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How many values can the integer p=x+3x3 assume?
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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.
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Originally posted by Zarrolou on 15 May 2013, 11:30.
Last edited by Zarrolou on 01 Jul 2013, 23:45, edited 2 times in total.
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Re: How many values can the integer p=x+3x3 assume?
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15 May 2013, 13:54
Good job arpanpatnaik, vinaymimani! Official explanationThe function \(x+3x3\) 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 CFor for clarity, below there is the graph of \(x+3x3\) that will make my explanation more clear.
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Re: How many values can the integer p=x+3x3 assume?
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15 May 2013, 11:47
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)! I'd go with [C]! For the expression \(p=x+3x3\) , one can think of 3 possible ranges for x to lie in: 1. x > 3 2. x belongs to {3,3} 3. x < 3 It can be calculated for x > 3 : P = 6 Also, for x < 3, P = 6 i.e. for the above ranges the value of x does not put an effect on P. But when it comes to the region x belongs to {3,3} P = 2x. Since P depends on x, and no definite rule has been defined for x, we have to assume x to be values so as to result 2x is a possible integer values. Hence x can be {2,1,0,1,2,1/2,1/2,3/2,3/2,5/2,5/2} For each of the above values the expression P has an integral value and all the above lie between 3 and 3. Hence the total number would be 11+2 = 13. Hope this is the right answer Regards, Arpan



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Re: How many values can the integer p=x+3x3 assume?
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15 May 2013, 12:01
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.
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Re: How many values can the integer p=x+3x3 assume?
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16 May 2013, 14:13
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 Dear Zarrolou, Here's some feedback. I like the question itself quite a bit. My only criticism are the answer choices. Obviously, the OA (C) need to be on the list, and (B) 7 is an excellent distractor (what folks get if they just plug in integer values of x). I don't know about (D) 14  who would pick that? If folks realize that the output is +6 for x > 3 and 6 for x < 3, I could see them saying, "Hmm, how many integers from 6 to 6? That must be 12." Forgetting inclusive counting in something like this  that's a huge predictable error. I think both 6 & 12 would be good to have on the answer list for this reason. I have no idea how anyone could possibly pick (A) 0  anyone who can plug in any single number and get an output will know there are now zero outputs. I see that as a wasted incorrect answer: one that no one will pick. Choice (E) is debatable. I would recommend 3, 6, 7, 12, 13 as the answer choices myself. What do you think? Mike
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Re: How many values can the integer p=x+3x3 assume?
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16 May 2013, 14:21
mikemcgarry wrote: Dear Zarrolou, Here's some feedback. I like the question itself quite a bit. My only criticism are the answer choices. Obviously, the OA (C) need to be on the list, and (B) 7 is an excellent distractor (what folks get if they just plug in integer values of x). I don't know about (D) 14  who would pick that? If folks realize that the output is +6 for x > 3 and 6 for x < 3, I could see them saying, "Hmm, how many integers from 6 to 6? That must be 12." Forgetting inclusive counting in something like this  that's a huge predictable error. I think both 6 & 12 would be good to have on the answer list for this reason. I have no idea how anyone could possibly pick (A) 0  anyone who can plug in any single number and get an output will know there are now zero outputs. I see that as a wasted incorrect answer: one that no one will pick. Choice (E) is debatable. I would recommend 3, 6, 7, 12, 13 as the answer choices myself. What do you think? Mike Many thanks for the feedback first of all, it's always good to hear your opinion. I think that your advise is great, I will update the original question. Just I will not use that order, otherwise folks who got it right and pressed C will have a wrong answer on their workbook (I think...) 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



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Re: How many values can the integer p=x+3x3 assume?
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30 May 2013, 11:45
Can you tell me if my explanation is correct? I have a good deal of difficultly with absolute value!
For problems like this, we get a range of numbers to test by seeing what values with make x=0. For example, 3 makes x+3 =0 and 3 makes x3 = 0. So we test those cases and get two results. Then, we test cases where X is greater and less than 3, which get's us the same result for every integer greater than + x and less than  x. Then, we test integers between 3 and 3 getting us the remainder of the possible answers.
Out of curiosity, why is P = (x+3)(x3) when X>0 and p = (x+3)(x3) when x<0? Is it because a negative value of x will get a negative value for x3 and for all x = (x)?
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Re: How many values can the integer p=x+3x3 assume?
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30 May 2013, 11:55
WholeLottaLove wrote: Can you tell me if my explanation is correct? I have a good deal of difficultly with absolute value!
For problems like this, we get a range of numbers to test by seeing what values with make x=0. For example, 3 makes x+3 =0 and 3 makes x3 = 0. So we test those cases and get two results. Then, we test cases where X is greater and less than 3, which get's us the same result for every integer greater than + x and less than  x. Then, we test integers between 3 and 3 getting us the remainder of the possible answers.
Out of curiosity, why is P = (x+3)(x3) when X>0 and p = (x+3)(x3) when x<0? Is it because a negative value of x will get a negative value for x3 and for all x = (x)?
Thanks!
The explanation above is correct, but I don't think that is doable. I mean that testing each value is not a good way to find the answer, you could miss something; and more important if the answers involved bigger numbers as 150 for example, I dare you to count all the possibilities... "Out of curiosity, why is P = (x+3)(x3) when X>0 and p = (x+3)(x3) when x<0? Is it because a negative value of x will get a negative value for x3 and for all x = (x)?" This is not correct. if x>3 we have P=(x+3)(x3)=6: a straight line if 3<x<3 we have P=(x+3)+(x3)=2x if x<3 we have P=(x3)+(x3)=6: a straight line We have three functions in P for the intervals above described. The quickest way to get an asnwer is to see that P will assume all values form 6 to 6 inclusive. Let me know if you have more doubts



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Re: How many values can the integer p=x+3x3 assume?
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30 May 2013, 12:00
Hello, thanks for the response. My reasoning came from this: findthevalueof135023.html post, which I thought was a similar situation. In it, a1 can be a positive or negative value, thus causing the signs to flip accordingly. I do not understand why if 3<x<3, P=(x+3)+(x3)=2x and I am also not understanding the relation to the graph. Sorry...absolute value problems and concepts have proven to be very difficult to understand.



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Re: How many values can the integer p=x+3x3 assume?
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30 May 2013, 12:13
WholeLottaLove wrote: Hello, thanks for the response. My reasoning came from this: findthevalueof135023.html post, which I thought was a similar situation. In it, a1 can be a positive or negative value, thus causing the signs to flip accordingly. I do not understand why if 3<x<3, P=(x+3)+(x3)=2x and I am also not understanding the relation to the graph. Sorry...absolute value problems and concepts have proven to be very difficult to understand. That question is similar and I am gonna apply the same method here, so: How many values can the integer \(p=x+3x3\) assume? I think that we both agree that if \(x>3\), both \(x+3x3\) will be positive so we can eliminate the  signs and keep \(x+3(x3)=6\). Same reasoning for \(x<3\): if that's the case then both \(x+3x3\) will be negative so we must "flip" the sign according to the rule \(value=+value\). We obtain \((x+3)()(x3)=6\). I hope that this so far makes sense. Now, what if 3<x<3? In this case the term \(x+3\) will be positive (you can try with values 2+3>0 for example). The other term will be negative \(x3\) (23<0 for example). The first one will be positive=> the abs value will not affect the sign; The second one will be negative=> the abs value will turn it positive. Lets write this down: \(p=(x+3)()(x3)=2x\)  if 3<x<3 Now we have for x>3 we have P=(x+3)(x3)=6, for 3<x<3 P=(x+3)+(x3)=2x, for x<3 P=(x3)+(x3)=6 the graph just represents those equations. For example if x>3 the graph shows you a line P=6



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Re: How many values can the integer p=x+3x3 assume?
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30 May 2013, 12:24
But how does 2x translate into an additional 11 answers? If you plug in values for 2x of 2≤x≤2 then I see only 5!



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Re: How many values can the integer p=x+3x3 assume?
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30 May 2013, 12:33
WholeLottaLove wrote: But how does 2x translate into an additional 11 answers? If you plug in values for 2x of 2≤x≤2 then I see only 5! If this is your question, I assume that the abs value problem now is clear. Let me know if I assume correctly First of all it's "values for 2x of 3≤x≤3 not 2≤x≤2 ". for \(3<x<3\) \(2x\) can have all values between 2*(3) and 2*(3) or in the range \(6,6\). How many integers are there between 6 and 6? 5,4,...,0,...4,5 so 11 To those we sum 6 and 6 and the total reaches 11+2=13 You see only 5 values I think because you are not considering cases as \(x=\frac{1}{2}\)(fractions), for this \(p=2x\) will be an integer for example. Same case for \(x=\frac{1}{2}\) and so on



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Re: How many values can the integer p=x+3x3 assume?
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30 May 2013, 12:37
Oooh. Because P must be an integer. So x can be any integer or fraction between 3 and 3 that yields an integer result for x...i.e. 2, 2.....1/2, 1/2? Zarrolou wrote: WholeLottaLove wrote: But how does 2x translate into an additional 11 answers? If you plug in values for 2x of 2≤x≤2 then I see only 5! If this is your question, I assume that the abs value problem now is clear. Let me know if I assume correctly First of all it's "values for 2x of 3≤x≤3 not 2≤x≤2 ". for \(3<x<3\) \(2x\) can have all values between 2*(3) and 2*(3) or in the range \(6,6\). How many integers are there between 6 and 6? 5,4,...,0,...4,5 so 11 To those we sum 6 and 6 and the total reaches 11+2=13 You see only 5 values I think because you are not considering cases as \(x=\frac{1}{2}\)(fractions), for this \(p=2x\) will be an integer for example. Same case for \(x=\frac{1}{2}\) and so on



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Re: How many values can the integer p=x+3x3 assume?
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30 May 2013, 12:42
WholeLottaLove wrote: Oooh. Because P must be an integer. So x can be any integer or fraction between 3 and 3 that yields an integer result for x...i.e. 2, 2.....1/2, 1/2?
Yes, almost correct! Remeber that we are considering the range 3<x<3 so the list of values is 5/2 (2.5), 2,3/2,1,1/2,0,1/2,1,3/2,2,5/2 => 11 values I said "almost correct" because in "x...i.e. 2, 2.....1/2, 1/2" you missed 2.5 at the beginning, but everything else is fine



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Re: How many values can the integer p=x+3x3 assume?
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17 Jun 2013, 06:54
How many values can the integer p=x+3x3 assume?
A)6 B)7 C)13 D)12 E)Cannot be determined
Hi all.
I thought I understood this problem but upon further review, I am still a bit uncertain.
You say that the range is between 6 and 6, and for any x value that holds true. For example, if x=20 then p=x+3x3 ==> p=20+3203 = 2317 = 6.
One explanation that I liked (because it was simple haha) was that P could equal any integer between 6 and 6 inclusive. That seems too simple however so I want to see if I can understand the problem further.
P has to = to an integer, so that limits values (I suppose it would be nearly impossible to solve in a few minutes if P could = any number) but how can we figure out exactly which integers it can equal. For example, isn't it possible that a problem similar to this might not be valid for every number in the range? (i.e. between 6 and 6)
Thanks!



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Re: How many values can the integer p=x+3x3 assume?
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17 Jun 2013, 07:14
WholeLottaLove wrote: How many values can the integer p=x+3x3 assume?
A)6 B)7 C)13 D)12 E)Cannot be determined
Hi all.
I thought I understood this problem but upon further review, I am still a bit uncertain.
You say that the range is between 6 and 6, and for any x value that holds true. For example, if x=20 then p=x+3x3 ==> p=20+3203 = 2317 = 6.
One explanation that I liked (because it was simple haha) was that P could equal any integer between 6 and 6 inclusive. That seems too simple however so I want to see if I can understand the problem further.
P has to = to an integer, so that limits values (I suppose it would be nearly impossible to solve in a few minutes if P could = any number) but how can we figure out exactly which integers it can equal. For example, isn't it possible that a problem similar to this might not be valid for every number in the range? (i.e. between 6 and 6)
Thanks! Good question! In this question every integer between 6 and 6 is a valid result. How can we be sure of that? Because in the range 6,6 the function equals \(2x\) [(x+3)+(x3)=2x], we can be sure that it will assume every integer value between 6 and 6. I can say so because the funcion is defined in every point ( it's a line) so does not have any invalid point, you can draw it and control yourself: it is defined in every point. To give you an example of a function that is NOT defined in every point I could use \(p=\frac{3}{3+x}\). This is beyond the question itself, but that function is NOT defined for \(x=3\) : \(p=\frac{3}{0}\) is not a denfined value. But since 2x is a line, it will assume every value between 6 and 6. Hope it's clear



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Re: How many values can the integer p=x+3x3 assume?
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17 Jun 2013, 07:27
So,
This equation will always yield a value between 6 and 6 for any value of x. Any value of x (positive or negative) will result in (x+3)  (x3)
But then how does that = 2x?Scratch that. For any x value greater than 3, P=6 For any x value less than 3, P=6 Those are the boundaries. For x=3,3 =2x, but how are you left with 2x when you plug in values for both x's? Zarrolou wrote: WholeLottaLove wrote: How many values can the integer p=x+3x3 assume?
A)6 B)7 C)13 D)12 E)Cannot be determined
Hi all.
I thought I understood this problem but upon further review, I am still a bit uncertain.
You say that the range is between 6 and 6, and for any x value that holds true. For example, if x=20 then p=x+3x3 ==> p=20+3203 = 2317 = 6.
One explanation that I liked (because it was simple haha) was that P could equal any integer between 6 and 6 inclusive. That seems too simple however so I want to see if I can understand the problem further.
P has to = to an integer, so that limits values (I suppose it would be nearly impossible to solve in a few minutes if P could = any number) but how can we figure out exactly which integers it can equal. For example, isn't it possible that a problem similar to this might not be valid for every number in the range? (i.e. between 6 and 6)
Thanks! Good question! In this question every integer between 6 and 6 is a valid result. How can we be sure of that? Because in the range 6,6 the function equals \(2x\) [(x+3)+(x3)=2x], we can be sure that it will assume every integer value between 6 and 6. I can say so because the funcion is defined in every point ( it's a line) so does not have any invalid point, you can draw it and control yourself: it is defined in every point. To give you an example of a function that is NOT defined in every point I could use \(p=\frac{3}{3+x}\). This is beyond the question itself, but that function is NOT defined for \(x=3\) : \(p=\frac{3}{0}\) is not a denfined value. But since 2x is a line, it will assume every value between 6 and 6. Hope it's clear



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Re: How many values can the integer p=x+3x3 assume?
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17 Jun 2013, 07:39
WholeLottaLove wrote: So,
This equation will always yield a value between 6 and 6 for any value of x. Any value of x (positive or negative) will result in (x+3)  (x3)
But then how does that = 2x?
The original function is \(p= x+3x3\). For values of x>3 it will equal 6 \(p=x+3x+3=6\) For values of x<3 it will equal 6 \(p=x3+x3=6\) For values \(6\leq{x}\leq{6}\) it equals \(p = (x+3)+(x3)=2x\), any value of x between \(6\leq{x}\leq{6}\) will result in p=2x. Some posts back we enstablished that "Yes, almost correct! Remeber that we are considering the range 3<x<3 so the list of values is 5/2 (2.5), 2,3/2,1,1/2,0,1/2,1,3/2,2,5/2 => 11 values
I said "almost correct" because in "x...i.e. 2, 2.....1/2, 1/2" you missed 2.5 at the beginning, but everything else is fine "what I am trying to say is that p=2x is a line, so I am asking you: "How many integer values does the function p=2x assume in the interval 6,6?" Answer: if X=5/2 P=5, if X=2 P= 4,... Plus 6 and 6 that are the "edge" values: total 13.



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Re: How many values can the integer p=x+3x3 assume?
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17 Jun 2013, 07:55
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=6When x=4 p=(x+3) (x3) p=x3  x+3 p=x3 + x3 p=6For any value in between: x=1 P=(x+3) (x3) P=x+3  x+3 P=x+3 +x3 P=2xSo, 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? Zarrolou wrote: WholeLottaLove wrote: So,
This equation will always yield a value between 6 and 6 for any value of x. Any value of x (positive or negative) will result in (x+3)  (x3)
But then how does that = 2x?
The original function is \(p= x+3x3\). For values of x>3 it will equal 6 \(p=x+3x+3=6\) For values of x<3 it will equal 6 \(p=x3+x3=6\) For values \(6\leq{x}\leq{6}\) it equals \(p = (x+3)+(x3)=2x\), any value of x between \(6\leq{x}\leq{6}\) will result in p=2x. Some posts back we enstablished that "Yes, almost correct! Remeber that we are considering the range 3<x<3 so the list of values is 5/2 (2.5), 2,3/2,1,1/2,0,1/2,1,3/2,2,5/2 => 11 values
I said "almost correct" because in "x...i.e. 2, 2.....1/2, 1/2" you missed 2.5 at the beginning, but everything else is fine "what I am trying to say is that p=2x is a line, so I am asking you: "How many integer values does the function p=2x assume in the interval 6,6?" Answer: if X=5/2 P=5, if X=2 P= 4,... Plus 6 and 6 that are the "edge" values: total 13.



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Re: How many values can the integer p=x+3x3 assume?
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17 Jun 2013, 08:07
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




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