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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 ## What is the biggest positive integer which will always divid tagged by: GMATinsight ##### This topic has 2 expert replies and 3 member replies ### Top Member ## What is the biggest positive integer which will always divid ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult What is the biggest positive integer which will always divide n(n+1)(n+2) evenly for any even integer value of n? a) 2 b) 3 c) 6 d) 12 e) 24 Source: www.GMATinsight.com _________________ Bhoopendra Singh & Sushma Jha - Founder "GMATinsight" Testimonials e-mail: info@GMATinsight.com I Mobile: +91-9999687183 / +91-9891333772 To register for One-on-One FREE ONLINE DEMO Class Call/e-mail One-On-One Private tutoring fee - US$40 per hour & for FULL COURSE (38 LIVE Sessions)-US$1000 ### Top Member Master | Next Rank: 500 Posts Joined 15 Oct 2009 Posted: 325 messages Upvotes: 27 GMATinsight wrote: What is the biggest positive integer which will always divide n(n+1)(n+2) evenly for any even integer value of n? a) 2 b) 3 c) 6 d) 12 e) 24 Source: www.GMATinsight.com To find the biggest number, we must minimize the expression, so let's use 1, 2 and 3. 1x2x3 = 6 is the largest number that will ALWAYS divide the expression. We know that 6 will always divide the expression regardless of choice of n as follows; To be divisible by 6 it needs to be divisible by both 2 and 3. Let's assume n is even. That means it can be written as n=2K. Rewriting the expression: 2K(2K+1)(2K+2) > this is clearly divisible by 2. But is it divisible by 3 ? . Let's assume 2K is not divisible by 3, because if it were, then we would be done. So 2K equals 3 times some number X plus a remainder of 1 or 2, or 3X + 1 or 2 Try 3X+1: (3X+1)(3X+2)(3X+3) = 3(3X+1)(3X+2)(X+1) > clearly divisible by 3 Now 3X+2: (3X+2)(3X+3)(3X+4). Notice that same 3X+3 as above, so also divisible by 3. So now we know that with n being even, the expression is divisible by 3 and 2, or 6. Testing n being odd means n = 2K+1 . Rewriting the original expression: (2K+1)(2K+2)(2K+3) = 2(2K+1)(K+1)(2k+3) > Clearly divisible by 2. But is it divisible by 3 ? Following logic above, (2K+1)= 3X plus a remainder of 1 or 2. and the expression is the same as above, establishing that with an odd n, it is also divisible by 3 So n(n+1)(n+2) is always divisible by 2 and 3, or 6 ### Top Member Legendary Member Joined 10 May 2014 Posted: 1074 messages Followed by: 23 members Upvotes: 205 regor60 wrote: GMATinsight wrote: What is the biggest positive integer which will always divide n(n+1)(n+2) evenly for any even integer value of n? a) 2 b) 3 c) 6 d) 12 e) 24 Source: www.GMATinsight.com To find the biggest number, we must minimize the expression, so let's use 1, 2 and 3. 1x2x3 = 6 is the largest number that will ALWAYS divide the expression. We know that 6 will always divide the expression regardless of choice of n as follows; To be divisible by 6 it needs to be divisible by both 2 and 3. Let's assume n is even. That means it can be written as n=2K. Rewriting the expression: 2K(2K+1)(2K+2) > this is clearly divisible by 2. But is it divisible by 3 ? . Let's assume 2K is not divisible by 3, because if it were, then we would be done. So 2K equals 3 times some number X plus a remainder of 1 or 2, or 3X + 1 or 2 Try 3X+1: (3X+1)(3X+2)(3X+3) = 3(3X+1)(3X+2)(X+1) > clearly divisible by 3 Now 3X+2: (3X+2)(3X+3)(3X+4). Notice that same 3X+3 as above, so also divisible by 3. So now we know that with n being even, the expression is divisible by 3 and 2, or 6. Testing n being odd means n = 2K+1 . Rewriting the original expression: (2K+1)(2K+2)(2K+3) = 2(2K+1)(K+1)(2k+3) > Clearly divisible by 2. But is it divisible by 3 ? Following logic above, (2K+1)= 3X plus a remainder of 1 or 2. and the expression is the same as above, establishing that with an odd n, it is also divisible by 3 So n(n+1)(n+2) is always divisible by 2 and 3, or 6 Read the question carefully... n is even integer hence answer is Option E _________________ Bhoopendra Singh & Sushma Jha - Founder "GMATinsight" Testimonials e-mail: info@GMATinsight.com I Mobile: +91-9999687183 / +91-9891333772 To register for One-on-One FREE ONLINE DEMO Class Call/e-mail One-On-One Private tutoring fee - US$40 per hour & for FULL COURSE (38 LIVE Sessions)-US$1000 ### Top Member Master | Next Rank: 500 Posts Joined 15 Oct 2009 Posted: 325 messages Upvotes: 27 GMATinsight wrote: Read the question carefully... n is even integer hence answer is Option E Yes, thanks, read the two evens in an odd manner ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12663 messages Followed by: 1246 members Upvotes: 5254 GMAT Score: 770 GMATinsight wrote: What is the biggest positive integer which will always divide n(n+1)(n+2) evenly for any even integer value of n? a) 2 b) 3 c) 6 d) 12 e) 24 One approach is to list integers in the form n, n+1 and n+2 such that n is even, and look for a pattern... 2, 3, 4 Product = 24, which is divisible by 2, 3, 6, 12 and 24 4, 5, 6 Product = 120, which is divisible by 2, 3, 6, 12 and 24 6, 7, 8 Product = 336, which is divisible by 2, 3, 6, 12 and 24 8, 9, 10 Product = 720, which is divisible by 2, 3, 6, 12 and 24 We probably have enough information to conclude that the correct answer is E Cheers, Brent _________________ Brent Hanneson – Creator of GMATPrepNow.com Use our video course along with Sign up for our free Question of the Day emails And check out all of our free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1296 messages Followed by: 29 members Upvotes: 59 GMATinsight wrote: What is the biggest positive integer which will always divide n(n+1)(n+2) evenly for any even integer value of n? a) 2 b) 3 c) 6 d) 12 e) 24 Source: www.GMATinsight.com $n = 2M\,\,,\,\,\,\,M\,\,\operatorname{int}$ $\frac{{2M\left( {2M + 1} \right)\left( {2M + 2} \right)}}{{? = \max \,\,\operatorname{int} }}\,\,\, = \operatorname{int}$ (1) Exactly one of the factors among 2M , (2M+1) and (2M+2) is divisible by 3. (They are three consecutive integers!) We guarantee (at least) one factor 3. (Nine could be 2M+1, therefore more than one factor 3 is possible...) (2) 2M , 2M+1 , 2(M+1) is even, odd, even :: in the product of these 3 factors there are: two factors 2 plus another factor 2 (we have M and M+1 consecutive integers... one of them is even)! We guarantee (at least) three factors 2. We have already found 24 (one factor 3, three factors 2), and the maximum available alternative choice is exactly 24... we are done! This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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