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15 Sep 2014, 23:18
Official Solution:If \(x\) and \(y\) are positive integers, is \(x\) a prime number? (1) \(x  2 \lt 2  y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2  y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(2y \le{0}\) and it cannot be greater than \(x  2\)). Next, since given that \(y\) is a positive integer, then \(y=1\). So, we have that: \(x  2 \lt 1\), which implies that \(1 \lt x2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient. (2) \(x + y  3 = 1y\). Since \(y\) is a positive integer, then \(1y \le {0}\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient. Answer: D
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27 Nov 2014, 11:44
[quote="Bunuel"]Official Solution:
(if \(y\) is more than or equal to 2, then \(y2 \le{0}\) Since \(y\) is a positive integer, then \(1y \le {0}\),
I have not understood the above two things. Could you please help me on this? How can we get these equation?



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28 Nov 2014, 04:33
Raihanuddin wrote: Bunuel wrote: Official Solution:
(if \(y\) is more than or equal to 2, then \(y2 \le{0}\) Since \(y\) is a positive integer, then \(1y \le {0}\),
I have not understood the above two things. Could you please help me on this? How can we get these equation? I tried to explain this in my solution. 1. (1) \(x  2 \lt 2  y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2  y\), thus \(y \lt 2\). 2. For (2): we know that y is a positive integer 1, 2, 3, ... so 1  y must be less than or equal to 0.
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Re: M0225
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28 Nov 2014, 05:30
positive integer means we should not consider zero??



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28 Nov 2014, 05:45
sudd1 wrote: positive integer means we should not consider zero?? Yes, positive integers can't be zero. It will start from 1



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20 Apr 2015, 22:00
Bunuel, from what I understand about absolute values, the number/expression within the absolute number sign could either be positive or negative (just like the way the square root of a number could either be positive or negative), so I didn't understand how the least value in the LHS is zero. Please explain? Thanks



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21 Apr 2015, 04:25
Kigunda wrote: Bunuel, from what I understand about absolute values, the number/expression within the absolute number sign could either be positive or negative (just like the way the square root of a number could either be positive or negative), so I didn't understand how the least value in the LHS is zero. Please explain? Thanks I think this is explained here: m02183573.html#p1448632The point is that some expression >= 0. So, when we have some expression = x, it means that x must be more than or equal to 0 too. Also, the square root of a number cannot be negative \(\sqrt{4}=2\), not +/2. When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is: \(\sqrt{9} = 3\), NOT +3 or 3; \(\sqrt[4]{16} = 2\), NOT +2 or 2; Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and 3. Because \(x^2 = 9\) means that \(x =\sqrt{9}=3\) or \(x=\sqrt{9}=3\).
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16 Oct 2015, 07:18
I think this is a highquality question and I agree with explanation. This question is the same as M2858



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20 Nov 2015, 18:38
Can someone help me understand this part in detail
2)x+y−3=1−y. Since y is a positive integer, then 1−y≤0, thus 1−y=−(1−y).
why we are not considering the 1−y=(1−y) ??



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21 Nov 2015, 01:46



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23 Dec 2015, 08:43
Bunuel wrote: If \(x\) and \(y\) are positive integers, is \(x\) a prime number?
(1) \(x  2 \lt 2  y\)
(2) \(x + y  3 = 1y\) I've solved this problem in a little different way. 1) x  2 < 2  y It means (2  y) < x  2 < 2  y => y  2 < x  2 < 2  y => y < x < 4  y Try pick up numbers. It's said that x and y are positive integers. Let's say y is 1, then x could be 2 or 3. y can't be 2, 3 or any other integer, because it violates the inequality. In both cases, whether x is 2 or 3, x is a prime number. Therefore, first statement is sufficient. 2) x + y  3 = 1  y It means x + y  3 = 1  y or x + y  3 = (1  y), x + y  3 = y  1 x = 4  2y or x = 2 . 2 is a prime number. Therefore, second statement is sufficient too. Thus, correct answer is D: both statements are sufficient.



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17 Jul 2016, 09:37
I think this the explanation isn't clear enough, please elaborate. for option 1 my logic is since x is a positive integer, lx2l>0, therefore x2<2y x+y<4 so, X can be 1 or 2, cannot determine can someone correct me, if my logic is wrong?



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17 Jul 2016, 09:40
r19 wrote: I think this the explanation isn't clear enough, please elaborate. for option 1 my logic is since x is a positive integer, lx2l>0, therefore x2<2y x+y<4 so, X can be 1 or 2, cannot determine can someone correct me, if my logic is wrong? (1) \(x  2 \lt 2  y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2  y\), thus \(y \lt 2\).
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04 Sep 2016, 03:28
then 1−y≤01−y≤0, thus 1−y=−(1−y)1−y=−(1−y)
How is that 1−y=−(1−y)1−y=−(1−y) always?
How about the case when y =1 , shouldn't it be (1y) only then??



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04 Sep 2016, 05:12
Hi Bunnel, got it.. so is it that for <=0 inequalities, we can always take the negative?
Thanks.



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12 Dec 2016, 19:04
Bunuel wrote: Official Solution:
If x and y are positive integers, is x a prime number?
(1)\(x  2 \lt 2  y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2  y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \le{0}\) and it cannot be greater than \(x  2\)). Next, since given that \(y\) is a positive integer, then \(y=1\). So, we have that: \(x  2 \lt 1\), which implies that \(1 \lt x2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient. (2)\(x + y  3 = 1y\). Since \(y\) is a positive integer, then \(1y \le {0}\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D Thanks for the solution. But, i can`t understand one point. In statement 1, you mentioned if y is more than or equal to 2, then y−2≤0y−2≤0 and it cannot be greater than x−2x−2. Could you please explain this point in more detail? Thanks in advance.



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13 Dec 2016, 00:55
jahidhassan wrote: Bunuel wrote: Official Solution:
If x and y are positive integers, is x a prime number?
(1)\(x  2 \lt 2  y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2  y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y2 \le{0}\) and it cannot be greater than \(x  2\)). Next, since given that \(y\) is a positive integer, then \(y=1\). So, we have that: \(x  2 \lt 1\), which implies that \(1 \lt x2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient. (2)\(x + y  3 = 1y\). Since \(y\) is a positive integer, then \(1y \le {0}\), thus \(1y=(1y)\). So, we have that \(x + y  3 = (1y)\), which gives \(x=2=prime\). Sufficient.
Answer: D Thanks for the solution. But, i can`t understand one point. In statement 1, you mentioned if y is more than or equal to 2, then y−2≤0y−2≤0 and it cannot be greater than x−2x−2. Could you please explain this point in more detail? Thanks in advance. If y is greater than 2 then the right hand side of the inequality (2y) becomes negative. The left hand side if the inequality is an absolute value (x  2) which cannot be negative, thus the inequality (\(x  2 \lt 2  y\)) will not hold true. Please reread the discussion above.
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