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Re M3021 [#permalink]
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13 Oct 2014, 06:57
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Official Solution:If \(y=x+5x5\), then \(y\) can take how many integer values? A. 5 B. 10 C. 11 D. 20 E. 21 When \(x\leq{5}\), then \(x+5=(x+5)=x5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x5(5x)=10\). 1 integer value of \(y\) for this range. When \(5 < x < 5\), then \(x+5=x+5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x+5(5x)=2x\). Therefore for this range \(10 < (y=2x) < 10\). 19 integer values of \(y\) for this range (from 9 to 9, inclusive). When \(x\geq{5}\), then \(x+5=x+5\) and \(x5=x5\). Hence in this case \(y=x+5x5=x+5(x5)=10\). 1 integer value of \(y\) for this range. Total = 1 + 19 + 1 = 21. If anyone interested here is a graph of \(y=x+5−x−5:\) As you can see y is a continuous function from 10 to 10, inclusive. Answer: E
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Re: M3021 [#permalink]
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02 Dec 2014, 21:34
when 5<x< 5
Why do you multiply I X5 I by 1 ? ? I am trying to understand the logic behind this calculation When 5<x<5, then x+5=x+5 and x5=(x5)=5x.



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Re: M3021 [#permalink]
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03 Dec 2014, 04:16
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rsamant wrote: when 5<x< 5
Why do you multiply I X5 I by 1 ? ? I am trying to understand the logic behind this calculation When 5<x<5, then x+5=x+5 and x5=(x5)=5x. Absolute value properties: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\). Therefore, when 5 < x < 5, then x + 5 > 0 and x  5 < 0, thus x + 5 = x + 5 and x  5 = (x  5) = 5  x.Hope it helps.
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Re: M3021 [#permalink]
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17 Dec 2014, 22:00
Can you please explain that in condition 5<x<5 , how we calulated that which side should have +ve and which side should have ve.
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Re: M3021 [#permalink]
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20 Nov 2015, 08:52
Hello Bunuel,
Could you please explain the solution in some other way??
Thanks



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21 Nov 2015, 02:47



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Re: M3021 [#permalink]
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02 Oct 2016, 14:14
Bunuel, I fell for the trap and assumed x was an integer as well.
If you can add, "x does not have to be an integer" in the explanation, I think it will help.
Thanks for a great question!



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Re: M3021 [#permalink]
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23 Oct 2016, 06:27
Bunuel wrote: Official Solution:If \(y=x+5x5\), then \(y\) can take how many integer values? A. 5 B. 10 C. 11 D. 20 E. 21 When \(x\leq{5}\), then \(x+5=(x+5)=x5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x5(5x)=10\). 1 integer value of \(y\) for this range. When \(5 < x < 5\), then \(x+5=x+5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x+5(5x)=2x\). Therefore for this range \(10 < (y=2x) < 10\). 19 integer values of \(y\) for this range (from 9 to 9, inclusive). When \(x\geq{5}\), then \(x+5=x+5\) and \(x5=x5\). Hence in this case \(y=x+5x5=x+5(x5)=10\). 1 integer value of \(y\) for this range. Total = 1 + 19 + 1 = 21. If anyone interested here is a graph of \(y=x+5−x−5:\) As you can see y is a continuous function from 10 to 10, inclusive. Answer: E Dear Bunuel, Are the ranges correctly defined , shouldn't the ranges be For critical points [5,5] the solution range would be 1) x< 5 2) 5 <= x < 5 3) x>=5 Thanks



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Re: M3021 [#permalink]
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23 Oct 2016, 06:41
rt1601 wrote: Bunuel wrote: Official Solution:If \(y=x+5x5\), then \(y\) can take how many integer values? A. 5 B. 10 C. 11 D. 20 E. 21 When \(x\leq{5}\), then \(x+5=(x+5)=x5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x5(5x)=10\). 1 integer value of \(y\) for this range. When \(5 < x < 5\), then \(x+5=x+5\) and \(x5=(x5)=5x\). Hence in this case \(y=x+5x5=x+5(5x)=2x\). Therefore for this range \(10 < (y=2x) < 10\). 19 integer values of \(y\) for this range (from 9 to 9, inclusive). When \(x\geq{5}\), then \(x+5=x+5\) and \(x5=x5\). Hence in this case \(y=x+5x5=x+5(x5)=10\). 1 integer value of \(y\) for this range. Total = 1 + 19 + 1 = 21. If anyone interested here is a graph of \(y=x+5−x−5:\) As you can see y is a continuous function from 10 to 10, inclusive. Answer: E Dear Bunuel, Are the ranges correctly defined , shouldn't the ranges be For critical points [5,5] the solution range would be 1) x< 5 2) 5 <= x < 5 3) x>=5 Thanks You have the same ranges  it does not matter in which range you include = sign.
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Re: M3021 [#permalink]
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13 Jan 2017, 08:50
I always get a bit lost with regards to >= versus >.
Could someone please clarify the following: if I have X5 which of the following is correct:
X<=5 > X + 5
X<5 > X + 5



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Re: M3021 [#permalink]
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19 Jul 2017, 21:42
hi brunel when x>5, y = 10 when x<5, y = 10 when 5<x<5, y= 2x...These 3 steps are clear to me.But my concern is as per your solution you have taken the following range:
Therefore for this range −10<(y=2x)<10. still not clear with this part as why we are taking value from 10 to 10.What is the logic behind this



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19 Jul 2017, 21:52



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Re M3021 [#permalink]
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26 Aug 2017, 07:55
I think this is a poorquality question and I don't agree with the explanation. possible mistake? in the range of 5<x<5, according to the explanation, y can get 19 values. however, if we assign x= 4,3,2,1,0,1,2,3,4 it results return exactly 9 options. respectively y=2x as mentioned.



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Re: M3021 [#permalink]
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26 Aug 2017, 07:59
gleshem wrote: I think this is a poorquality question and I don't agree with the explanation. possible mistake? in the range of 5<x<5, according to the explanation, y can get 19 values. however, if we assign x= 4,3,2,1,0,1,2,3,4 it results return exactly 9 options. respectively y=2x as mentioned. You should read the question more carefully. We are not told that x is an integer so there are more integer values of y possible when x is not integer.
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This is a min/max and count of set problem in disguise.
For convenience, let x+5 = a and x5 = b, so y = ab, where a≥0 and b≥0 (modulus' result can't be less than 0)
y is minimized when a=0, and maximized when b=0, so set a and b to these numbers: a = x+5 = 0 > x1=5 b = x5 = 0 > x2=5
Sub x1 & x2 into the original question for the min/max of y Min(y) = 5+5  55 = 10 Max(y) = 5+5  55 = 10
Thus the subset of y containing only integers has 10  (10) + 1 = 21 members in it



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Re M3021 [#permalink]
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18 Nov 2017, 01:29
I think this is a highquality question and I agree with explanation. How do we get to know about the range which we have to check?
Like x<5 5<x<5 and x>5
Please explain @bunnel



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18 Nov 2017, 01:54



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Re M3021 [#permalink]
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04 Feb 2018, 20:17
I think this is a poorquality question and the explanation isn't clear enough, please elaborate. How exactly did y=2x equation have its values between 10 and 10? There is nothing mentioned.



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