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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 ## How many trailing zeroes would be found in 63! upon ##### This topic has 2 expert replies and 0 member replies ### Top Member ## How many trailing zeroes would be found in 63! upon ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult How many trailing zeroes would be found in 63! upon expansion? A. 6 B. 12 C. 14 D. 53 E. 57 OA C Source: e-GMAT ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2791 messages Followed by: 18 members Upvotes: 43 Top Reply BTGmoderatorDC wrote: How many trailing zeroes would be found in 63! upon expansion? A. 6 B. 12 C. 14 D. 53 E. 57 OA C Source: e-GMAT The number of trailing zeros in 63! is the number of 2-and-5 pairs it contains since each such pair produces a trailing zero (notice that 2 x 5 = 10). However, since there are more factors of 2 than factors of 5, the number of trailing zeros really depends on the number of factors of 5 in 63!. To find the number of factors of 5 in n!, we can use the following trick: Divide n by 5, then divide the nonzero quotient (ignore any nonzero remainder) by 5, and continue this process until the quotient become 0. Lastly, add these nonzero quotients up, and the sum will be the number of factors of 5 in n!. Letâ€™s use this trick for 63!: 63/5 = 12 R 3 (ignore remainder 3) 12/5 = 2 R 2 (ignore remainder 2) Since 2/5 = 0 R 2, we can stop. Therefore, the number of factors of 5 in 63! is 12 + 2 = 14, and hence there are 14 trailing zeros in 63!. 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 ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15344 messages Followed by: 1864 members Upvotes: 13060 GMAT Score: 790 Top Reply BTGmoderatorDC wrote: How many trailing zeroes would be found in 63! upon expansion? A. 6 B. 12 C. 14 D. 53 E. 57 TRAILING 0's = the number of 0's at the end of a large product. 63! = 63*62*61*....*3*2*1. Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 63! will yield a 0 at the end of the integer representation of 63!. The prime-factorization of 63! includes FAR MORE 2'S than 5's. Thus, the number of 0's depends on the NUMBER OF 5's contained within 63!. To count the number of 5's, simply divide increasing POWERS OF 5 into 63. Every multiple of 5 within 63! provides at least one 5: 63/5 = 12 --> twelve 5's. Every multiple of 5Â² within 63! provides a SECOND 5: 63/5Â² = 2 --> two more 5's. Thus, the total number of 5's contained within 63! = 12+2 = 14. Since each of these 14 5's can serve to produce a trailing zero, the total number of trailing zeros = 14. The correct answer is C. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. • Award-winning private GMAT tutoring Register now and save up to$200

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