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If \(x\) and \(y\) are positive integers, is \(x\) a prime number? (1) \(x  2 \lt 2  y\) (2) \(x + y  3 = 1y\)
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16 Sep 2014, 00:18
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, 12:44
[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, 05:33
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, 06:30
positive integer means we should not consider zero??



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28 Nov 2014, 08:25
sudd1 wrote: positive integer means we should not consider zero?? Positive integers are integers which are to the right of 0 on the number line: 1, 2, 3, 4, ...
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20 Apr 2015, 23:00
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, 05:25
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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20 Nov 2015, 19:38
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, 02:46
sajib2126 wrote: 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) ?? Let me ask you if y is a positive integer what are the possible values of 1  y ?
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23 Dec 2015, 09:43
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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12 Dec 2016, 20:04
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, 01:55
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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27 Jun 2017, 20:50
Hi Bunuel, So from statement 1 what I understand is that x−2<2−y will only happen if x being positive will be equal to 2. Since it is absolute value and will be positive so for RHS Y cannot be >2 since then RHS<LHS (ve value). so y has to be only 1 (positive integer) and hence x has to be 2 as no other value will make it less than RHS (1). Also alternatively, now since x−2 < 1 (value with y=1), 1<x2<1 and hence 1<x<3 so x=2 which is prime. Kindly confirm if the above approach for statement 1 is correct.
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27 Jun 2017, 23:08
kreel11 wrote: Hi Bunuel,
So from statement 1 what I understand is that x−2<2−y will only happen if x being positive will be equal to 2. Since it is absolute value and will be positive so for RHS Y cannot be >2 since then RHS<LHS (ve value). so y has to be only 1 (positive integer) and hence x has to be 2 as no other value will make it less than RHS (1).
Also alternatively, now since x−2 < 1 (value with y=1), 1<x2<1 and hence 1<x<3 so x=2 which is prime.
Kindly confirm if the above approach for statement 1 is correct. ______________ Yes, that's correct.
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25 Dec 2017, 09:37
Hi BunuelPlease let me know if my approach makes sense Given: x, y > +ve x > prime? (1) x2 < 2y Opening up the Absolute Value: when +vex2 < 2y x+y < 4 OR when vex+2 < 2y x<y x>y Test Cases: x=2 y=1 for x+y < 4 or = 3 [Given x, y > 0 & x>y, so total has to be 3] and so x is prime Sufficient(2) x+y3 = 1y when +vex+y3 = 1y x+2y = 4 Testing Cases: x = 2, y = 1: Only option that works OR when vex+y3 = 1+y x=2 > prime SufficientAnswer: DThanks Prathamesh



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25 Dec 2017, 11:26
ppnimkar wrote: Hi BunuelPlease let me know if my approach makes sense Given: x, y > +ve x > prime? (1) x2 < 2y Opening up the Absolute Value: when +vex2 < 2y x+y < 4 OR when vex+2 < 2y x<y x>y Test Cases: x=2 y=1 for x+y < 4 or = 3 [Given x, y > 0 & x>y, so total has to be 3] and so x is prime Sufficient(2) x+y3 = 1y when +vex+y3 = 1y x+2y = 4 Testing Cases: x = 2, y = 1: Only option that works OR when vex+y3 = 1+y x=2 > prime SufficientAnswer: DThanks Prathamesh I guess when you test values for (1), you consider x + y < 4 and x > y as simultaneous equations, and that is why you test only x = 2 and y = 1, whereas x + y < 4 and x > y are two different cases (true in certain conditions).
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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 \(2y \le{0}\) 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 HI BunuelFor S1:Since x & y > 0 then the expression x  2< 2  y can be simplified as follows: x 2 < 2  y (if x>0 , then x = x) x < 4  y Now, for y to be a positive integer y should be < 4 (i.e. y can be 1, 2 and 3)Hence, if y = 1, x < 4  y x < 4 1 x < 3 so x can be 1 or 2 ( 1 is NOT A PRIME & 2 IS A PRIME ) if y = 2x < 4  2 x < 2 so x= 1 (1 is NOT A PRIME) if y = 3x < 4  3 x < 1 (this is not possible since x is a positive integer) Can you please let me know where am i going wrong?? THANKS!



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01 Apr 2019, 05:41
JIAA 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 \(2y \le{0}\) 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 HI BunuelFor S1:Since x & y > 0 then the expression x  2< 2  y can be simplified as follows: x 2 < 2  y (if x>0 , then x = x) x < 4  y Now, for y to be a positive integer y should be < 4 (i.e. y can be 1, 2 and 3)Hence, if y = 1, x < 4  y x < 4 1 x < 3 so x can be 1 or 2 ( 1 is NOT A PRIME & 2 IS A PRIME ) if y = 2x < 4  2 x < 2 so x= 1 (1 is NOT A PRIME) if y = 3x < 4  3 x < 1 (this is not possible since x is a positive integer) Can you please let me know where am i going wrong?? THANKS! As you correctly wrote x = x, when x >= 0 (and x = x, when x < 0). So, x  2 = x  2, when x  2 >= 0, or which is the same, when x >= 2. In this case, so when x >= 2, x  2 < 2  y. If y = 1, then x < 3 > since we are looking for x >= 2, then x = 2.If y = 2, then x < 2 > since we are looking for x >= 2, then here we have no solution. If y = 3, then x < 1 > since we are looking for x >= 2, then here we have no solution. Next, the second case would be when x  2 = (x  2) (x = x, when x < 0). This would happen when x  2 < 0, or which is the same, when x < 2. In this case, so when x < 2, (x  2) < 2  y. If y = 1, then x > 1 > since we are looking for x < 2, then here we have no solution. If y = 2, then x > 2 > since we are looking for x > 2, then here we have no solution. If y = 3, then x > 3 > since we are looking for x > 2, then here we have no solution. As you can see we have only one solution: x = 2 = prime. Hope it's clear.
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01 Apr 2019, 06:59
Bunuel wrote: JIAA 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 \(2y \le{0}\) 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 HI BunuelFor S1:Since x & y > 0 then the expression x  2< 2  y can be simplified as follows: x 2 < 2  y (if x>0 , then x = x) x < 4  y Now, for y to be a positive integer y should be < 4 (i.e. y can be 1, 2 and 3)Hence, if y = 1, x < 4  y x < 4 1 x < 3 so x can be 1 or 2 ( 1 is NOT A PRIME & 2 IS A PRIME ) if y = 2x < 4  2 x < 2 so x= 1 (1 is NOT A PRIME) if y = 3x < 4  3 x < 1 (this is not possible since x is a positive integer) Can you please let me know where am i going wrong?? THANKS! As you correctly wrote x = x, when x >= 0 (and x = x, when x < 0). So, x  2 = x  2, when x  2 >= 0, or which is the same, when x >= 2. In this case, so when x >= 2, x  2 < 2  y. If y = 1, then x < 3 > since we are looking for x >= 2, then x = 2.If y = 2, then x < 2 > since we are looking for x >= 2, then here we have no solution. If y = 3, then x < 1 > since we are looking for x >= 2, then here we have no solution. Next, the second case would be when x  2 = (x  2) (x = x, when x < 0). This would happen when x  2 < 0, or which is the same, when x < 2. In this case, so when x < 2, (x  2) < 2  y. If y = 1, then x > 1 > since we are looking for x < 2, then here we have no solution. If y = 2, then x > 2 > since we are looking for x > 2, then here we have no solution. If y = 3, then x > 3 > since we are looking for x > 2, then here we have no solution. As you can see we have only one solution: x = 2 = prime. Hope it's clear. THANKS BunuelJust one more thing ! It's explicitly stated in Q stem that x is positive i.e. x> 0, shouldn't we consider second case ONLY when we don't know if x is positive or negative?? Why are we even considering second case here (i.e.x = x, when x < 0)?? Would really appreciate your feedback!







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