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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote If integer C is randomly selected from 20 to 99, inclusive. This topic has 2 expert replies and 0 member replies Top Member If integer C is randomly selected from 20 to 99, inclusive. Timer 00:00 Your Answer A B C D E Global Stats Difficult If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ? A. 1/2 B. 2/3 C. 3/4 D. 4/5 E. 1/3 OA C Source: Manhattan Prep GMAT/MBA Expert GMAT Instructor Joined 22 Aug 2016 Posted: 1979 messages Followed by: 30 members Upvotes: 470 BTGmoderatorDC wrote: If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ? A. 1/2 B. 2/3 C. 3/4 D. 4/5 E. 1/3 OA C Source: Manhattan Prep We have to determine whether C^3 - C is divisible by 12. C^3 - C = C(C^2 - 1) = C(C - 1)(C + 1) = (C - 1)C(C + 1) = Product of three consecutive integers This question can be solved taking few sample values of C from 20 to 99, inclusive. Note that from 20 to 99, inclusive, there are 80 integers -- 40 even and 40 odd. 1. Say C = 20; thus, (C - 1)C(C + 1) = 19*20*21 = 19*4*5*3*7 = Multiple of 12; 2. Say C = 21; thus, (C - 1)C(C + 1) = 20*21*22 = 4*5*3*7*2*11 = Multiple of 12; 3. Say C = 22; thus, (C - 1)C(C + 1) = 21*22*23 = 3*7*2*11*23 = Multiple of 6, not 12; 4. Say C = 23; thus, (C - 1)C(C + 1) = 22*23*24 = 2*11**23*24 = Multiple of 12; If you take another set of four values of C: {23, 24, 25, 26}, you would find that three values would ensure that C^3 - C is divisible by 12 and one would not. So out of 80 integer values of C, 3/4th values would ensure that C^3 - C is divisible by 12 and 1/4th would not. The correct answer: C Hope this helps! -Jay _________________ Manhattan Review GMAT Prep Locations: GRE Manhattan | ACT Tutoring Tampa | GRE Prep Courses Seattle | Boston IELTS Tutoring | and many more... Schedule your free consultation with an experienced GMAT Prep Advisor! Click here. GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2801 messages Followed by: 18 members Upvotes: 43 BTGmoderatorDC wrote: If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ? A. 1/2 B. 2/3 C. 3/4 D. 4/5 E. 1/3 OA C Source: Manhattan Prep We are given that an integer C is to be chosen at random from the integers 20 to 99 inclusive, and we need to determine the probability that C3 - C will be divisible by 12. We should recall that when a number is divisible by 12, it is divisible by 4 and 3. We should also recognize that C3 - C = C(C2 - 1) = C(C - 1)(C + 1) = (C - 1)(C)(C + 1) is a product of three consecutive integers. Furthermore, we should recognize that any product of three consecutive integers is divisible by 3! = 6; thus, it’s divisible by 3. We have to make sure it’s also divisible by 4. Case 1: C is odd If C is odd, then both C - 1 and C + 1 will be even. Moreover, either C - 1 or C + 1 will be divisible by 4. Since C(C - 1)(C + 1) is already divisible by 3 and now we know it’s also divisible by 4, C(C - 1)(C + 1) will be divisible by 12. Since there are 80 integers between 20 and 99 inclusive, and half of those integers are odd, there are 40 odd integers (i.e., 21, 23, 25, …, 99) from 20 to 99 inclusive. Thus, when C is odd, there are 40 instances in which C(C - 1)(C + 1) will be divisible by 12. Case 2: C is even If C is even, the both C - 1 and C + 1 will be odd. If C is a multiple of 4 but not a multiple of 3, then either C - 1 or C + 1 will be divisible by 3 (for example, if C = 28, then C - 1 = 27 is divisible by 3, and if C = 44, then C + 1 = 45 is divisible by 3). In this case, C(C - 1)(C + 1) will be divisible by 12. If C is a multiple 4 and also a multiple of 3, then C is a multiple of 12 and of course C(C - 1)(C + 1) will be divisible by 12. Therefore, if C is even and a multiple of 4, then C(C - 1)(C + 1) will be divisible by 12. So, let’s determine the number of multiples of 4 between 20 and 99 inclusive. Number of multiples of 4 = (96 - 20)/4 + 1 = 76/4 + 1 = 20. Thus, when C is even, there are 20 instances in which C(C - 1)(C + 1) will be divisible by 12. In total, there are 40 + 20 = 60 outcomes in which C(C - 1)(C + 1) will be divisible by 12. Thus, the probability that C(C - 1)(C + 1) will be divisible by 12 is: 60/80 = 6/8 = 3/4. Answer: C • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

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