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M11-16

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M11-16  [#permalink]

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New post 16 Sep 2014, 00:44
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If \(x\) is an integer and \(|1-x| \lt 2\) 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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Re M11-16  [#permalink]

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New post 16 Sep 2014, 00:44
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Official Solution:

If \(x\) is an integer and \(|1-x| \lt 2\) 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


\(|1-x|\) is just the distance between 1 and \(x\) on the number line. We are told that this distance is less than 2:

--(-1)----1----3-- 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: M11-16  [#permalink]

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New post 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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Re: M11-16  [#permalink]

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New post 15 Nov 2015, 09:10
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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: M11-16  [#permalink]

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New post 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: M11-16  [#permalink]

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New post 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: M11-16  [#permalink]

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New post 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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Re: M11-16  [#permalink]

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New post 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: M11-16  [#permalink]

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New post 29 Jun 2016, 17:56
X could be 0,1,2.
Now only option D gives correct answer
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Re M11-16  [#permalink]

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New post 02 Aug 2016, 03:03
I think this is a high-quality question and I agree with explanation.
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Re: M11-16  [#permalink]

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New post 03 Aug 2016, 16:46
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Here is my two cents for those who need a more detailed approach and credit to Ron from Veritas

https://www.veritasprep.com/blog/2013/0 ... s-of-zero/
>> !!!

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Re: M11-16  [#permalink]

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New post 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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Re: M11-16  [#permalink]

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New post 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: M11-16  [#permalink]

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New post 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: M11-16  [#permalink]

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New post 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:

|1-x| < 2

gives us:
a) 1-x < 2 for 1-x > 0 leads to -1 < x < 1
b) 1-x > -2 for 1-x < 0 leads to 1 < x < 3

see how 1 was not included in the range because I also took the equation 1-x >/< 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.

Best
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Re: M11-16  [#permalink]

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New post 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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New post 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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Re M11-16  [#permalink]

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New post 07 May 2018, 04:06
I think this is a high-quality question and I agree with explanation.
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Re: M11-16  [#permalink]

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New post 25 May 2018, 14:34
Good question. Here is how I approached it:

\(|1-x| < 2\) --> 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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Re: M11-16 &nbs [#permalink] 25 May 2018, 14:34
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