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How many values can the integer p=|x+3|-|x-3| assume?

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Re: How many values can the integer p=|x+3|-|x-3| assume?  [#permalink]

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New post 06 Jan 2017, 07:14
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Re: How many values can the integer p=|x+3|-|x-3| assume?  [#permalink]

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New post 13 Feb 2017, 06:53
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Re: How many values can the integer p=|x+3|-|x-3| assume?  [#permalink]

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New post 06 Apr 2018, 06:37
Zarrolou wrote:
Good job arpanpatnaik, vinaymimani!

Official explanation

The function \(|x+3|-|x-3|\) for values \(\geq{}3\) equals \(6\), and for values \(\leq{}-3\) equals \(-6\)
For the middle values it follows the equation \(2x\) (as the users above correctly say)

However there is a quicker way to get to the answer than counting the possible values.

Its upper limit is \(6\), its lower limit is \(-6\) and the function \(2x\) is monotonic and increasing (and continuous), so will assume all values between 6 and -6 included.
(This is not theory necessary for the GMAT, but if notice the fact that \(2x\) must pass for all values between 6 and -6, you can save time)

So the values that the integer p can assume are \(-6,-5,...,0,...,5,6\) TOT=\(13\)

The correct answer is C

For for clarity, below there is the graph of \(|x+3|-|x-3|\) that will make my explanation more clear.


Can you please explain how you drew the graph with two variables. sort of struggling with that.
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Re: How many values can the integer p=|x+3|-|x-3| assume?  [#permalink]

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New post 17 Sep 2019, 22:09
missed the last step saying p is an integer... but this is a nice question!
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How many values can the integer p=|x+3|-|x-3| assume?  [#permalink]

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New post 03 Nov 2019, 09:25
Zarrolou wrote:
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.


Asked: How many values can the integer \(p=|x+3|-|x-3|\) assume?

Region 1: x<-3
p = -x-3 - (3-x) = -x - 3 - 3 + x = -6

Region 2: -3<=x<=3
p = x+3 - (3-x) = x + 3 - 3 +x = 2x
p can take integer values between -6 and 6
p = {-6,,-5,-4,-3,-2,-1,0,1,2,3,4,5,6}

Region 3: x>3
p = x+ 3 - (x-3) = x + 3 - x + 3 = 6

Combining results of all regions: -
p = {-6,,-5,-4,-3,-2,-1,0,1,2,3,4,5,6}
p takes 13 different values

IMO C
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How many values can the integer p=|x+3|-|x-3| assume?   [#permalink] 03 Nov 2019, 09:25

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