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Re: How many values can the integer p=x+3x3 assume?
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17 Jun 2013, 08:32
Zarrolou wrote: WholeLottaLove wrote: 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+3x3 doesn't change. For example:
when x=4 P=x+3x+3 p=6
When x=4 p=(x+3) (x3) p=x3  x+3 p=x3 + x3 p=6
For any value in between: x=1 P=(x+3) (x3) P=x+3  x+3 P=x+3 +x3 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 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.



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Re: How many values can the integer p=x+3x3 assume?
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17 Jun 2013, 08:40
WholeLottaLove wrote: 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. 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



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Re: How many values can the integer p=x+3x3 assume?
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01 Jul 2013, 08:40
How many values can the integer p=x+3x3 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+3x3 (x+3) (x3) x3  (x+3) x3 + x3 P=6 For 3<x<3 x+3x3 (x+3)  (x3) (x+3)  (x+3) (x+3) + x 3 P=2x For x>3: x+3x3 (x+3)  (x3) (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!



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Re: How many values can the integer p=x+3x3 assume?
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01 Jul 2013, 09:05
WholeLottaLove wrote: How many values can the integer p=x+3x3 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+3x3 (x+3) (x3) x3  (x+3) x3 + x3 P=6 For 3<x<3 x+3x3 (x+3)  (x3) (x+3)  (x+3) (x+3) + x 3 P=2x For x>3: x+3x3 (x+3)  (x3) (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! Well, if you are not sure, you can plug in a value greater than 3 or less than 3 and see what you find. \(p=10+3103=137=6\) 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 \(p=(x+3)(x3)=6\) so 6 is a possible value same thing for values less than 3.



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Re: How many values can the integer p=x+3x3 assume?
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01 Jul 2013, 12:32
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 (x3) cases).



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Re: How many values can the integer p=x+3x3 assume?
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01 Jul 2013, 12:39
nancerella wrote: 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 (x3) cases). I think the quickest way is the one explained here: howmanyvaluescantheintegerpx3x3assume152859.html#p1225488You find that the function \(p=x+3x3\) 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



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Re: How many values can the integer p=x+3x3 assume?
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01 Jul 2013, 12:51
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 "goto" strategy/process that cuts down on computation time.



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Re: How many values can the integer p=x+3x3 assume?
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01 Jul 2013, 20:35
Zarrolou wrote: 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 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  x3 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...
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Re: How many values can the integer p=x+3x3 assume?
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01 Jul 2013, 20:36
Zarrolou wrote: How many values can the integer \(p=x+3x3\) 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. Also, it is not a 600700 level question. It is certainly 700+ level.
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Re: How many values can the integer p=x+3x3 assume?
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02 Jul 2013, 08:35
Zarrolou wrote: How many values can the integer \(p=x+3x3\) 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. 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



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Re: How many values can the integer p=x+3x3 assume?
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10 Jul 2013, 08:35
How many values can the integer p=x+3x3 assume?
Checkpoints at 3, 3
x<3, 3<x<3, x>3
x<3: p=x+3x3 p=(x+3)  (x3) p= x3  (x+3) p= x3 + x 3 p= 6
3<x<3 p=x+3x3 p= (x+3)  (x3) p= x+3 + x 3 p= 2x
x>3 p=x+3x3 p=(x+3)(x3) 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



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Re: How many values can the integer p=x+3x3 assume?
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26 Dec 2014, 02:03
How many values can the integer p=x+3x3 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 x3>0 x+3>0 and x3<0 x+3<0 and x3>0 x+3<0 and x3<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



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How many values can the integer p=x+3x3 assume?
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28 Dec 2014, 20:48
sehosayho wrote: How many values can the integer p=x+3x3 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 x3>0 x+3>0 and x3<0 x+3<0 and x3>0 x+3<0 and x3<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 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 x3 >0 x+3 >0 and x3 <0 x+3 <0 and x3 >0 x+3 <0 and x3<0 Case 1: x+3 >0 and x3 >0 x > 3 and x > 3 This is the case of x > 3 Case 2: x+3 >0 and x3 <0 x > 3 and x < 3 This is the case of 3 < x< 3 Case 3: x+3 <0 and x3 >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 x3<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: http://www.veritasprep.com/blog/2014/06 ... thegmat/
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Re: How many values can the integer p=x+3x3 assume?
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01 Jan 2015, 15:18
mau5 wrote: Zarrolou wrote: How many values can the integer \(p=x+3x3\) 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)(x3) = 6 For all 3<x<3, we have the value of p = (x+3)(3x) = 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 = (x3)(3x) = 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



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Re: How many values can the integer p=x+3x3 assume?
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01 Jan 2015, 15:18
mau5 wrote: Zarrolou wrote: How many values can the integer \(p=x+3x3\) 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)(x3) = 6 For all 3<x<3, we have the value of p = (x+3)(3x) = 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 = (x3)(3x) = 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



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Re: How many values can the integer p=x+3x3 assume?
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01 Jan 2015, 15:19
mau5 wrote: Zarrolou wrote: How many values can the integer \(p=x+3x3\) 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)(x3) = 6 For all 3<x<3, we have the value of p = (x+3)(3x) = 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 = (x3)(3x) = 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



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How many values can the integer p=x+3x3 assume?
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27 Jul 2015, 17:07
I did it differently.
p = x+3x3; 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? 105 = 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(x3)^2 ((x+3)(x3))((x+3)+(x3)) = 6*2x = 12x p = 12x = the difference that results when subtracting the two absolute values. 12 possibilities + 1 = 13.



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Re: How many values can the integer p=x+3x3 assume?
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27 Jul 2015, 20:16
Salvetor wrote: mau5 wrote: Zarrolou wrote: How many values can the integer \(p=x+3x3\) 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)(x3) = 6 For all 3<x<3, we have the value of p = (x+3)(3x) = 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 = (x3)(3x) = 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 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+3x3 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.
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Re: How many values can the integer p=x+3x3 assume?
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24 Nov 2015, 03:19
Zarrolou wrote: How many values can the integer \(p=x+3x3\) 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. 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



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Re: How many values can the integer p=x+3x3 assume?
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06 Jan 2017, 06:00
What is the level of this question. Above 650?




Re: How many values can the integer p=x+3x3 assume?
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