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Re M1523
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16 Sep 2014, 00:56



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07 Jan 2015, 10:51
Bunuel wrote: Official Solution:
(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer. (2) \(x + y\) is divisible by \(8\). Now, \(x^2  y^2=(x+y)(xy)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2  y^2\) is not. Not sufficient. (1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(xy)\) is divisible by \(8\). Sufficient.
Answer: C what if X=Y
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07 Jan 2015, 11:07
him1985 wrote: Bunuel wrote: Official Solution:
(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer. (2) \(x + y\) is divisible by \(8\). Now, \(x^2  y^2=(x+y)(xy)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2  y^2\) is not. Not sufficient. (1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(xy)\) is divisible by \(8\). Sufficient.
Answer: C what if X=Y 0 is divisible by every integer (except 0 itself). I think you should go through the basics and brush up fundamentals once more. For a start check here: divisibilitymultiplesfactorstipsandhints174998.htmlHope it helps.
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Re: M1523
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07 Jan 2015, 12:00
Bunuel wrote: Official Solution:
(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer. (2) \(x + y\) is divisible by \(8\). Now, \(x^2  y^2=(x+y)(xy)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2  y^2\) is not. Not sufficient. (1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(xy)\) is divisible by \(8\). Sufficient.
Answer: C Hi Bunuel If \(x=4.8\) and \(y=3.2\) then also we can write \(4.8^2  3.2^2=(4.8+3.2)(4.83.2)\) and 8.0(1.6) can be divisible by 8 as well (if not than why?) I know you have mentioned that if one of the multiples is divisible by 8 then so is the product: true for integers, but it seems that with \(x=4.8\) and \(y=3.2\) , its working in this case. puzzled
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07 Jan 2015, 12:12
Ankur9 wrote: Bunuel wrote: Official Solution:
(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer. (2) \(x + y\) is divisible by \(8\). Now, \(x^2  y^2=(x+y)(xy)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2  y^2\) is not. Not sufficient. (1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(xy)\) is divisible by \(8\). Sufficient.
Answer: C Hi Bunuel If \(x=4.8\) and \(y=3.2\) then also we can write \(4.8^2  3.2^2=(4.8+3.2)(4.83.2)\) and 8.0(1.6) can be divisible by 8 as well (if not than why?) I know you have mentioned that if one of the multiples is divisible by 8 then so is the product: true for integers, but it seems that with \(x=4.8\) and \(y=3.2\) , its working in this case. puzzled 4.8^2  3.2^2 = 12.8 is not divisible by 8 because 12.8 is not an integer.
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Re: M1523
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11 Apr 2015, 20:32
Bunuel wrote: Ankur9 wrote: Bunuel wrote: Official Solution:
(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer. (2) \(x + y\) is divisible by \(8\). Now, \(x^2  y^2=(x+y)(xy)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2  y^2\) is not. Not sufficient. (1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(xy)\) is divisible by \(8\). Sufficient.
Answer: C Hi Bunuel If \(x=4.8\) and \(y=3.2\) then also we can write \(4.8^2  3.2^2=(4.8+3.2)(4.83.2)\) and 8.0(1.6) can be divisible by 8 as well (if not than why?) I know you have mentioned that if one of the multiples is divisible by 8 then so is the product: true for integers, but it seems that with \(x=4.8\) and \(y=3.2\) , its working in this case. puzzled 4.8^2  3.2^2 = 12.8 is not divisible by 8 because 12.8 is not an integer. we have x = 4.8 and y = 3.2, in which case (x+y)(xy) = (4.8+3.2)(4.83.2) = (8)(1.6) = 12.8 Now We know that a no. is divisible by another no. only if the remainder is an integer. Here, if we accept your expalantion, then remainder is 1.6, which is not an integer. Therefore not divisible by 8.



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12 Apr 2015, 04:39
abhi1785 wrote: Bunuel wrote: Ankur9 wrote: Hi Bunuel If \(x=4.8\) and \(y=3.2\) then also we can write \(4.8^2  3.2^2=(4.8+3.2)(4.83.2)\) and 8.0(1.6) can be divisible by 8 as well (if not than why?) I know you have mentioned that if one of the multiples is divisible by 8 then so is the product: true for integers, but it seems that with \(x=4.8\) and \(y=3.2\) , its working in this case. puzzled 4.8^2  3.2^2 = 12.8 is not divisible by 8 because 12.8 is not an integer. we have x = 4.8 and y = 3.2, in which case (x+y)(xy) = (4.8+3.2)(4.83.2) = (8)(1.6) = 12.8 Now We know that a no. is divisible by another no. only if the remainder is an integer. Here, if we accept your expalantion, then remainder is 1.6, which is not an integer. Therefore not divisible by 8. Sorry, but your question is not clear. Please elaborate. As for the red part: an integer is divisible by an integer if the remainder is 0.
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03 May 2015, 03:11
the definition of being divisible has to be mentioned: A is divisible by B if A/B = C is an integer. 12,8/8 = 1,6 which isn't an integer.



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Re: M1523
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28 May 2015, 21:06
Bunuel wrote: Official Solution:
(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer. (2) \(x + y\) is divisible by \(8\). Now, \(x^2  y^2=(x+y)(xy)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2  y^2\) is not. Not sufficient. (1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(xy)\) is divisible by \(8\). Sufficient.
Answer: C But even if we add 3.2 and 4.8 it equals 8. And if we expand x^2y^2, it will be divisible by 8. So irrespective of x & y being integer or rational or decimal, if it gives x+y or xy as 8 or a multiple of 8, i guess it should work. That's why I marked B.



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29 May 2015, 05:02
usadude05 wrote: Bunuel wrote: Official Solution:
(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer. (2) \(x + y\) is divisible by \(8\). Now, \(x^2  y^2=(x+y)(xy)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2  y^2\) is not. Not sufficient. (1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(xy)\) is divisible by \(8\). Sufficient.
Answer: C But even if we add 3.2 and 4.8 it equals 8. And if we expand x^2y^2, it will be divisible by 8. So irrespective of x & y being integer or rational or decimal, if it gives x+y or xy as 8 or a multiple of 8, i guess it should work. That's why I marked B. Please read the whole thread: m15184055.html#p1466250
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19 Jun 2015, 07:19
Great question Bunnel.. Totally Stumped



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17 Mar 2016, 13:56
Cadaver wrote: Great question Bunnel.. Totally Stumped Totally agree with Cadaver on this one. Hat off Bunuel. Lovely question. Really awesome solution/explanation.



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01 Jun 2016, 20:41
desiwolverine wrote: Cadaver wrote: Great question Bunnel.. Totally Stumped Totally agree with Cadaver on this one. Hat off Bunuel. Lovely question. Really awesome solution/explanation. Yep, agreed! I was trapped as well  what a good question!
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24 Jul 2016, 19:10
hi! when (x+y)(xy) = xsq ysq... then one of the multiple that is x+y is divisible by 8.. why do we care about second one. does it matter we write xsq  y sq or (x+y)(xy)? thanks



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24 Jul 2016, 20:31
Celestial09 wrote: hi! when (x+y)(xy) = xsq ysq... then one of the multiple that is x+y is divisible by 8.. why do we care about second one. does it matter we write xsq  y sq or (x+y)(xy)? thanks Celestial09 : the key here is whether x & y are Integer or not, because divisibility is defined by remainder = 0. So in the case that, x=3.8, y=4.2 > \(x+y=8 (divisible by 8), xy=0.4 > (x^2y^2) = 4.2*0.4=1.68\) > this number is not divisible by 8 (since remainder is #0)
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20 Aug 2016, 04:40
I think this is a poorquality question and I don't agree with the explanation. if x+y is a multiple of 8 as stated by statement 2, then x^2  y^2 is divisible by 8 as well since the 8 divides the (x+y) in (x+y)(xy) form of the expression. Why is this logic wrong , this is not addressed in this thread.



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hi 12.8/8=4*3.2/8=3.2/2=32/10*2=16/10=1.6
8 is completely divided in 12.8. Why should we care whether (xy) is integer or not?



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