GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 27 Jun 2019, 01:52

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# M11-16

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55804

### Show Tags

16 Sep 2014, 00:44
3
6
00:00

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 55804

### Show Tags

16 Sep 2014, 00:44
1
2
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.

_________________
Intern
Joined: 26 Oct 2014
Posts: 22

### Show Tags

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!
Math Expert
Joined: 02 Sep 2009
Posts: 55804

### Show Tags

15 Nov 2015, 09:10
2
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.
_________________
Intern
Joined: 26 Oct 2014
Posts: 22

### Show Tags

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!!!
Intern
Joined: 20 Jan 2016
Posts: 12

### Show Tags

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!!
Math Expert
Joined: 02 Sep 2009
Posts: 55804

### Show Tags

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.
_________________
Manager
Joined: 08 Jul 2015
Posts: 53
GPA: 3.8
WE: Project Management (Energy and Utilities)

### Show Tags

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!
_________________
[4.33] In the end, what would you gain from everlasting remembrance? Absolutely nothing. So what is left worth living for?
This alone: justice in thought, goodness in action, speech that cannot deceive, and a disposition glad of whatever comes, welcoming it as necessary, as familiar, as flowing from the same source and fountain as yourself. (Marcus Aurelius)
Intern
Joined: 14 Mar 2015
Posts: 16
Location: India
Concentration: General Management, Strategy
Schools: ISB '18, IIMA , IIMB
GPA: 3.53
WE: Engineering (Other)

### Show Tags

29 Jun 2016, 17:56
X could be 0,1,2.
Now only option D gives correct answer
Senior Manager
Joined: 31 Mar 2016
Posts: 376
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

### Show Tags

02 Aug 2016, 03:03
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 23 Jun 2009
Posts: 176
Location: Brazil
GMAT 1: 470 Q30 V20
GMAT 2: 620 Q42 V33

### Show Tags

03 Aug 2016, 16:46
1
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/
>> !!!

You do not have the required permissions to view the files attached to this post.

Intern
Joined: 25 Jul 2016
Posts: 2

### Show Tags

04 Apr 2017, 12:49
isnt't 0 a positive intger..in that case shouldn't Option 3 not be true?
Math Expert
Joined: 02 Sep 2009
Posts: 55804

### Show Tags

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/math-number- ... 88376.html

Hope it helps.
_________________
Intern
Joined: 20 Feb 2017
Posts: 1

### Show Tags

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).

Thank you
Retired Moderator
Joined: 22 Aug 2013
Posts: 1437
Location: India

### Show Tags

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).

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)
Intern
Joined: 29 Jan 2018
Posts: 6

### Show Tags

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
Intern
Joined: 08 Aug 2017
Posts: 31

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 55804

### Show Tags

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.
_________________
Manager
Joined: 26 Feb 2018
Posts: 51
Location: India
GMAT 1: 640 Q45 V34
GPA: 3.9
WE: Web Development (Computer Software)

### Show Tags

07 May 2018, 04:06
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 29 Sep 2017
Posts: 117
Location: United States

### Show Tags

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
_________________
If this helped, please give kudos!
Re: M11-16   [#permalink] 25 May 2018, 14:34

Go to page    1   2    Next  [ 21 posts ]

Display posts from previous: Sort by

# M11-16

Moderators: chetan2u, Bunuel