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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 ## A lecture course consists of 595 students. The students are tagged by: BTGmoderatorLU ##### This topic has 2 expert replies and 0 member replies ### Top Member ## A lecture course consists of 595 students. The students are ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Source: Veritas Prep A lecture course consists of 595 students. The students are to be divided into discussion sections, each with an equal number of students. Which of the following cannot be the number of students in a discussion section? A. 17 B. 35 C. 45 D. 85 E. 119 The OA is C ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12985 messages Followed by: 1249 members Upvotes: 5254 GMAT Score: 770 BTGmoderatorLU wrote: Source: Veritas Prep A lecture course consists of 595 students. The students are to be divided into discussion sections, each with an equal number of students. Which of the following cannot be the number of students in a discussion section? A. 17 B. 35 C. 45 D. 85 E. 119 The OA is C In order to have an EQUAL number of students in each section, the number of students per section MUST BE A FACTOR of 595 Let's do some prime factorization 595 = (5)(7)(17) From the prime factorization, we can see that answer choice A, B, D and E are all factors of 595 Answer: C Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2801 messages Followed by: 18 members Upvotes: 43 BTGmoderatorLU wrote: Source: Veritas Prep A lecture course consists of 595 students. The students are to be divided into discussion sections, each with an equal number of students. Which of the following cannot be the number of students in a discussion section? A. 17 B. 35 C. 45 D. 85 E. 119 The OA is C Any number whose digits sum to a number divisible by 3 is itself divisible by 3. For example, 3,912 is divisible by 3 because the sum of its digits is 3 + 9 + 1 + 2 = 15, which is divisible by 3. Since 5 + 9 + 5 = 19, we see that 595 is not a multiple of 3. Since 45 is a multiple of 3, we cannot have 45 students in a discussion section. Alternate solution: Since 595 = 5 x 119 = 5 x 7 x 17, we see that 17 and 119 obviously can be the number of students in a discussion section, and so can 35 (which is 5 x 7) and 85 (which is 5 x 17). So, by process of elimination, it canâ€™t be 45. Answer: C _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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