Last visit was: 19 Nov 2025, 14:50 It is currently 19 Nov 2025, 14:50
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3   
User avatar
WholeLottaLove
User avatar
Joined: 13 May 2013
Last visit: 13 Jan 2014
Posts: 305
Own Kudos:
627
Given Kudos: 134
Posts: 305
Kudos: 627
How many values can the integer p=|x+3|-|x-3| assume? 17 Jun 2013, 09:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Zarrolou
WholeLottaLove
I believe I understand 2x and how it functions - 2x must = an integer, so x can be any value so long as it produces an integer result between -6 and 6.

Here is where I continue to get tripped up, though. You say that for every value of x > 3, P=6. This I see, as I have plugged in a few integers > 3 for x and the result is always x. The same goes for integers less than negative 3.

Wait a second, I think I am making the mistake of looking for actual values to plug in for x rather than look for where the equation is positive and negative.

When x > 3, p=|x+3|-|x-3| doesn't change. For example:

when x=4
P=x+3-x+3
p=6

When x=-4
p=-(x+3)- -(x-3)
p=-x-3 - -x+3
p=-x-3 + x-3
p=-6

For any value in between:
x=1
P=(x+3)- -(x-3)
P=x+3 - -x+3
P=x+3 +x-3
P=2x

So, again, 2x must e an integer between -6 and 6 so any value of x is sufficient as long as it satisfies the constraints of -6 and 6?

99% correct.

The red part contains an error, the correct version is:
So, again, P(not 2x) must be an integer between -6 and 6 so any value of x is sufficient as long as it satisfies the constraints of -3 and 3 (-3<x<3) AND for it P=INTEGER,

so x=3/2 it's valid option: it's in the interval -3,3 and p=2*3/2=3 (integer between -6 and 6)
or x=3/4 it's a valid option: it's in the interval -3,3 BUT p=2*3/4=3/2 (not an integer)

But from there, from the red part, it's easier to count all the integer between 6 and -6, rather than to find the values of x that generate them (as I did above for x=3/2 and 3/4)

Hope it's clear
Show more

Ok,

So 2x must be an integer and the result must lie within -6 and 6 when the values of x lie within -3 and 3?

And (I know I asked this before but I'm not 100% sure) how do I know if it's a trap and not all values between -6 and 6 are valid?

Again, thank you for your patience and help.
User avatar
Zarrolou
User avatar
Joined: 02 Sep 2012
Last visit: 11 Dec 2013
Posts: 846
Own Kudos:
5,145
 [1]
Given Kudos: 219
Status:Far, far away!
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Posts: 846
Kudos: 5,145
 [1]
How many values can the integer p=|x+3|-|x-3| assume? 17 Jun 2013, 09:40
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
WholeLottaLove

Ok,

So 2x must be an integer and the result must lie within -6 and 6 when the values of x lie within -3 and 3?

And (I know I asked this before but I'm not 100% sure) how do I know if it's a trap and not all values between -6 and 6 are valid?

Again, thank you for your patience and help.
Show more

Yes, correct.

You know that there are no trap values because \(2x\) is a straight line, so is defined for every \(x\).
Lines in general are defined for every value of x. With this I mean that you can draw a line, and for whatever value of x you pick, you'll always find a corresponding value on the line.
You can ask yourself: is there any value of \(x\) for which \(2x\) is not defined? The answer is no.

You're welcome :)
User avatar
WholeLottaLove
User avatar
Joined: 13 May 2013
Last visit: 13 Jan 2014
Posts: 305
Own Kudos:
627
Given Kudos: 134
Posts: 305
Kudos: 627
How many values can the integer p=|x+3|-|x-3| assume? 01 Jul 2013, 09:40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
How many values can the integer p=|x+3|-|x-3| assume?

A)6
B)7
C)13
D)12
E)Cannot be determined

We're not looking for how many valid solutions of "x" there are...we are looking for how many integers "p" there are (that's what was tripping me up before!!!)

We can do this by finding the range of values of x (i.e. what numbers, if any, does x lie between)

Find the check points: -3, 3

We have three ranges to test for: -3< x, -3<x<3, 3>x

For x<-3: |x+3|-|x-3| -(x+3)- -(x-3) -x-3 - (-x+3) -x-3 + x-3 P=-6
For -3<x<3 |x+3|-|x-3| (x+3) - -(x-3) (x+3) - (-x+3) (x+3) + x -3 P=2x
For x>3: |x+3|-|x-3| (x+3) - (x-3) (x+3) -x+3 P=6

So, the range of P is from -6 ≤ P ≤ 6. There are 13 integers between -6 and 6 inclusive.

Just one question - how do I know the values are inclusive (-6 ≤ p ≤ 6) as opposed to not (-6 < p < 6)?

Thanks!
User avatar
Zarrolou
User avatar
Joined: 02 Sep 2012
Last visit: 11 Dec 2013
Posts: 846
Own Kudos:
5,145
 [2]
Given Kudos: 219
Status:Far, far away!
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Posts: 846
Kudos: 5,145
 [2]
How many values can the integer p=|x+3|-|x-3| assume? 01 Jul 2013, 10:05
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
WholeLottaLove
How many values can the integer p=|x+3|-|x-3| assume?

A)6
B)7
C)13
D)12
E)Cannot be determined

We're not looking for how many valid solutions of "x" there are...we are looking for how many integers "p" there are (that's what was tripping me up before!!!)

We can do this by finding the range of values of x (i.e. what numbers, if any, does x lie between)

Find the check points: -3, 3

We have three ranges to test for: -3< x, -3<x<3, 3>x

For x<-3: |x+3|-|x-3| -(x+3)- -(x-3) -x-3 - (-x+3) -x-3 + x-3 P=-6
For -3<x<3 |x+3|-|x-3| (x+3) - -(x-3) (x+3) - (-x+3) (x+3) + x -3 P=2x
For x>3: |x+3|-|x-3| (x+3) - (x-3) (x+3) -x+3 P=6

So, the range of P is from -6 ≤ P ≤ 6. There are 13 integers between -6 and 6 inclusive.

Just one question - how do I know the values are inclusive (-6 ≤ p ≤ 6) as opposed to not (-6 < p < 6)?

Thanks!
Show more

Well, if you are not sure, you can plug in a value greater than 3 or less than -3 and see what you find.

\(p=|10+3|-|10-3|=|13|-|7|=6\) so 6 is a possible value, same thing for x=-10

from a more methodical point of view, if x is greater than 3, the whole expression becomes

\(p=(x+3)-(x-3)=6\) so 6 is a possible value

same thing for values less than -3.
avatar
nancerella
avatar
Joined: 09 Jun 2013
Last visit: 14 Feb 2016
Posts: 4
Posts: 4
Kudos: 0
How many values can the integer p=|x+3|-|x-3| assume? 01 Jul 2013, 13:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
thanks to everyone's explanations, i think i've finally understood the solution to this problem.

now my question is, what's the best way to be solving this and similar type (multiple mods) questions to keep within the 2 mins mark?
would it be first, identifying the key ref points (e.g. 3 and -3, in this case) and then plugging in numbers within the ranges?
or is this just a concept that you need to get really good at and be able to quickly recognize the +/- setups of each mod for each of the respective scenarios? (e.g. if x<3, then setup equation with neg (x+3) and pos (x-3) cases).
User avatar
Zarrolou
User avatar
Joined: 02 Sep 2012
Last visit: 11 Dec 2013
Posts: 846
Own Kudos:
5,145
 [1]
Given Kudos: 219
Status:Far, far away!
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Posts: 846
Kudos: 5,145
 [1]
How many values can the integer p=|x+3|-|x-3| assume? 01 Jul 2013, 13:39
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nancerella
thanks to everyone's explanations, i think i've finally understood the solution to this problem.

now my question is, what's the best way to be solving this and similar type (multiple mods) questions to keep within the 2 mins mark?
would it be first, identifying the key ref points (e.g. 3 and -3, in this case) and then plugging in numbers within the ranges?
or is this just a concept that you need to get really good at and be able to quickly recognize the +/- setups of each mod for each of the respective scenarios? (e.g. if x<3, then setup equation with neg (x+3) and pos (x-3) cases).
Show more

I think the quickest way is the one explained here: how-many-values-can-the-integer-p-x-3-x-3-assume-152859.html#p1225488

You find that the function \(p=|x+3|-|x-3|\) ranges from 6 to -6 : will assume every value in that range. The question asks for the number of INTEGER values p can have, so just count the integers between -6 and 6 included.

Hope it makes sense :)
avatar
nancerella
avatar
Joined: 09 Jun 2013
Last visit: 14 Feb 2016
Posts: 4
Posts: 4
Kudos: 0
How many values can the integer p=|x+3|-|x-3| assume? 01 Jul 2013, 13:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
thanks for the response, zarrolou, and yes, i agree. that explanation makes it quite easy and quick to solve the problem.
i was just asking for these type of multi mod type questions - 2 or 3 mods on both sides of the equal sign - if there's a "go-to" strategy/process that cuts down on computation time.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,000
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,000
 [1]
How many values can the integer p=|x+3|-|x-3| assume? 01 Jul 2013, 21:35
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Zarrolou

Just one thing before I proceed, my option E "cannot be determined" sounds good as possible answer.
Some users wrote to me saying that is the first thing that came into their mind when they saw the question. What do you think?

Thanks again Mike :)
Show more

I agree with Mike. 'Cannot be determined' is not a valid GMAT option. You probably wanted to say 'Infinite values'.

By the way, it's a very nice question. I think it has many subtle takeaways

p = |x+3| - |x-3|
First thing to realize here is that p needs to be an integer, not x.
Another thing, when you subtract two mods, the result takes the same value over a wide range.


____________________ -3 ___________x ___________3_________________________

<----------------------------

<-----------------------------------------------------------------


These are the two points -3 and 3 on the number line. We need to find
'the distance from -3' - 'the distance from 3' = p
i.e red line - green line.

Notice that the red line will cancel the part of green line to the left of -3 and hence red line - green line will always be -6 for all value to the left of -3.

Similarly, red line - green line will be 6 for all values to the right of 3.

The tricky values are the ones lying in between -3 and 3. When x = -3, we get p = -6. For some point between -3 and 3, we will get p = -5, -4, -3, -2.... 6. So there will be 13 values.
e.g. p = 5
If you move 0.5 to the right of -3, distance from -3 will be 0.5 and distance from 3 will be 5.5.
0.5 - 5.5 = -5

and so on...
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,000
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,000
 [1]
How many values can the integer p=|x+3|-|x-3| assume? 01 Jul 2013, 21:36
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Zarrolou
How many values can the integer \(p=|x+3|-|x-3|\) assume?

A)6
B)7
C)13
D)12
E)Cannot be determined

My own question, as always any feedback is appreciated
Click here for the OE.
Show more



Also, it is not a 600-700 level question. It is certainly 700+ level.
avatar
krrish
avatar
Joined: 04 Mar 2013
Last visit: 03 Feb 2014
Posts: 46
Own Kudos:
15
Given Kudos: 6
Location: India
Concentration: General Management, Marketing
Schools: Stanford '14 Kellogg '14 Sloan '14 Booth '16
GPA: 3.49
WE:Web Development (Computer Software)
Schools: Stanford '14 Kellogg '14 Sloan '14 Booth '16
Posts: 46
Kudos: 15
How many values can the integer p=|x+3|-|x-3| assume? 02 Jul 2013, 09:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Zarrolou
How many values can the integer \(p=|x+3|-|x-3|\) assume?

A)6
B)7
C)13
D)12
E)Infinite values

My own question, as always any feedback is appreciated
Click here for the OE.
Show more

3



here's my take :
from -3 to 3, we have values from -6 to 6, so there are 13 in limit check for that option, it done :)

simple logic
Mod is always positive or negative and the value will always lie beteween the limits
xii maths :)
User avatar
WholeLottaLove
User avatar
Joined: 13 May 2013
Last visit: 13 Jan 2014
Posts: 305
Own Kudos:
627
 [1]
Given Kudos: 134
Posts: 305
Kudos: 627
 [1]
How many values can the integer p=|x+3|-|x-3| assume? 10 Jul 2013, 09:35
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
How many values can the integer p=|x+3|-|x-3| assume?

Checkpoints at -3, 3

x<-3, -3<x<3, x>3

x<-3:
p=|x+3|-|x-3|
p=-(x+3) - -(x-3)
p= -x-3 - (-x+3)
p= -x-3 + x -3
p= -6

-3<x<3
p=|x+3|-|x-3|
p= (x+3) - -(x-3)
p= x+3 + x -3
p= 2x

x>3
p=|x+3|-|x-3|
p=(x+3)-(x-3)
p= 6

So the range of p is from -6 to 6. There are a total of 13 integers between -6 and 6

(C) 13
User avatar
sehosayho
User avatar
Joined: 26 Aug 2013
Last visit: 04 Jan 2016
Posts: 11
Own Kudos:
14
Given Kudos: 19
Posts: 11
Kudos: 14
How many values can the integer p=|x+3|-|x-3| assume? 26 Dec 2014, 03:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
How many values can the integer p=|x+3|-|x-3| assume?

A)6
B)7
C)13
D)12
E)Infinite values

Many folks used folloing approach
1. x >or= 3
2. x belongs to {-3,3}
3. x < or = -3
leaving 3 possible valuse for P ( 6,-6, and 2x)

I understand this approach, but I am confused with another approach of soloving inequalities questions like this one
the other approach is following
|x+3|>0 and |x-3|>0
|x+3|>0 and |x-3|<0
|x+3|<0 and |x-3|>0
|x+3|<0 and |x-3|<0
this case there are 4 possible valuse for P
P= 6, -6, 2x, or -2x
what am I missing here??
If I cannot use the second approach could you explain why please.
Thank you
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,000
 [2]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,000
 [2]
How many values can the integer p=|x+3|-|x-3| assume? 28 Dec 2014, 21:48
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
sehosayho
How many values can the integer p=|x+3|-|x-3| assume?

A)6
B)7
C)13
D)12
E)Infinite values

Many folks used folloing approach
1. x >or= 3
2. x belongs to {-3,3}
3. x < or = -3
leaving 3 possible valuse for P ( 6,-6, and 2x)

I understand this approach, but I am confused with another approach of soloving inequalities questions like this one
the other approach is following
|x+3|>0 and |x-3|>0
|x+3|>0 and |x-3|<0
|x+3|<0 and |x-3|>0
|x+3|<0 and |x-3|<0
this case there are 4 possible valuse for P
P= 6, -6, 2x, or -2x
what am I missing here??
If I cannot use the second approach could you explain why please.
Thank you
Show more

The first approach is the logical conclusion of the second approach (with modification - the second approach as given by you in incorrect). The absolute value can never be negative. The four cases are
x+3 >0 and x-3 >0
x+3 >0 and x-3 <0
x+3 <0 and x-3 >0
x+3 <0 and x-3<0

Case 1:
x+3 >0 and x-3 >0
x > -3 and x > 3
This is the case of x > 3

Case 2:
x+3 >0 and x-3 <0
x > -3 and x < 3
This is the case of -3 < x< 3

Case 3:
x+3 <0 and x-3 >0
x < -3 and x > 3
Is this possible? Can x be less than -3 and greater than 3 simultaneously? No.

Case 4:
x+3 <0 and x-3<0
x<-3 and x < 3
This is the case of x < -3

Note that the cases come from the definition of absolute value:
|x| = x when x > 0 and -x when x < 0

You will find this post helpful: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/06 ... -the-gmat/
avatar
Salvetor
avatar
Joined: 23 Dec 2014
Last visit: 20 Dec 2021
Posts: 41
Own Kudos:
46
Given Kudos: 52
Posts: 41
Kudos: 46
How many values can the integer p=|x+3|-|x-3| assume? 01 Jan 2015, 16:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mau5
Zarrolou
How many values can the integer \(p=|x+3|-|x-3|\) assume?

A)0
B)7
C)13
D)14
E)Cannot be determined

My own question, as always any feedback is appreciated
Kudos to the first correct solution(s)!

For x=3/-3 we have the value of p = 6/-6.
For all x>3, we have the value of p = (x+3)-(x-3) = 6
For all -3<x<3, we have the value of p = (x+3)-(3-x) = 2x. We could have x = 1/2,1,3/2,2,5/2 ,0 and the same set of negatives--> 11 values
For all x<-3, we have the value of p = (-x-3)-(3-x) = -6.
Thus, the number of unique values that p can have are 13.
C.
Show more

sorry for the noob question,
first why you chose X>3 and X<-3 , but why not x>-3 and X<3

second, beside the 11 values 1/2,1,3/2,2,5/2 ,0 , what are the two values you added to make it 13, are they 6 and -6 ? A detailed an easy explanation would be great :)
avatar
Salvetor
avatar
Joined: 23 Dec 2014
Last visit: 20 Dec 2021
Posts: 41
Own Kudos:
46
Given Kudos: 52
Posts: 41
Kudos: 46
How many values can the integer p=|x+3|-|x-3| assume? 01 Jan 2015, 16:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mau5
Zarrolou
How many values can the integer \(p=|x+3|-|x-3|\) assume?

A)0
B)7
C)13
D)14
E)Cannot be determined

My own question, as always any feedback is appreciated
Kudos to the first correct solution(s)!

For x=3/-3 we have the value of p = 6/-6.
For all x>3, we have the value of p = (x+3)-(x-3) = 6
For all -3<x<3, we have the value of p = (x+3)-(3-x) = 2x. We could have x = 1/2,1,3/2,2,5/2 ,0 and the same set of negatives--> 11 values
For all x<-3, we have the value of p = (-x-3)-(3-x) = -6.
Thus, the number of unique values that p can have are 13.
C.
Show more

sorry for the noob question,
first why you chose X>3 and X<-3 , but why not x>-3 and X<3

second, beside the 11 values 1/2,1,3/2,2,5/2 ,0 , what are the two values you added to make it 13, are they 6 and -6 ? A detailed an easy explanation would be great :)
avatar
Salvetor
avatar
Joined: 23 Dec 2014
Last visit: 20 Dec 2021
Posts: 41
Own Kudos:
46
Given Kudos: 52
Posts: 41
Kudos: 46
How many values can the integer p=|x+3|-|x-3| assume? 01 Jan 2015, 16:19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mau5
Zarrolou
How many values can the integer \(p=|x+3|-|x-3|\) assume?

A)0
B)7
C)13
D)14
E)Cannot be determined

My own question, as always any feedback is appreciated
Kudos to the first correct solution(s)!

For x=3/-3 we have the value of p = 6/-6.
For all x>3, we have the value of p = (x+3)-(x-3) = 6
For all -3<x<3, we have the value of p = (x+3)-(3-x) = 2x. We could have x = 1/2,1,3/2,2,5/2 ,0 and the same set of negatives--> 11 values
For all x<-3, we have the value of p = (-x-3)-(3-x) = -6.
Thus, the number of unique values that p can have are 13.
C.
Show more

sorry for the noob question,
first why you chose X>3 and X<-3 , but why not x>-3 and X<3

second, beside the 11 values 1/2,1,3/2,2,5/2 ,0 , what are the two values you added to make it 13, are they 6 and -6 ? A detailed an easy explanation would be great :)
User avatar
iPen
User avatar
Joined: 08 Jun 2015
Last visit: 20 Dec 2020
Posts: 83
Own Kudos:
109
Given Kudos: 40
Posts: 83
Kudos: 109
How many values can the integer p=|x+3|-|x-3| assume? 27 Jul 2015, 18:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I did it differently.

p = |x+3|-|x-3|; what are the possible values for p?

Absolute values of x have two times the range of values as x. So, this is a difference of squares. The difference tells us the range of values. However, this difference value doesn't count the first value you subtract from (e.g. how many values inclusive are between 10 and 5? 10-5 = 5; but, you have to add 1 because there are 6 values: 5, 6, 7, 8, 9, and 10).

P takes on the form a^2 - b^2.
(x+3)^2-(x-3)^2
((x+3)-(x-3))((x+3)+(x-3)) = 6*2x = 12x
p = 12x = the difference that results when subtracting the two absolute values.
12 possibilities + 1 = 13.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,000
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,000
 [1]
How many values can the integer p=|x+3|-|x-3| assume? 27 Jul 2015, 21:16
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Salvetor
mau5
Zarrolou
How many values can the integer \(p=|x+3|-|x-3|\) assume?

A)0
B)7
C)13
D)14
E)Cannot be determined

My own question, as always any feedback is appreciated
Kudos to the first correct solution(s)!

For x=3/-3 we have the value of p = 6/-6.
For all x>3, we have the value of p = (x+3)-(x-3) = 6
For all -3<x<3, we have the value of p = (x+3)-(3-x) = 2x. We could have x = 1/2,1,3/2,2,5/2 ,0 and the same set of negatives--> 11 values
For all x<-3, we have the value of p = (-x-3)-(3-x) = -6.
Thus, the number of unique values that p can have are 13.
C.

sorry for the noob question,
first why you chose X>3 and X<-3 , but why not x>-3 and X<3

second, beside the 11 values 1/2,1,3/2,2,5/2 ,0 , what are the two values you added to make it 13, are they 6 and -6 ? A detailed an easy explanation would be great :)
Show more

Note that there are no constraints on the value of x. So x can take ANY value on the number line from - infinity to infinity. We can divide the number line into 3 parts:
x <= -3
-3 < x< 3 (which is the same as x>-3 and x<3)
x >= 3

These 3 cover all values of x. The function |x+3|-|x-3| will behave differently in these three different ranges because -3 and 3 are transition points.

When you solve it, you get that when x <= -3, p is always -6.
When x is between -3 and 3, p takes 11 different values.
When x >= 3, p is always 6.

So you get total 13 different values that p can take depending on the value of x.
User avatar
BrainLab
User avatar
Current Student
Joined: 10 Mar 2013
Last visit: 26 Jan 2025
Posts: 345
Own Kudos:
3,131
Given Kudos: 200
Location: Germany
Concentration: Finance, Entrepreneurship
Schools: WHU MBA"20 (A$)
GMAT 1: 580 Q46 V24
GPA: 3.7
WE:Marketing (Telecommunications)
Schools: WHU MBA"20 (A$)
GMAT 1: 580 Q46 V24
Posts: 345
Kudos: 3,131
How many values can the integer p=|x+3|-|x-3| assume? 24 Nov 2015, 04:19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Zarrolou
How many values can the integer \(p=|x+3|-|x-3|\) assume?

A)6
B)7
C)13
D)12
E)Infinite values

My own question, as always any feedback is appreciated
Click here for the OE.
Show more

can someone Elaborate more on the range -3≤x≤3. Why to use x non integer values if any number in this range let's say 0 or -2 yields value 6, so we have only to extreme values -6 and 6 and the range is 6-(-6)+1=13 Answer (C)
x=0: 3-(-3)=6
x=-2: 1-(-5)=6
avatar
Salvetor
avatar
Joined: 23 Dec 2014
Last visit: 20 Dec 2021
Posts: 41
Own Kudos:
46
Given Kudos: 52
Posts: 41
Kudos: 46
How many values can the integer p=|x+3|-|x-3| assume? 06 Jan 2017, 07:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
What is the level of this question. Above 650?
   1   2   3   
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts