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Re: New Set: Number Properties!!!
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15 Sep 2013, 07:34
fido wrote: Quote: 6. Set S consists of more than two integers. Are all the integers in set S negative?
(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.
(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.
(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.
Answer: C. Dear Bunuel, First of all,i would like to thank you for all your sincere efforts that help innumerable gMAT takers like me. In the explanation provided for statement 2,i did not quite understand how all the numbers of the set would have the same sign,if the smallest and largest terms have the same sign.The set contains>2 terms and from statement,we can comment only about 2 terms.Can you help me understand? Thank you, Fido (2) says: The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. So either: smallest * greatest = negative * negative and in this case as both the smallest and the greatest are negative then ALL integers in the list are negative OR smallest * greatest = positive * positive and in this case as both the smallest and the greatest are positive then ALL integers in the list are positive. Hope it's clear.
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Re: New Set: Number Properties!!!
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26 Sep 2013, 06:09
Bunuel wrote: SOLUTIONS:
1. If x is an integer, what is the value of x?
(1) \(23x\) is a prime number. From this statement it follows that x=1 or x=1. Not sufficient.
(2) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=1. Not sufficient.
(1)+(2) x could be 1 or 1. Not sufficient.
Answer: E. Bunuel A question, in Statement 1 if we take the value of x = 1, than 23x = 23 I think 23 is not a prime. So x can only and only be +1 Therefore, Statement (1) is sufficient. What's your take on it. Regards



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Re: New Set: Number Properties!!!
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11 Nov 2013, 21:35
Hi, In question 3 above, it says that X and Y have 3 postive factors. But it does not say only positive factors. Could the number not have other factors such as 3,1,9 apart from 3,1,9? What am I missing out? It will be great if someone can point what I am doing incorrectly. Thanks!



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Re: New Set: Number Properties!!!
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11 Nov 2013, 22:27
adeel2000 wrote: Hi, In question 3 above, it says that X and Y have 3 postive factors. But it does not say only positive factors. Could the number not have other factors such as 3,1,9 apart from 3,1,9? What am I missing out? It will be great if someone can point what I am doing incorrectly. Thanks! In GMAT, factors are always positive. On the other hand, multiples can be positive as well as negative.
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Re: New Set: Number Properties!!!
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02 Jan 2014, 09:52
Bunuel wrote: 7. Is x the square of an integer?
The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square).
Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.
(1) When x is divided by 12 the remainder is 6. Given that \(x=12q+6=6(2q+1)=2*3*(2q+1)\). Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient.
(2) When x is divided by 14 the remainder is 2. Given that \(x=14p+2\). So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient.
Answer: A. first and foremost, thanks for this problem set. it's helped me immenselyCan you offer more explanation on St. (1) and St. (2), specifically: St. (1): which portion of the equation represents any specific "power" that is distributed to 2 ? All I can gather is that 2 will be multiplied by 3 (odd) and then multiplied by some odd number (2q + 1). In effect you are multiplying an even (6) by some odd number. How does this tell us the power that 2 is raised to? Example: x = 2*3*[2*2 +1) = 18 Prime factors of 18 = 2^1 and 3^2. How does the equation (2q+1) ensure that 2 will always be raised to an odd power? St. (2): Does it make sense/is it possible to break it out in similar fashion as you did to statement 1? x = 2(7p+1) ? The (7p+1) could yield an odd or even number, depending on what q is (as displayed in your example). Also, what program do you use for equation writing?



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03 Jan 2014, 03:53
mrwells2 wrote: Bunuel wrote: 7. Is x the square of an integer?
The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square).
Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.
(1) When x is divided by 12 the remainder is 6. Given that \(x=12q+6=6(2q+1)=2*3*(2q+1)\). Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient.
(2) When x is divided by 14 the remainder is 2. Given that \(x=14p+2\). So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient.
Answer: A. first and foremost, thanks for this problem set. it's helped me immenselyCan you offer more explanation on St. (1) and St. (2), specifically: St. (1): which portion of the equation represents any specific "power" that is distributed to 2 ? All I can gather is that 2 will be multiplied by 3 (odd) and then multiplied by some odd number (2q + 1). In effect you are multiplying an even (6) by some odd number. How does this tell us the power that 2 is raised to? Example: x = 2*3*[2*2 +1) = 18 Prime factors of 18 = 2^1 and 3^2. How does the equation (2q+1) ensure that 2 will always be raised to an odd power? St. (2): Does it make sense/is it possible to break it out in similar fashion as you did to statement 1? x = 2(7p+1) ? The (7p+1) could yield an odd or even number, depending on what q is (as displayed in your example). Also, what program do you use for equation writing? For (1): \(x=2*3*(2q+1)=2*odd*odd\) > two odd multiples there obviously do not have 2 as their factors, thus x has 2 in first power only (we have only one 2 there). Fro (2) your reasoning is correct. As for formatting, check here: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628Hope this helps.
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Re: New Set: Number Properties!!!
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13 Mar 2014, 17:47
Bunuel wrote: SOLUTIONS:
1. If x is an integer, what is the value of x?
(1) \(23x\) is a prime number. From this statement it follows that x=1 or x=1. Not sufficient.
(2) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=1. Not sufficient.
(1)+(2) x could be 1 or 1. Not sufficient.
Answer: E. I thought on the GMAT, you can only have positive roots as answers ? Following the post here: http://www.manhattangmat.com/forums/neg ... 14759.htmlIs Statement 2 something we can expect to see on the GMAT ?



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Re: New Set: Number Properties!!!
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13 Mar 2014, 19:43
mrwells2 wrote: Bunuel wrote: SOLUTIONS:
1. If x is an integer, what is the value of x?
(1) \(23x\) is a prime number. From this statement it follows that x=1 or x=1. Not sufficient.
(2) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=1. Not sufficient.
(1)+(2) x could be 1 or 1. Not sufficient.
Answer: E. I thought on the GMAT, you can only have positive roots as answers ? Following the post here: http://www.manhattangmat.com/forums/neg ... 14759.htmlIs Statement 2 something we can expect to see on the GMAT ? Yes, you do have only positive roots on GMAT. \(\sqrt{4} = 2\) only (not 2) \(\sqrt{x}\) is never negative Bot note here that the expression is different: \(\sqrt{x^2}\) \(\sqrt{x^2} = x\) (absolute value of x) x can be 1 or 1 here because it is squared. \(\sqrt{1}\) will be 1 only. So, if \(x^2 = 4\), x = 2 or 2 But \(\sqrt{4} = 2\) only on the GMAT
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Re: New Set: Number Properties!!!
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26 Mar 2014, 09:04
Bunuel wrote: 8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient. (2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient. Answer: B. Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Note that if denominator already has only 2s and/or 5s then it doesn't matter whether the fraction is reduced or not. For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal. We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced. Questions testing this concept: doesthedecimalequivalentofpqwherepandqare89566.htmlanydecimalthathasonlyafinitenumberofnonzerodigits101964.htmlifabcdandeareintegersandp2a3bandq2c3d5eispqaterminatingdecimal125789.html700question94641.htmlisrs2isaterminatingdecimal91360.htmlplexplain89566.htmlwhichofthefollowingfractions88937.html Hi Bunuel Quick question on Statement 1 Here Let us assume SET S with 10 elements and let 5 th and 6th element be 1/a and 1/b where a and b are both prime. Now the median of the set would be a+b/2ab and it is given that reciprocal of the median is also prime.which means that 2(ab/a+b) will also be prime. So we have to achieve this ab=a+b and that is only possible when a and b are equal to 2.And this proves that Statement 1 is sufficient to answer the question. Can you please point where did I went wrong.
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26 Mar 2014, 09:16
282552 wrote: Bunuel wrote: 8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient. (2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient. Answer: B. Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Note that if denominator already has only 2s and/or 5s then it doesn't matter whether the fraction is reduced or not. For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal. We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced. Questions testing this concept: doesthedecimalequivalentofpqwherepandqare89566.htmlanydecimalthathasonlyafinitenumberofnonzerodigits101964.htmlifabcdandeareintegersandp2a3bandq2c3d5eispqaterminatingdecimal125789.html700question94641.htmlisrs2isaterminatingdecimal91360.htmlplexplain89566.htmlwhichofthefollowingfractions88937.html Hi Bunuel Quick question on Statement 1 Here Let us assume SET S with 10 elements and let 5 th and 6th element be 1/a and 1/b where a and b are both prime. Now the median of the set would be a+b/2ab and it is given that reciprocal of the median is also prime.which means that 2(ab/a+b) will also be prime. So we have to achieve this ab=a+b and that is only possible when a and b are equal to 2.And this proves that Statement 1 is sufficient to answer the question. Can you please point where did I went wrong. 2ab/(a+b) will be prime if a=b=2 or a=b=5 (in all case when a=b=prime).
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Re: New Set: Number Properties!!!
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31 Mar 2014, 20:49
Bunuel wrote: 6. Set S consists of more than two integers. Are all the integers in set S negative?
(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.
(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.
(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.
Answer: C. Hi Bunuel, Thanks for your questions, you're the Benchmark. In this particular question I was wondering if we are assuming the S is a set of consecutive integers. Let me explain, If S is [1,2,3] C will be insufficient. So what I'm missing? Thanks



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01 Apr 2014, 00:37
ArnauAnglerill wrote: Bunuel wrote: 6. Set S consists of more than two integers. Are all the integers in set S negative?
(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.
(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.
(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.
Answer: C. Hi Bunuel, Thanks for your questions, you're the Benchmark. In this particular question I was wondering if we are assuming the S is a set of consecutive integers. Let me explain, If S is [1,2,3] C will be insufficient. So what I'm missing? Thanks Your set {3, 2, 1} does not satisfy any of the two statements: (1) The product of any three integers in the set is negative > the product of the three terms of your set is positive: (3)(2)1=6. (2) The product of the smallest and largest integers in the set is a prime number > the product of the smallest and largest integers in your set is NOT a prime: (3)1=3. Hope it helps.
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Re: New Set: Number Properties!!!
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09 Apr 2014, 23:41
Hi Bunnel
Could you please explain question 5 on the part "Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. "?
Thanks



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23 May 2014, 09:57
Bunuel wrote: 6. Set S consists of more than two integers. Are all the integers in set S negative?
(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.
(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.
(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.
Answer: C. a doubt t on explanation for statement 1 if a set S={A(+ve) , B(ve), C(+ve) , D (+ve)} and we take product ACD>it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing



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Re: New Set: Number Properties!!!
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23 May 2014, 10:02
Bunuel wrote: SOLUTIONS:
1. If x is an integer, what is the value of x?
(1) \(23x\) is a prime number. From this statement it follows that x=1 or x=1. Not sufficient.
(2) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=1. Not sufficient.
(1)+(2) x could be 1 or 1. Not sufficient.
Answer: E. a doubt on the statement 2 \sqrt{x*x}=1 0r 1 but since its given that 2*(sqrtX^2)=prime; x can only take positive value as 2 would not be a prime no.. thus according to me statement two should be sufficient



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23 May 2014, 10:17
somyakalra wrote: Bunuel wrote: SOLUTIONS:
1. If x is an integer, what is the value of x?
(1) \(23x\) is a prime number. From this statement it follows that x=1 or x=1. Not sufficient.
(2) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=1. Not sufficient.
(1)+(2) x could be 1 or 1. Not sufficient.
Answer: E. a doubt on the statement 2 \sqrt{x*x}=1 0r 1 but since its given that 2*(sqrtX^2)=prime; x can only take positive value as 2 would not be a prime no.. thus according to me statement two should be sufficient \(2\sqrt{x^2}\) won't give negative value for any x, because \(\sqrt{nonnegative \ number}\geq{0}\): the square root function cannot give negative result. If x=1 or x=1, then \(2\sqrt{x^2}=2\sqrt{1}=2=prime\). Writing Mathematical Formulas on the Forum: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628
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Re: New Set: Number Properties!!!
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23 May 2014, 10:26
somyakalra wrote: Bunuel wrote: 6. Set S consists of more than two integers. Are all the integers in set S negative?
(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.
(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.
(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.
Answer: C. a doubt t on explanation for statement 1 if a set S={A(+ve) , B(ve), C(+ve) , D (+ve)} and we take product ACD>it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing You misinterpreted the statement. It says: the product of ANY three integers in the set is negative. So, the set cannot be {positive, positive, positive, negative} because the product of the first 3 terms there is NOT negative. Basically if the set has 4 or more terms, and not all of them are negative, then this statement will be violated, because we'll be able to pick two negative terms and one nonnegative, which will not give negative product. Hope it's clear.
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Re: New Set: Number Properties!!!
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23 May 2014, 10:34
somyakalra wrote: Bunuel wrote: 6. Set S consists of more than two integers. Are all the integers in set S negative?
(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.
(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.
(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.
Answer: C. a doubt t on explanation for statement 1 if a set S={A(+ve) , B(ve), C(+ve) , D (+ve)} and we take product ACD>it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing As for the second statement. (2) says: The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. So either: smallest * greatest = negative * negative and in this case as both the smallest and the greatest are negative then ALL integers in the list are negative OR smallest * greatest = positive * positive and in this case as both the smallest and the greatest are positive then ALL integers in the list are positive. Hope it's clear.
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Re: New Set: Number Properties!!! &nbs
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23 May 2014, 10:34



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