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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 the positive integer n is greater than 6, what is the ##### This topic has 2 expert replies and 0 member replies ### Top Member ## If the positive integer n is greater than 6, what is the ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult If the positive integer n is greater than 6, what is the remainder when n is divided by 6? (1) When n is divided by 9, the remainder is 2. (2) When n is divided by 4, the remainder is 1. OA C Source: Manhattan Prep ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12838 messages Followed by: 1247 members Upvotes: 5254 GMAT Score: 770 Top Reply BTGmoderatorDC wrote: If the positive integer n is greater than 6, what is the remainder when n is divided by 6? (1) When n is divided by 9, the remainder is 2. (2) When n is divided by 4, the remainder is 1. Here's a somewhat long approach (but it works!): Target question: What is the remainder when n is divided by 6? Given: Positive integer n is greater than 6 Statement 1: When n is divided by 9, the remainder is 2. ------ASIDE---------------- When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. ------------------------------ So, from statement 1, we can conclude that some possible values of n are: 11, 20, 29, 38, 47, 56, 65. . . (since we're told n > 6, we can rule out n = 2) . Let's test some possible values of n: Case a: n = 11. When we divide 11 by 6, the remainder is 5 Case b: n = 20. When we divide 20 by 6, the remainder is 2 Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2: When n is divided by 4, the remainder is 1. from statement 2, we can conclude that some possible values of n are: 9, 13, 17, 21, 25, 29, 33, 37, 41, . . . (since we're told n > 6, we can rule out n = 1 and n = 5) . Let's test some possible values of n: Case a: n = 9. When we divide 9 by 6, the remainder is 3 Case b: n = 13. When we divide 13 by 6, the remainder is 1 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Statements 1 and 2 combined Statement 1 tells us that some possible values of n are: 11, 20, 29, 38, 47, 56, 65, . . . Statement 2 tells us that some possible values of n are: 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69 . . . From the above, we can see that some values of n that satisfy BOTH statements are: 29, 65, etc We might also notice that if we keep adding 36 (the least common multiple of 9 and 4) to these possible n-values, we'll get more possible n-values such as 101, 137, 173, etc Let's test some possible values of n: Case a: n = 29. When we divide 29 by 6, the remainder is 5 Case b: n = 65. When we divide 65 by 6, the remainder is 5 Case c: n = 101. When we divide 101 by 6, the remainder is 5 Case d: n = 137. When we divide 137 by 6, the remainder is 5 So, it certainly looks like the remainder will always be 5. Since we can answer the target question with certainty, the combined statements are SUFFICIENT Answer: C Cheers, Brent _________________ Brent Hanneson â€“ 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 GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 Top Reply BTGmoderatorDC wrote: If the positive integer n is greater than 6, what is the remainder when n is divided by 6? (1) When n is divided by 9, the remainder is 2. (2) When n is divided by 4, the remainder is 1. Source: Manhattan Prep $n \geqslant 7\,\,\,\operatorname{int} \,\,\,\,\left( * \right)$ $$\left. \matrix{ n = 6Q + R\,\,\, \hfill \cr Q\,\,\mathop \ge \limits^{\left( * \right)} \,\,1\,\,{\mathop{\rm int}} \hfill \cr} \right\}\,\,\,\,\,? = R\,\,\,\,\,\,\left( {0 \le R \le 5\,\,{\mathop{\rm int}} } \right)$$ $\left( 1 \right)\,\,\,n = 9K + 2\,\,\,,\,\,\,K\,\mathop \geqslant \limits^{\left( * \right)} \,1\,\,\operatorname{int} \,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,K = 1\,\,\,\, \Rightarrow \,\,\,n = 11\,\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{5}}\,\,\,\,\,\left( {Q = 1} \right)\, \hfill \\ \,{\text{Take}}\,\,K = 2\,\,\,\, \Rightarrow \,\,\,n = 20\,\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{2}}\,\,\,\,\left( {Q = 3} \right) \hfill \\ \end{gathered} \right.$ $\left( 2 \right)\,\,\,n = 4J + 1\,\,\,,\,\,\,J\,\mathop \geqslant \limits^{\left( * \right)} \,2\,\,\operatorname{int} \,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,J = 2\,\,\,\, \Rightarrow \,\,\,n = 9\,\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{3}}\,\,\,\,\,\left( {Q = 1} \right)\, \hfill \\ \,{\text{Take}}\,\,J = 3\,\,\,\, \Rightarrow \,\,\,n = 13\,\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{1}}\,\,\,\,\left( {Q = 2} \right) \hfill \\ \end{gathered} \right.$ $$\left( {1 + 2} \right)\,\,\,\,9K = n - 2\,\,\, = \,\,4J - 1\,\,\,\,\, \Rightarrow \,\,\,\,\,n - 2\,\,\,\,\left\{ \matrix{ \,{\rm{odd}}\,\,\,\, \Rightarrow \,\,\,n = {\rm{odd}}\,\,\,\, \Rightarrow \,\,\,R \in \left\{ {1,3,5} \right\} \hfill \cr \,{\rm{multiple}}\,\,{\rm{of}}\,\,9,\,\,{\rm{hence}}\,\,{\rm{of}}\,\,3\,\,\,\, \Rightarrow \,\,\,n = 3M + 2 \hfill \cr} \right.\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,\,\,? = 5$$ $$\left( {**} \right)\,\,\,\,\left\{ \matrix{ \,R = 1\,\,\,\, \Rightarrow \,\,\,\,n = 6Q + 1\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 3\left( {2Q} \right) + 1\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 3M + 2\,\,\,\,\,{\rm{impossible}} \hfill \cr \,R = 3\,\,\,\, \Rightarrow \,\,\,\,n = 6Q + 3\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 3\left( {2Q + 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 3M + 2\,\,\,\,\,{\rm{impossible}} \hfill \cr} \right.\,\,\,\,\,\,\,\,$$ 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 • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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