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Data Sufficiency Pack 4, Question 2) Z is a positive integer...
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20 Nov 2015, 15:21
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QUANT 4PACK SERIES Data Sufficiency Pack 4 Question 2 Z is a positive integer...Z is a positive integer greater than 3. How many distinct prime factors does (Z + 1)(Z – 1) have? 1) Z is not even 2) Z is not a multiple of 5 48 Hour Window Answer & Explanation WindowEarn KUDOS! Post your answer and explanation. OA, and explanation will be posted after the 48 hour window closes. This question is part of the Quant 4Pack seriesScroll Down For Official Explanation
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Re: Data Sufficiency Pack 4, Question 2) Z is a positive integer...
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20 Nov 2015, 15:43
EMPOWERgmatRichC wrote: Z is a positive integer greater than 3. How many distinct prime factors does (Z + 1)(Z – 1) have?
1) Z is not even 2) Z is not a multiple of 5
Target question: How many distinct prime factors does (Z + 1)(Z – 1) have? Given: Z is a positive integer greater than 3 Statement 1: Z is not even This statement doesn't FEEL sufficient, so I'll TEST some values. There are several values of Z that satisfy statement 1. Here are two: Case a: Z = 9, in which case (Z + 1)(Z  1) = (10)(8) = 80 = (2)(2)(2)(2)(5). In this case, we have TWO distinct prime factorsCase b: Z = 11, in which case (Z + 1)(Z  1) = (12)(10) = 120 = (2)(2)(2)(3)(5). In this case, we have THREE distinct prime factorsSince we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: http://www.gmatprepnow.com/articles/dat ... lugvalues Statement 2: Z is not a multiple of 5 This statement doesn't FEEL sufficient either, so I'll TEST some values. There are several values of Z that satisfy statement 2. Here are two: Case a: Z = 9, in which case (Z + 1)(Z  1) = (10)(8) = 80 = (2)(2)(2)(2)(5). In this case, we have TWO distinct prime factorsCase b: Z = 11, in which case (Z + 1)(Z  1) = (12)(10) = 120 = (2)(2)(2)(3)(5). In this case, we have THREE distinct prime factorsSince we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Statements 1 and 2 combined Notice that I used the same values of Z for the first 2 statements. This means that the same Zvalues satisfy BOTH statements. That is: Case a: Z = 9, in which case (Z + 1)(Z  1) = (10)(8) = 80 = (2)(2)(2)(2)(5). In this case, we have TWO distinct prime factorsCase b: Z = 11, in which case (Z + 1)(Z  1) = (12)(10) = 120 = (2)(2)(2)(3)(5). In this case, we have THREE distinct prime factorsSince we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT Answer = E Cheers, Brent
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Re: Data Sufficiency Pack 4, Question 2) Z is a positive integer...
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24 Nov 2015, 23:28
EMPOWERgmatRichC wrote: QUANT 4PACK SERIES Data Sufficiency Pack 4 Question 2 Z is a positive integer...
Z is a positive integer greater than 3. How many distinct prime factors does (Z + 1)(Z – 1) have?
1) Z is not even 2) Z is not a multiple of 5
Hi All, This question can be solved by TESTing VALUES. We're told that Z is a POSITIVE INTEGER greater than 3. We're asked for the number of DISTINCT (meaning 'different') prime factors in (Z + 1)(Z – 1). To start, we can rewrite (Z + 1)(Z – 1) as \(Z^{2}\)  1 1) Z is NOT even We already know that Z is an INTEGER greater than 3... IF... Z = 7, then 491 = 48 = (2)(2)(2)(2)(3) and has 2 distinct prime factors. IF... Z = 11, then 1211 = 120 = (2)(2)(2)(3)(5) and has 3 distinct prime factors. Fact 1 is INSUFFICIENT 2) Z is not a multiple of 5 The same 2 TESTs that we used in Fact 1 will 'fit' Fact 2 as well... IF... Z = 7, then 491 = 48 = (2)(2)(2)(2)(3) and has 2 distinct prime factors. IF... Z = 11, then 1211 = 120 = (2)(2)(2)(3)(5) and has 3 distinct prime factors. Fact 2 is INSUFFICIENT Combined, we can see that the same two TESTs 'fit' both Facts, but yield two different answers. Combined, INSUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Data Sufficiency Pack 4, Question 2) Z is a positive integer...
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07 Jun 2017, 03:49
As given in question statement Z>3. So we can not take Z=3. Regards MathRevolution wrote: Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Z is a positive integer greater than 3. How many distinct prime factors does (Z + 1)(Z – 1) have?
1) Z is not even 2) Z is not a multiple of 5 There is one variable (z) and 2 equations are given by the 2 conditions, increasing the chance (D) will be our answer. For condition 1, z=3, (3+1)(31)=8=2^3, the no. of distinct prime factors:1 z=7, (7+1)(71)=(2^4)3, the no. of distinct prime factors: 2 For condition 2, z=3, (3+1)(31)=8=2^3, the no. of distinct prime factors:1 z=7, (7+1)(71)=(2^4)3, the no. of distinct prime factors: 2 Looking at the conditions together, z=3, (3+1)(31)=8=2^3, the no. of distinct prime factors:1 z=7, (7+1)(71)=(2^4)3, the no. of distinct prime factors: 2. Therefore, the answer becomes (E).
For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.



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Re: Data Sufficiency Pack 4, Question 2) Z is a positive integer...
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22 Nov 2015, 01:42
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Z is a positive integer greater than 3. How many distinct prime factors does (Z + 1)(Z – 1) have? 1) Z is not even 2) Z is not a multiple of 5 There is one variable (z) and 2 equations are given by the 2 conditions, increasing the chance (D) will be our answer. For condition 1, z=3, (3+1)(31)=8=2^3, the no. of distinct prime factors:1 z=7, (7+1)(71)=(2^4)3, the no. of distinct prime factors: 2 For condition 2, z=3, (3+1)(31)=8=2^3, the no. of distinct prime factors:1 z=7, (7+1)(71)=(2^4)3, the no. of distinct prime factors: 2 Looking at the conditions together, z=3, (3+1)(31)=8=2^3, the no. of distinct prime factors:1 z=7, (7+1)(71)=(2^4)3, the no. of distinct prime factors: 2. Therefore, the answer becomes (E). For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Data Sufficiency Pack 4, Question 2) Z is a positive integer...
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Re: Data Sufficiency Pack 4, Question 2) Z is a positive integer...
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