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Math Expert V
Joined: 02 Sep 2009
Posts: 55804

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Difficulty:   95% (hard)

Question Stats: 47% (01:27) correct 53% (01:27) wrong based on 163 sessions

HideShow timer Statistics 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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Math Expert V
Joined: 02 Sep 2009
Posts: 55804

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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.

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Intern  B
Joined: 26 Oct 2014
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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!
Math Expert V
Joined: 02 Sep 2009
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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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Intern  B
Joined: 26 Oct 2014
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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!!!
Intern  Joined: 20 Jan 2016
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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!!
Math Expert V
Joined: 02 Sep 2009
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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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Thanks Bunuel, I was overlooked "1" is also a factor and 0 is a multiple of all odd - good question!
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X could be 0,1,2.
Now only option D gives correct answer
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I think this is a high-quality question and I agree with explanation.
Manager  B
Joined: 23 Jun 2009
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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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Intern  B
Joined: 25 Jul 2016
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isnt't 0 a positive intger..in that case shouldn't Option 3 not be true?
Math Expert V
Joined: 02 Sep 2009
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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/math-number- ... 88376.html

Hope it helps.
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Intern  B
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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).

Thank you
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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).

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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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
Intern  B
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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.
Math Expert V
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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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I think this is a high-quality question and I agree with explanation.
Manager  S
Joined: 29 Sep 2017
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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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