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How many different values of positive integer x, for which [#permalink]
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07 Aug 2014, 15:14
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How many different values of positive integer x, for which x+8<x, are there? A. 0 B. 2 C. 3 D. 8 E. 16
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Re: How many different values of positive integer x, for which [#permalink]
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07 Aug 2014, 20:02
Since it states that \(x\) is a positive integer, then \(x+8\) will always be positive.
The question stem can be simplified to: \(x+8<x\) which is simply not possible, no solutions are possible.
Answer is A.




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Re: How many different values of positive integer x, for which [#permalink]
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23 Aug 2015, 05:18
Here the only transition point is 8 and the value of x belongs to positive integer greater than 8 .Since x+8<x is the ultimate inequalities.So there is no answer available if I pick any number of x. So the Correct answer is A
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Re: How many different values of positive integer x, for which [#permalink]
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15 Oct 2015, 21:23
why cannot we split this inequality into 2 parts?
x+8<x and (x+8)<x
then the second one solves to x<4
or do we dismiss the 've situation because it says only positive values of x?



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Re: How many different values of positive integer x, for which [#permalink]
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23 Oct 2015, 04:40
GMATDemiGod wrote: why cannot we split this inequality into 2 parts?
x+8<x and (x+8)<x
then the second one solves to x<4
or do we dismiss the 've situation because it says only positive values of x? You can do it this way but you are interpreting the cases incorrectly. You are asked how many integer values satisfy x+8)<x . Now you have 2 cases: Case 1 : x \(\geq\)8 > x+8 = x+8 > x+8<x > 8<0 Not possible. Thus x \(\geq\)8 is not an acceptable case. Case 2 : x \(<\)8 > x+8 =( x+8) > x8<x > 2x>8 > x>4 but as x<8 , x>4 is not an acceptable case. Hence, you get 0 integer values satisfying the given equality. Hope this helps.



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Re: How many different values of positive integer x, for which [#permalink]
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23 Oct 2015, 11:02
Engr2012 wrote: GMATDemiGod wrote: why cannot we split this inequality into 2 parts?
x+8<x and (x+8)<x
then the second one solves to x<4
or do we dismiss the 've situation because it says only positive values of x? You can do it this way but you are interpreting the cases incorrectly. You are asked how many integer values satisfy x+8)<x . Now you have 2 cases: Case 1 : x \(\geq\)8 > x+8 = x+8 > x+8<x > 8<0 Not possible. Thus x \(\geq\)8 is not an acceptable case. Case 2 : x \(<\)8 > x+8 =( x+8) > x8<x > 2x>8 > x>4 but as x<8 , x>4 is not an acceptable case. Hence, you get 0 integer values satisfying the given equality. Hope this helps. Is this because in the x>4 case, if we put 2 in the main it doesn't hold? 2+8<(2) 6 < 2 clearly false



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Re: How many different values of positive integer x, for which [#permalink]
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23 Oct 2015, 11:12
GMATDemiGod wrote: Engr2012 wrote: GMATDemiGod wrote: why cannot we split this inequality into 2 parts?
x+8<x and (x+8)<x
then the second one solves to x<4
or do we dismiss the 've situation because it says only positive values of x? You can do it this way but you are interpreting the cases incorrectly. You are asked how many integer values satisfy x+8)<x . Now you have 2 cases: Case 1 : x \(\geq\)8 > x+8 = x+8 > x+8<x > 8<0 Not possible. Thus x \(\geq\)8 is not an acceptable case. Case 2 : x \(<\)8 > x+8 =( x+8) > x8<x > 2x>8 > x>4 but as x<8 , x>4 is not an acceptable case. Hence, you get 0 integer values satisfying the given equality. Hope this helps. Is this because in the x>4 case, if we put 2 in the main it doesn't hold? 2+8<(2) 6 < 2 clearly false Yes, that is one way of looking at it.



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Re: How many different values of positive integer x, for which [#permalink]
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19 Mar 2016, 21:14
goodyear2013 wrote: How many different values of positive integer x, for which x+8<x, are there?
A. 0 B. 2 C. 3 D. 8 E. 16 Answer A I opted to put the random value option. i used 0 , 8 , 8 and the the extreme of 20 and 20.. i was able to solve it in 1:09



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Re: How many different values of positive integer x, for which [#permalink]
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17 May 2018, 03:39
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