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Re M3131 [#permalink]
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14 Jun 2015, 14:27
Official Solution:If \(x\) is an integer, what is the remainder when \(1  x^2\) is divided by 4? Notice that if \(x\) is odd, then \(1  x^2\) is a multiple of 4. For example: If \(x=1\), \(1  x^2 = 0\); If \(x=3\), \(1  x^2 = 8\); If \(x=5\), \(1  x^2 = 24\). ... (1) The sum of any two factors of \(x\) is even. For the sum of ANY two factors of \(x\) to be even all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum), which means that \(x\) is an odd number. Sufficient. (2) The product of any two factors of \(x\) is odd. Basically the same here: for the product of ANY two factors of \(x\) to be odd all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with any other factor and we'd get even product), which means that \(x\) is an odd number. Sufficient. Answer: D
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Re: M3131 [#permalink]
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31 Oct 2015, 07:07
Bunuel wrote: Official Solution:
If \(x\) is an integer, what is the remainder when \(1  x^2\) is divided by 4? Notice that if \(x\) is odd, then \(1  x^2\) is a multiple of 4. For example: If \(x=1\), \(1  x^2 = 0\); If \(x=3\), \(1  x^2 = 8\); If \(x=5\), \(1  x^2 = 24\). ... (1) The sum of any two factors of \(x\) is even. For the sum of ANY two factors of \(x\) to be even all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum), which means that \(x\) is an odd number. Sufficient. (2) The product of any two factors of \(x\) is odd. Basically the same here: for the product of ANY two factors of \(x\) to be odd all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with any other factor and we'd get even product), which means that \(x\) is an odd number. Sufficient.
Answer: D >> even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum), which means that \(x\) is an odd number Could you please give some example to illustrate the point? I am having difficulty comprehending this point.



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Bunuel wrote: Official Solution:
If \(x\) is an integer, what is the remainder when \(1  x^2\) is divided by 4? Notice that if \(x\) is odd, then \(1  x^2\) is a multiple of 4. For example: If \(x=1\), \(1  x^2 = 0\); If \(x=3\), \(1  x^2 = 8\); If \(x=5\), \(1  x^2 = 24\). ... (1) The sum of any two factors of \(x\) is even. For the sum of ANY two factors of \(x\) to be even all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum), which means that \(x\) is an odd number. Sufficient. (2) The product of any two factors of \(x\) is odd. Basically the same here: for the product of ANY two factors of \(x\) to be odd all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with any other factor and we'd get even product), which means that \(x\) is an odd number. Sufficient.
Answer: D Bunuel, is this a correct line of reasoning for your explaination on statement 1?: If \(x = 4\), we have three factors; \(1, 2\) and \(2\). From S1, we get that the sum of ANY two factors of \(x = even\), which includes \(1\). So, we have two cases: 1) \(1 + 2 = 3 = odd\) 2) \(2 + 2 = 4 = even\) Thus, not all/ANY factors of 4, when added together, gives an even number. However, if \(x = 5\), we only have two factors; \(1\) and \(5\). This only gives one case: \(1 + 5 = 6 = even\). Since \(5\) only has these two factors, we will always have an even number when any two factors of \(5\) are added. Conclusion: \(x\) must be an odd number for ANY two factors of \(x\) to be \(even\). Side note, these are our even/odd rules for addition: 1) \(Even + even = even\) 2) \(Odd + odd = even\) 3) \(Even + odd = odd\) 4) \(Odd + even = odd\)



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Re: M3131 [#permalink]
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18 May 2016, 12:38
Sallyzodiac wrote: Bunuel wrote: Official Solution:
If \(x\) is an integer, what is the remainder when \(1  x^2\) is divided by 4? Notice that if \(x\) is odd, then \(1  x^2\) is a multiple of 4. For example: If \(x=1\), \(1  x^2 = 0\); If \(x=3\), \(1  x^2 = 8\); If \(x=5\), \(1  x^2 = 24\). ... (1) The sum of any two factors of \(x\) is even. For the sum of ANY two factors of \(x\) to be even all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum), which means that \(x\) is an odd number. Sufficient. (2) The product of any two factors of \(x\) is odd. Basically the same here: for the product of ANY two factors of \(x\) to be odd all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with any other factor and we'd get even product), which means that \(x\) is an odd number. Sufficient.
Answer: D Bunuel, is this a correct line of reasoning for your explaination on statement 1?: If \(x = 4\), we have three factors; \(1, 2\) and \(2\). From S1, we get that the sum of ANY two factors of \(x = even\), which includes \(1\). So, we have two cases: 1) \(1 + 2 = 3 = odd\) 2) \(2 + 2 = 4 = even\) Thus, not all/ANY factors of 4, when added together, gives an even number. However, if \(x = 5\), we only have two factors; \(1\) and \(5\). This only gives one case: \(1 + 5 = 6 = even\). Since \(5\) only has these two factors, we will always have an even number when any two factors of \(5\) are added. Conclusion: \(x\) must be an odd number for ANY two factors of \(x\) to be \(even\). Side note, these are our even/odd rules for addition: 1) \(Even + even = even\) 2) \(Odd + odd = even\) 3) \(Even + odd = odd\) 4) \(Odd + even = odd\) ______________ Yes that's correct.
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Re M3131 [#permalink]
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08 Jul 2016, 09:02
I think this is a highquality question and the explanation isn't clear enough, please elaborate. For choice 1 to be sufficient, there is no constraint in the question that rules out considering 0 which is also an even number. Doesn't matter whether you state "any factors" but that does not necessarily rule out "0" which is an even number as well in which case remainder will be 0. Can you explain?



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Re: M3131 [#permalink]
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Re: M3131 [#permalink]
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16 May 2017, 10:59
Hi Bunuel,
My analysis is exactly same as yours. But I chose the option E. Since we are not getting a unique odd number. We are just concluding that x is an odd from both the statements. From Stat 1, x could be 1 and from stat 2 x could be 3 and the mentioned numbers satisfy the given respective statements.
I think you have marked D as the correct answer since GMAT doesn't contradict with the statement i.e. two different cases can't be correct. ?
Can you please throw some light on this.



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Re: M3131 [#permalink]
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16 May 2017, 11:06
msk0657 wrote: Hi Bunuel,
My analysis is exactly same as yours. But I chose the option E. Since we are not getting a unique odd number. We are just concluding that x is an odd from both the statements. From Stat 1, x could be 1 and from stat 2 x could be 3 and the mentioned numbers satisfy the given respective statements.
I think you have marked D as the correct answer since GMAT doesn't contradict with the statement i.e. two different cases can't be correct. ?
Can you please throw some light on this. The question asks: what is the remainder when 1−x^2 is divided by 4? NOT what is the value of x. We concluded that if x is odd number (any odd number), then the reminder when 1−x^2 is divided by 4 is 0. From each statement we got that x must be odd, thus from each statement we have that the remainder must be 0. Therefore each statement gives unique answer to the question (0) and thus each is sufficient. Answer D. Hope it's clear.
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Re: M3131 [#permalink]
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17 Jul 2017, 09:42
Bunuel wrote: Official Solution:
If \(x\) is an integer, what is the remainder when \(1  x^2\) is divided by 4? Notice that if \(x\) is odd, then \(1  x^2\) is a multiple of 4. For example: If \(x=1\), \(1  x^2 = 0\); If \(x=3\), \(1  x^2 = 8\); If \(x=5\), \(1  x^2 = 24\). ... (1) The sum of any two factors of \(x\) is even. For the sum of ANY two factors of \(x\) to be even all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum), which means that \(x\) is an odd number. Sufficient.
Answer: D Hi, The line of reasoning given in the first statement is that becuase sum of any two factors of x is even, all the factors of x should be odd. Why are we assuming this? It could also be that all the factors of x is even, in that case also the sum of any two factors of x would be even as well, right? Secondly, I did not understand the '(even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum)' part. If we are getting one pair of sum as odd, isn't that rendering the statement untrue? Can someone explain an alternate method of explaining statement 1? Thanks!
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Re: M3131 [#permalink]
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17 Jul 2017, 09:47
ashikaverma13 wrote: Bunuel wrote: Official Solution:
If \(x\) is an integer, what is the remainder when \(1  x^2\) is divided by 4? Notice that if \(x\) is odd, then \(1  x^2\) is a multiple of 4. For example: If \(x=1\), \(1  x^2 = 0\); If \(x=3\), \(1  x^2 = 8\); If \(x=5\), \(1  x^2 = 24\). ... (1) The sum of any two factors of \(x\) is even. For the sum of ANY two factors of \(x\) to be even all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum), which means that \(x\) is an odd number. Sufficient.
Answer: D Hi, The line of reasoning given in the first statement is that becuase sum of any two factors of x is even, all the factors of x should be odd. Why are we assuming this? It could also be that all the factors of x is even, in that case also the sum of any two factors of x would be even as well, right? Secondly, I did not understand the '(even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum)' part. If we are getting one pair of sum as odd, isn't that rendering the statement untrue? Can someone explain an alternate method of explaining statement 1? Thanks! An integer cannot have only even factors because 1, an odd integer, is a factor of all integers. About you second point: exactly, the fact that even one even factor will make the statement untrue means that x cannot have any even factors. Hope it's clear.
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Re: M3131 [#permalink]
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17 Jul 2017, 12:21
Bunuel wrote: ashikaverma13 wrote: Bunuel wrote: Official Solution:
If \(x\) is an integer, what is the remainder when \(1  x^2\) is divided by 4? Notice that if \(x\) is odd, then \(1  x^2\) is a multiple of 4. For example: If \(x=1\), \(1  x^2 = 0\); If \(x=3\), \(1  x^2 = 8\); If \(x=5\), \(1  x^2 = 24\). ... (1) The sum of any two factors of \(x\) is even. For the sum of ANY two factors of \(x\) to be even all factors of \(x\) must be odd (even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum), which means that \(x\) is an odd number. Sufficient.
Answer: D Hi, The line of reasoning given in the first statement is that becuase sum of any two factors of x is even, all the factors of x should be odd. Why are we assuming this? It could also be that all the factors of x is even, in that case also the sum of any two factors of x would be even as well, right? Secondly, I did not understand the '(even if one of the factors is even then we could pair that even factor with 1, which is a factor of every integer, and we'd get odd sum)' part. If we are getting one pair of sum as odd, isn't that rendering the statement untrue? Can someone explain an alternate method of explaining statement 1? Thanks! An integer cannot have only even factors because 1, an odd integer, is a factor of all integers. About you second point: exactly, the fact that even one even factor will make the statement untrue means that x cannot have any even factors. Hope it's clear. yes, I think I understand now. Thanks
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what if I consider the factors of 21 as 3 and 7 and then proceed for option 2. its 121^2, which yields 440 leading us to a remainder of 0. where am I missing. Pls help..



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Re: M3131 [#permalink]
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18 Jul 2017, 09:49
Yes yes ..... My bad. My wrong interpretation. Thanks Bunuel !!!!!



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Re: M3131 [#permalink]
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02 Oct 2017, 10:51
Bunuel . Vyshak What if x is 10 ? Factors 1,2,5,10. 5*1 = 5 is odd but remainder will be 1 1x*x = 1100 = 99/4
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Re: M3131 [#permalink]
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02 Oct 2017, 11:00
Bunuel wrote: mbsingh wrote: Bunuel . Vyshak What if x is 10 ? Factors 1,2,5,10. 5*1 = 5 is odd but remainder will be 1 1x*x = 1100 = 99/4 x cannot be 10 for neither of the statements. Please read above for more. Hi Bunuel, i don't understand why x cannot be 10. Is it because when statement 2 says product of any two factors is odd, does that imply that thats the only possible product for x's factors ?
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Re: M3131 [#permalink]
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02 Oct 2017, 11:34
mbsingh wrote: Bunuel wrote: mbsingh wrote: Bunuel . Vyshak What if x is 10 ? Factors 1,2,5,10. 5*1 = 5 is odd but remainder will be 1 1x*x = 1100 = 99/4 x cannot be 10 for neither of the statements. Please read above for more. Hi Bunuel, i don't understand why x cannot be 10. Is it because when statement 2 says product of any two factors is odd, does that imply that thats the only possible product for x's factors ? The factor of 10 are 1, 2, 5, and 10. (1) says: The sum of ANY two factors of \(x\) is even. This is not true for 10. For example, 1 + 2 = 3 = odd. (2) The product of ANY two factors of \(x\) is odd. This is not true for 10. For example, 1*2 = 2 = even.
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