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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 n is a positive integer, and k is the product of all integer This topic has 2 expert replies and 0 member replies Top Member n is a positive integer, and k is the product of all integer Timer 00:00 Your Answer A B C D E Global Stats Difficult n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is A. 8 B. 12 C. 16 D. 18 E. 24 OA A Source: Magoosh GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2791 messages Followed by: 18 members Upvotes: 43 BTGmoderatorDC wrote: n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is A. 8 B. 12 C. 16 D. 18 E. 24 OA A Source: Magoosh Let’s break 1440 into prime factors: 1440 = 144 x 10 = 12 x 12 x 10 = 2^5 x 3^2 x 5^1 Thus, k/(2^5 x 3^2 x 5^1) = integer. We also know that k is the product of all integers from 1 to n, inclusive, or, in other words, k = n!. Let’s check our answer choices: A. If n = 8, then k = 8! and 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 = 2^3 x 7 x 3 x 2 x 5 x 2^2 x 3 x 2 = 7 x 5 x 3^2 x 2^7 does contain five 2s, two 3s and one 5. Answer: A GMAT/MBA Expert GMAT Instructor Joined 02 Jun 2008 Posted: 2475 messages Followed by: 350 members Upvotes: 1090 GMAT Score: 780 This is essentially a direct copy of an official question, with one number changed: https://gmatclub.com/forum/if-n-is-a-positive-integer-and-the-product-of-all-integers-90855.html though you might notice how much more elegant the wording of the official problem is (there's no need to introduce two separate letters in this kind of problem). Here, we know n! is divisible by 1440 = (12)^2 * (10) = (2^5)(3^2)(5). So we need n to be large enough so that we have one 5, two 3s, and five 2s among the divisors of the integers from 1 through n. It might be clear if you've done similar kinds of problems, or looked at prime factorizations of factorials, that we won't need an especially large value of n here, so you could just confirm that the smallest answer choice, n=8, works and be done here. But you could also solve without answer choices - as long as n is 5 or greater, we'll have a '5' in the product of n!, and as long as n is 6 or greater, we'll a 3 and a 6 in the product making up n!, so 3^2 will be a divisor of n!. Lastly, we just need to be sure to get five 2s. If n = 6, then n! = 6! is only divisible by 2^4, because we only get twos from 2, 4, and 6. But as long as n is 8 or greater, then 2^5 will easily divide n! (in fact, 2^7 will), so 8 is the smallest possible value of n. _________________ If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • Free Trial & Practice Exam 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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