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Question Stats:
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If \(x\) is an integer and \(1x \lt 2\) then which of the following must be true? A. \(x\) is not a prime number B. \(x^2+x\) is not a prime number C. \(x\) is positive D. Number of distinct positive factors of \(x+2\) is a prime number E. \(x\) is not a multiple of an odd prime number
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16 Sep 2014, 00:44
Official Solution:If \(x\) is an integer and \(1x \lt 2\) then which of the following must be true? A. \(x\) is not a prime number B. \(x^2+x\) is not a prime number C. \(x\) is positive D. Number of distinct positive factors of \(x+2\) is a prime number E. \(x\) is not a multiple of an odd prime number \(1x\) is just the distance between 1 and \(x\) on the number line. We are told that this distance is less than 2: (1) 13 so, \(1 \lt x \lt 3\). Since it is given that \(x\) is an integer, then \(x\) can be 0, 1 or 2. A. \(x\) is not a prime number. Not true if \(x=2\). B. \(x^2+x\) is not a prime number. Not true if \(x=1\). C. \(x\) is positive. Not true if \(x=0\). D. Number of distinct positive factors of \(x+2\) is a prime number. True for all three values of \(x\). E. \(x\) is not a multiple of an odd prime number. Not true if \(x=0\), since zero is a multiple of every integer except zero itself. Answer: D
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Re: M1116
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15 Nov 2015, 01:15
Dear Bunuel,
The correct answer D is confusing. If possible values of x are (0, 1, 2), then how is number of distinct positive factors of x+2 is a prime number for when x=2?
Many thanks and sorry asking maybe too basic question!



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15 Nov 2015, 09:10
Liza99 wrote: Dear Bunuel,
The correct answer D is confusing. If possible values of x are (0, 1, 2), then how is number of distinct positive factors of x+2 is a prime number for when x=2?
Many thanks and sorry asking maybe too basic question! D. Number of distinct positive factors of x + 2 is a prime number. If x = 2, then x + 2 = 4 and the number of distinct positive factors of 4 is 3 (1, 2, and 4). 3 is a prime number.
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Re: M1116
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15 Nov 2015, 10:01
Bunuel wrote: Liza99 wrote: Dear Bunuel,
The correct answer D is confusing. If possible values of x are (0, 1, 2), then how is number of distinct positive factors of x+2 is a prime number for when x=2?
Many thanks and sorry asking maybe too basic question! D. Number of distinct positive factors of x + 2 is a prime number. If x = 2, then x + 2 = 4 and the number of distinct positive factors of 4 is 3 (1, 2, and 4). 3 is a prime number. Many thanks Bunuel!!!



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Re: M1116
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15 Apr 2016, 07:57
Bunuel wrote: Liza99 wrote: Dear Bunuel,
The correct answer D is confusing. If possible values of x are (0, 1, 2), then how is number of distinct positive factors of x+2 is a prime number for when x=2?
Many thanks and sorry asking maybe too basic question! D. Number of distinct positive factors of x + 2 is a prime number. If x = 2, then x + 2 = 4 and the number of distinct positive factors of 4 is 3 (1, 2, and 4). 3 is a prime number. Bunel, I want to make sure I understand this problem. Is the following solution correct for testing other cases? x=0. x+2=2. 2 factors (1,2). 2 is prime #> GOOD! x=1. x+2=3. 2 factors (1,2). 2 is a prime #> GOOD! x=2. x+2=4. 3 factors (1,2,4). 3 is a prime #>GOOD! You are amazing Bunel, thank you so much!!



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Re: M1116
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15 Apr 2016, 08:02
NothingComes3asy wrote: Bunuel wrote: Liza99 wrote: Dear Bunuel,
The correct answer D is confusing. If possible values of x are (0, 1, 2), then how is number of distinct positive factors of x+2 is a prime number for when x=2?
Many thanks and sorry asking maybe too basic question! D. Number of distinct positive factors of x + 2 is a prime number. If x = 2, then x + 2 = 4 and the number of distinct positive factors of 4 is 3 (1, 2, and 4). 3 is a prime number. Bunel, I want to make sure I understand this problem. Is the following solution correct for testing other cases? x=0. x+2=2. 2 factors (1,2). 2 is prime #> GOOD! x=1. x+2=3. 2 factors ( 1,2). 2 is a prime #> GOOD! x=2. x+2=4. 3 factors (1,2,4). 3 is a prime #>GOOD! You are amazing Bunel, thank you so much!! All correct except small typo: 3 has two factors 1 and 3.
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23 May 2016, 21:51
Thanks Bunuel, I was overlooked "1" is also a factor and 0 is a multiple of all odd  good question!
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Re: M1116
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29 Jun 2016, 17:56
X could be 0,1,2. Now only option D gives correct answer



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02 Aug 2016, 03:03
I think this is a highquality question and I agree with explanation.



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03 Aug 2016, 16:46
Here is my two cents for those who need a more detailed approach and credit to Ron from Veritashttps://www.veritasprep.com/blog/2013/0 ... sofzero/
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Re: M1116
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04 Apr 2017, 12:49
isnt't 0 a positive intger..in that case shouldn't Option 3 not be true?



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04 Apr 2017, 12:55
rai0512 wrote: isnt't 0 a positive intger..in that case shouldn't Option 3 not be true? No. 0 is neither positive nor negative even integer. Check here for more: https://gmatclub.com/forum/mathnumber ... 88376.htmlHope it helps.
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Re: M1116
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11 Jun 2017, 18:42
E. x is not a multiple of an odd prime number. Not true if x=0, since zero is a multiple of every integer except zero itself.
Hello, I cannot understand the explanation of E. I agree it is not true. However, shouldn't it be that E is not true if x=1? 1 is not a multiple of an odd prime number. On the other hand 0 is a multiple of an odd prime number (i.e. 0*3=0). Please explain.
Thank you



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Re: M1116
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11 Jun 2017, 22:52
dimitratsia wrote: E. x is not a multiple of an odd prime number. Not true if x=0, since zero is a multiple of every integer except zero itself.
Hello, I cannot understand the explanation of E. I agree it is not true. However, shouldn't it be that E is not true if x=1? 1 is not a multiple of an odd prime number. On the other hand 0 is a multiple of an odd prime number (i.e. 0*3=0). Please explain.
Thank you Hi The question asks "Which of the following MUST be true". We have recognised that there are 3 possible values of x: 0, 1, 2. So if option E is true, it MUST be true for All the three values of x: 0, 1, 2. But its clear that option E is NOT true when x=0. Because 0 is actually a multiple of all odd prime numbers (0 is a multiple of 3, 0 is a multiple of 5, 0 is a multiple of 7, ... and so on) So since E is not true for x=0, we CANNOT say that E must be true. Thats why E cannot be our answer. (In your explanation when you say x=1, statement E actually becomes true for that value, because 1 is Not a multiple of any odd prime number. Also if you take x=2, statement E actually becomes true for that value too, because 2 is also Not a multiple of any odd prime number. But for x=0, statement E is Not true)



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Re: M1116
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05 Feb 2018, 08:08
Hey all,
I have one question regarding this problem. I do understand the solution but when solving the problem my solution of x did not involve the numer "1". Maybe you can help me finding my mistake. Here's what I did:
1x < 2
gives us: a) 1x < 2 for 1x > 0 leads to 1 < x < 1 b) 1x > 2 for 1x < 0 leads to 1 < x < 3
see how 1 was not included in the range because I also took the equation 1x >/< 0 ?
I have seen and used this approach in other problems so don't exactly know when to include this range as well. Here it seemed to be wrong as it excluded 1 from the solution range.
Would appreciate help on this.
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Re: M1116
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04 Mar 2018, 18:53
Hi Bunuel
Even I solved as gmatc2018 did. In the end I did not get 1 as the integer value to be considered. It would be great if you can point out the flaw in the approach. Thanks so much.



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04 Mar 2018, 21:11
pantera07 wrote: Hi Bunuel
Even I solved as gmatc2018 did. In the end I did not get 1 as the integer value to be considered. It would be great if you can point out the flaw in the approach. Thanks so much. 1  x < 2; O'pen modulus: 2 < 1  x < 2; Add 1 to all parts 1 <  x < 3; Multiply by 1 and flip sings: 1 > x > 3. You could solve the way is suggested by gmatc2018, but this way or the way shown in the OE is better. Still, if interested. 1  x < 2:If 1  x < 0 so if x > 1, then (1  x) < 2 > x < 3, so for this range: 1 < x < 3. If 1  x >= 0 so if x <= 1, then 1  x < 2 > x > 1, so for this range: 1 < x <= 1. Combining gives: 1 < x < 3. So, to summarize, you should include = sign in either of the ranges.
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07 May 2018, 04:06
I think this is a highquality question and I agree with explanation.



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25 May 2018, 14:34
Good question. Here is how I approached it: \(1x < 2\) > this simply means that x must be 0, 1, or 2. Why? Plug in 3 and you get \(2 = 2\) which is not < 2. Same applies to negative numbers. With x in {0, 1, 2}, plug in answer choices: A. 2 is a prime number so reject B. \(x^2 + x\) is not a prime number > \(1^2 + 1\) = 2 which is prime. Reject C. 0 is neither positive or negative, but it is even FYI D. Distinct factors of 0 + 2 is 1 and 2 so 2 numbers is prime since 2 is prime. 1 + 2 = 3 which has 1 and 3 as factors so again 2. 2 + 2 = 4 which has 1, 2, 4 as distinct factors and 3 is prime. Sufficient. E. 0 is multiple of every integer (except 0) so reject
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