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If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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14 Jun 2017, 05:59
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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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14 Jun 2017, 06:49
Bunuel wrote: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?
(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on xaxis So the question basically asks if Y=0 (x^2  y^2)^(1/2) = x + y Squaring both sides (x+y)(xy) = (x+y)^2 xy = x+y y=0 (1) xy is not a square of an integer Since \(x \leq y \leq x\) and xy is not a square means x is not equal to y But x+y > 0 (\(\sqrt{x^2  y^2}\) is +ve) => x^2>y^2 If y = 0 then xy=0 which is a square of an integer hence y is not zero and x cannot be 0 because y will be negative then as x>y and then x+y becomes negative but we know that x+y>0 Hence Sufficient (2) Point (x,y) is not on x axis shows that y can be anything but 0. Sufficient Hence both can solve Answer D Please correct me if i am wrong
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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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14 Jun 2017, 11:16
Bunuel wrote: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?
(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on xaxis As per question stem  \(\sqrt{x^2  y^2} = x + y\). squaring it we get \(x^2  y^2= x^2 + y^2 + 2xy\) solving this we get \(y^2+xy=0\) OR y(y+x) = 0 So the question asks us IS y=0 or y=x Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to x. For e.g x=2, y=2, then xy = 4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to x e.g (2,3) or (2,2). Hence insufficient Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient. Option E



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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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14 Jun 2017, 11:52



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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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15 Jun 2017, 07:42
niks18 wrote: Bunuel wrote: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?
(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on xaxis As per question stem  \(\sqrt{x^2  y^2} = x + y\). squaring it we get \(x^2  y^2= x^2 + y^2 + 2xy\) solving this we get \(y^2+xy=0\) OR y(y+x) = 0 So the question asks us IS y=0 or y=x Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to x. For e.g x=2, y=2, then xy = 4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to x e.g (2,3) or (2,2). Hence insufficient Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient. Option EIt is mentioned in the stem that x<y<x. So y cannot be equal to x. Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app



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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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15 Jun 2017, 07:51
Mayur21 wrote: niks18 wrote: Bunuel wrote: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?
(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on xaxis As per question stem  \(\sqrt{x^2  y^2} = x + y\). squaring it we get \(x^2  y^2= x^2 + y^2 + 2xy\) solving this we get \(y^2+xy=0\) OR y(y+x) = 0 So the question asks us IS y=0 or y=x Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to x. For e.g x=2, y=2, then xy = 4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to x e.g (2,3) or (2,2). Hence insufficient Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient. Option EIt is mentioned in the stem that x<y<x. So y cannot be equal to x. Sent from my ONEPLUS A3003 using GMAT Club Forum mobile appIt's less than or equal sign there: \(x \leq y \leq x\), so y can equal to x, as well as x. In fact x=y=x=0 is also possible.
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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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15 Jun 2017, 18:53
Bunuel wrote: There are flaws in both solutions above, even in the one with correct answer. P.S. Above question and similar question HERE are brand new questions by me and they seem to be quite tricky. Hi Bunuel, Those questions are awesome. You tagged both question as GMAT Club tests. Have you add them in the tests? i.e should I face them if I take mock exam in GMAT Club tests?



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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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15 Jun 2017, 20:54
Bunuel wrote: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?
(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on xaxis \(x \leq y \leq x\) means x is positive or 0.. \(\sqrt{x^2  y^2} = x + y\) .. Square both sides.. \(x^2y^2=(x+y)^2=x^2+y^2+2xy.........2y^2+2xy=0......2y(y+x) =0\) So we have to find whether y=0 or x=y... I tells us that x is not equal to y.. But can be 0 or x=y Insuff II tells us y is not equal to 0.. Insuff.. Combined X can be = y or not Insufficient E
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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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15 Jun 2017, 22:09
Bunuel wrote: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?
(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on xaxis \(\sqrt{x^2  y^2} = x + y\)  > this result will come about when \(x^2  y^2\) is either 0 or 1. Let's jump on to the statement analysis. 1) xy is not a square of integer. This means x and y are either different integers or same integers with different signs or they both are 0. Taking all the 3 scenarios, When x=2 and y=2, \(\sqrt{x^2  y^2} = x + y\) is true. When x=3 and y=2, \(\sqrt{x^2  y^2} = x + y\) is not true. When x=0 and y=0, \(\sqrt{x^2  y^2} = x + y\) is true.
Hence insufficient.
2) Point (x, y) is NOT on xaxis This implies y is not 0, but does not tell anything about x. Hence this statement in itself is insufficient.
Combining Stmt 1 and Stmt 2: Below two scenarios still hold : When x=2 and y=2, \(\sqrt{x^2  y^2} = x + y\) is true. When x=3 and y=2, \(\sqrt{x^2  y^2} = x + y\) is not true. Hence E.
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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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16 Jun 2017, 02:46
chetan2u wrote: Bunuel wrote: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?
(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on xaxis \(x \leq y \leq x\) means x is positive or 0.. \(\sqrt{x^2  y^2} = x + y\) .. Square both sides.. \(x^2y^2=(x+y)^2=x^2+y^2+2xy.........2y^2+2xy=0......2y(y+x) =0\) So we have to find whether x=0 or x=y...I tells us that x is not equal to y.. But can be 0 or x=y Insuff II tells us y is not equal to 0.. Insuff.. Combined X can be = y or not Insufficient E I think you mean either y=o or x =y Also I have a question if y = 0, is not xy = 0 which is square of an integer zero??



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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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21 Jun 2017, 01:35
Statement 1: if y = 0, answer is yes if y is not 0, answer is no
Statement 1 only tells us that x is not equal to y
Insuff
Statement 2: tells us that y is not equal to 0 but ignores the possibility of x = y or x = y
Insuff.
Combining  we know that neither x = y and nor y is 0 however x = y is still not known
hence E



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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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21 Jun 2017, 01:50
Bunuel  please provide detailed solution for the answer when OA is revealed, This question is an absolute nightmare



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If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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23 Jun 2017, 12:33
OFFICIAL SOLUTION: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?First of all, \(x \leq y \leq x\) ensures two things: 1. \(x^2y^2\geq 0\), so the square root of this number will be defined. 2. \(x+y\geq 0\), so the square root won't be equal to negative number. Next, \(x \leq x\) implies that \(x \geq 0\). And finally, before moving to the statements, let's rephrase the question: Does \(\sqrt{x^2  y^2} = x + y\)? Square both sides: does \(x^2  y^2 = x^2+2xy + y^2\)? Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).Does \(y(x+y)=0\)? Does \(y=0\) or \(x=y\)? (1) xy is NOT a square of an integer. If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0. However, if x = y and x ≠ 0 were true, for example if x = y = 1, then xy would be x^2, so negative, which cannot be a square of an integer. So, x = y IS possible. Of course, x can be not equal to y, and the statemented still could hold true. For example, x = 2 and y =1. Thus, we have have two different answers to the question: Not sufficient. (2) Point (x, y) is NOT on xaxisIf y = 0 were true, then point (x, y) would be ON the xaxis. So, we know that y ≠ 0. However, if x = y and x ≠ 0 were true, for example if x = y = 1, then then point (x, y) would be (x, x), so (positive, negative), which would mean that it's below xaxis. So, x = y IS possible. Of course, x can be not equal to y, and the statemented still could hold true. For example, x = 2 and y =1. (1)+(2) We could have the same cases: if x = y and x ≠ 0, for example if x = y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient. Answer: E.
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If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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23 Jun 2017, 12:34



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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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23 Jun 2017, 13:34
Bunuel wrote: OFFICIAL SOLUTION: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?First of all, \(x \leq y \leq x\) ensures two things: 1. \(x^2y^2\geq 0\), so the square root of this number will be defined. 2. \(x+y\geq 0\), so the square root won't be equal to negative number. Next, \(x \leq x\) implies that \(x \geq 0\). And finally, before moving to the statements, let's rephrase the question: Does \(\sqrt{x^2  y^2} = x + y\)? Square both sides: does \(x^2  y^2 = x^2+2xy + y^2\)? Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).Does \(y(x+y)=0\)? Does \(y=0\) or \(x=y\)? (1) xy is NOT a square of an integer. If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0. However, if x = y and x ≠ 0 were true, for example if x = y = 1, then xy would be x^2, so negative, which cannot be a square of an integer. So, x = y IS possible. Of course, x can be not equal to y, and the statemented still could hold true. For example, x = 2 and y =1. Thus, we have have two different answers to the question: Not sufficient. (2) Point (x, y) is NOT on xaxisIf y = 0 were true, then point (x, y) would be ON the xaxis. So, we know that y ≠ 0. However, if x = y and x ≠ 0 were true, for example if x = y = 1, then then point (x, y) would be (x, x), so (positive, negative), which would mean that it's below xaxis. So, x = y IS possible. Of course, x can be not equal to y, and the statemented still could hold true. For example, x = 2 and y =1. (1)+(2) We could have the same cases: if x = y and x ≠ 0, for example if x = y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient. Answer: E.Kudos to you Bunuelit is really spot on. It is high quality question and answer. But do you plan to add in the GMATCLub tests to experience it in mock test? I hope so Thanks



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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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23 Jun 2017, 13:38
Mo2men wrote: Bunuel wrote: OFFICIAL SOLUTION: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?First of all, \(x \leq y \leq x\) ensures two things: 1. \(x^2y^2\geq 0\), so the square root of this number will be defined. 2. \(x+y\geq 0\), so the square root won't be equal to negative number. Next, \(x \leq x\) implies that \(x \geq 0\). And finally, before moving to the statements, let's rephrase the question: Does \(\sqrt{x^2  y^2} = x + y\)? Square both sides: does \(x^2  y^2 = x^2+2xy + y^2\)? Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).Does \(y(x+y)=0\)? Does \(y=0\) or \(x=y\)? (1) xy is NOT a square of an integer. If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0. However, if x = y and x ≠ 0 were true, for example if x = y = 1, then xy would be x^2, so negative, which cannot be a square of an integer. So, x = y IS possible. Of course, x can be not equal to y, and the statemented still could hold true. For example, x = 2 and y =1. Thus, we have have two different answers to the question: Not sufficient. (2) Point (x, y) is NOT on xaxisIf y = 0 were true, then point (x, y) would be ON the xaxis. So, we know that y ≠ 0. However, if x = y and x ≠ 0 were true, for example if x = y = 1, then then point (x, y) would be (x, x), so (positive, negative), which would mean that it's below xaxis. So, x = y IS possible. Of course, x can be not equal to y, and the statemented still could hold true. For example, x = 2 and y =1. (1)+(2) We could have the same cases: if x = y and x ≠ 0, for example if x = y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient. Answer: E.Kudos to you Bunuelit is really spot on. It is high quality question and answer. But do you plan to add in the GMATCLub tests to experience it in mock test? I hope so Thanks Yes, all new questions will be added to the GMAT Club Tests.
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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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21 Sep 2017, 01:24
Bunuel wrote: OFFICIAL SOLUTION: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?First of all, \(x \leq y \leq x\) ensures two things: 1. \(x^2y^2\geq 0\), so the square root of this number will be defined. 2. \(x+y\geq 0\), so the square root won't be equal to negative number. Next, \(x \leq x\) implies that \(x \geq 0\). And finally, before moving to the statements, let's rephrase the question: Does \(\sqrt{x^2  y^2} = x + y\)? Square both sides: does \(x^2  y^2 = x^2+2xy + y^2\)? Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).Does \(y(x+y)=0\)? Does \(y=0\) or \(x=y\)? (1) xy is NOT a square of an integer. If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0. However, if x = y and x ≠ 0 were true, for example if x = y = 1, then xy would be x^2, so negative, which cannot be a square of an integer. So, x = y IS possible. Of course, x can be not equal to y, and the statemented still could hold true. For example, x = 2 and y =1. Thus, we have have two different answers to the question: Not sufficient. (2) Point (x, y) is NOT on xaxisIf y = 0 were true, then point (x, y) would be ON the xaxis. So, we know that y ≠ 0. However, if x = y and x ≠ 0 were true, for example if x = y = 1, then then point (x, y) would be (x, x), so (positive, negative), which would mean that it's below xaxis. So, x = y IS possible. Of course, x can be not equal to y, and the statemented still could hold true. For example, x = 2 and y =1. (1)+(2) We could have the same cases: if x = y and x ≠ 0, for example if x = y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient. Answer: E.I get the answer as B. Can you tell me where I might be making a mistake here: (b) Point (x, y) is NOT on xaxis \(\sqrt{x^2  y^2} = x + y\) => \(\sqrt{(xy)(x+y)} = (x+y)\) Dividing both sides by \(\sqrt{(x+y)}\), we get \(\sqrt{(xy)} = \sqrt{(x+y)}\) If y = 0 , then xy = x+y, otherwise they are not equal. Since y != 0(since point (x,y) is not on x axis), xy not equal to to x+y, hence \(\sqrt{(xy)} != \sqrt{(x+y)}\) . So, went with B



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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y [#permalink]
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21 Sep 2017, 01:29
InfiniteLoop80 wrote: Bunuel wrote: OFFICIAL SOLUTION: If x and y are integers and \(x \leq y \leq x\), does \(\sqrt{x^2  y^2} = x + y\)?First of all, \(x \leq y \leq x\) ensures two things: 1. \(x^2y^2\geq 0\), so the square root of this number will be defined. 2. \(x+y\geq 0\), so the square root won't be equal to negative number. Next, \(x \leq x\) implies that \(x \geq 0\). And finally, before moving to the statements, let's rephrase the question: Does \(\sqrt{x^2  y^2} = x + y\)? Square both sides: does \(x^2  y^2 = x^2+2xy + y^2\)? Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).Does \(y(x+y)=0\)? Does \(y=0\) or \(x=y\)?(1) xy is NOT a square of an integer. If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0. However, if x = y and x ≠ 0 were true, for example if x = y = 1, then xy would be x^2, so negative, which cannot be a square of an integer. So, x = y IS possible. Of course, x can be not equal to y, and the statemented still could hold true. For example, x = 2 and y =1. Thus, we have have two different answers to the question: Not sufficient. (2) Point (x, y) is NOT on xaxisIf y = 0 were true, then point (x, y) would be ON the xaxis. So, we know that y ≠ 0. However, if x = y and x ≠ 0 were true, for example if x = y = 1, then then point (x, y) would be (x, x), so (positive, negative), which would mean that it's below xaxis. So, x = y IS possible.Of course, x can be not equal to y, and the statemented still could hold true. For example, x = 2 and y =1. (1)+(2) We could have the same cases: if x = y and x ≠ 0, for example if x = y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient. Answer: E.I get the answer as B. Can you tell me where I might be making a mistake here: (b) Point (x, y) is NOT on xaxis \(\sqrt{x^2  y^2} = x + y\) => \(\sqrt{(xy)(x+y)} = (x+y)\) Dividing both sides by \(\sqrt{(x+y)}\), we get\(\sqrt{(xy)} = \sqrt{(x+y)}\) If y = 0 , then xy = x+y, otherwise they are not equal. Since y != 0(since point (x,y) is not on x axis), xy not equal to to x+y, hence \(\sqrt{(xy)} != \sqrt{(x+y)}\) . So, went with B Please reread the solution, paying attention to the highlighted part. As for your solution, you cannot divide by \(\sqrt{(x+y)}\) because it can be 0 for x = y and we cannot divide by 0.
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Re: If x and y are integers and x < y < x, does (x^2  y^2)^(1/2) = x + y
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21 Sep 2017, 01:29






