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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 ## Is N Odd tagged by: ##### This topic has 1 expert reply and 12 member replies This is a GMATPrep problem Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n For some reason the spoiler feature is not working for me. So, OA is the answer choice # 3+5-1+2-7 Senior | Next Rank: 100 Posts Joined 13 Oct 2009 Posted: 43 messages yankee.musk wrote: This is a GMATPrep problem Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n For some reason the spoiler feature is not working for me. So, OA is the answer choice # 3+5-1+2-7 (1) is not enough since it could be either 3, 6, 9... (2) gonna do some tests 3*2=6. 6 is divisible by 1,2,3,6, 3 is divisible by 1,3 63*2=126 63 is divisible by 1, 3, 7, 9, 21, 63 1,2,3,7,9,21,63,126 so it doesn't tell us whether or not it's odd. this should also be impossible to tell with positive numbers. We can tell though, that it will only double the number of multiples. The total number of multiple will be 4 probably, because by multiplying a number by 2, you're adding 2 extra multiples (2, if 2 wasn't already a multiple) and the number that equals 2n. so let's take a look at the multiples of 3. 3, 6, 9, 12. lets look at their multiples. 1,3 1,2,3,6 1,3,9 1,2,3,4,6,12 together, with (1) and (2), the multiples of 2n are 1,2,3,6 1,2,3,4,6,12 1,2,3,6,9,18 1,2,3,4,6,8,12,24 as you can see, the answer could be either 9 or 3, so we can tell that together, we can't solve the problem, so E would be the answer. Legendary Member Joined 29 Dec 2009 Posted: 1460 messages Followed by: 7 members Upvotes: 135 From stmt1,n can be 3,6,9,12. So A insufficient. From stmt2,2n is divisible by twice as many positive integers as n. We need to check this with n as odd and n as even. 1)n is odd. n=21 n is divisible by 1,3,7,21 2n=42 = 2*(21)=2*(3*7) 42 is divisible by 1,2,3,6,7,14,21,42. 2n is divisible by twice as many positive integers as n when n is odd. 2) n is even n=30 n=2*3*5 30 is divisible by 1,2,3,5,6,10,15,30 2n=2*(30)=2*(2*3*5) 60 is divisible by 1,2,3,4,5,6,10,12,15,30,60. When n is even 2n ,2n is not divisible by twice as many positive integers as n. So n is odd. B is sufficient. Answer is B Junior | Next Rank: 30 Posts Joined 23 Feb 2007 Posted: 15 messages Test Date: September 2010 Target GMAT Score: 720 GMAT Score: 610 Thank you all for the explanations By the way, how do guys use spoiler feature in BTG. I'm having trouble with it. Master | Next Rank: 500 Posts Joined 28 May 2010 Posted: 186 messages Upvotes: 11 Talkativetree wrote: 3*2=6. 6 is divisible by 1,2,3,6, 3 is divisible by 1,3 63*2=126 63 is divisible by 1, 3, 7, 9, 21, 63 1,2,3,7,9,21,63,126 Hey, how'd you happen to pick on 63 as the no.? I tried till 21 & stmt 2 seemed to be working till there Master | Next Rank: 500 Posts Joined 28 May 2010 Posted: 186 messages Upvotes: 11 I'm wondering if we can do it like this? A & B are out of picture based on what Talkative tree said Considering both the statements means: n=3k 2n=6k for k=2: n=6 which has factors 1,2,3,6 i.e. 4 factors 2n=12 which has factors 1,2,3,4,6 i.e. 6 factors which contradicts stmnt 2 - hence the 2 statements taken together is void. therefore, answer is E Legendary Member Joined 29 Dec 2009 Posted: 1460 messages Followed by: 7 members Upvotes: 135 Jube, U tried the stmt 2 with n as even. with n as even,it contradicts stmt2. Let consider n as odd. n=9 9 is divisible by 1,3,9 2n=18=2*3*6 18 is divisible by 1,2,3,6,9,18 From this you can see that 2n is divisible by twice as many integers as n. We dont need n to be a multiple of 3. Take any odd and even number and you can check with stmt2 When n is odd only stmt2 can be true. Hence B is sufficient. Master | Next Rank: 500 Posts Joined 28 May 2010 Posted: 186 messages Upvotes: 11 selango wrote: Jube, U tried the stmt 2 with n as even. with n as even,it contradicts stmt2. Let consider n as odd. n=9 9 is divisible by 1,3,9 2n=18=2*3*6 18 is divisible by 1,2,3,6,9,18 From this you can see that 2n is divisible by twice as many integers as n. We dont need n to be a multiple of 3. Take any odd and even number and you can check with stmt2 When n is odd only stmt2 can be true. Hence B is sufficient. I see what you're saying. You look to be right! Thanks for pointing this out. Was just wondering if there's another way of solving this one other than plugging values because with values you don't know for sure that there's some value out there that you couldn't think of which would contradict this statement. Legendary Member Joined 13 Mar 2010 Posted: 576 messages Followed by: 1 members Upvotes: 97 From 1 clearly n can be either even or odd. From 2, 2n has twice as many positive factor as n ,so 2n must have one more prime factor than n as for any number N=p^a*q^b the number of factors is(a+1)(b+1) only prime factor that 2n can have other than all prime factors of n is 2..hence n is odd..sufficient Ans option B _________________ “If you don't know where you are going, any road will get you there.” Lewis Carroll Master | Next Rank: 500 Posts Joined 28 May 2010 Posted: 186 messages Upvotes: 11 liferocks wrote: 2n has twice as many positive factor as n ,so 2n must have one more prime factor than n as for any number N=p^a*q^b the number of factors is(a+1)(b+1) Two questions: 1) I got the formula for calculating the no. of factors but why must there be only 1 more prime factor? Why not more than 1? 2) How do you know so much about no. properties? Is it something you've built up over the years or do you have a resource which mortals like me can use too? Legendary Member Joined 13 Mar 2010 Posted: 576 messages Followed by: 1 members Upvotes: 97 jube wrote: liferocks wrote: 2n has twice as many positive factor as n ,so 2n must have one more prime factor than n as for any number N=p^a*q^b the number of factors is(a+1)(b+1) Two questions: 1) I got the formula for calculating the no. of factors but why must there be only 1 more prime factor? Why not more than 1? 2) How do you know so much about no. properties? Is it something you've built up over the years or do you have a resource which mortals like me can use too? for the question 1, let n=p^a*q^b so number of factors of n=(a+1)(b+1) and number of factors for 2n=2*(a+1)(b+1) or 2n=(1+1)*(a+1)(b+1) so 2n has one more prime factor than n has For question 2,I am not using any specific resource..just brushed up the basics..and different forum discussions really helps! _________________ “If you don't know where you are going, any road will get you there.” Lewis Carroll Master | Next Rank: 500 Posts Joined 28 May 2010 Posted: 186 messages Upvotes: 11 liferocks wrote: for the question 1, let n=p^a*q^b so number of factors of n=(a+1)(b+1) and number of factors for 2n=2*(a+1)(b+1) or 2n=(1+1)*(a+1)(b+1) so 2n has one more prime factor than n has For question 2,I am not using any specific resource..just brushed up the basics..and different forum discussions really helps! thanks! makes sense. what other forums do you use? Legendary Member Joined 13 Mar 2010 Posted: 576 messages Followed by: 1 members Upvotes: 97 jube wrote: liferocks wrote: for the question 1, let n=p^a*q^b so number of factors of n=(a+1)(b+1) and number of factors for 2n=2*(a+1)(b+1) or 2n=(1+1)*(a+1)(b+1) so 2n has one more prime factor than n has For question 2,I am not using any specific resource..just brushed up the basics..and different forum discussions really helps! thanks! makes sense. what other forums do you use? As of now only btg..I should rephrase my sentence 'different forum discussions really helps' as 'different discussions in the forum really helps' _________________ “If you don't know where you are going, any road will get you there.” Lewis Carroll ### GMAT/MBA Expert GMAT Instructor Joined 28 May 2009 Posted: 1578 messages Followed by: 34 members Upvotes: 128 GMAT Score: 760 Is there a number properties rule that will help me think about this problem or is there an algebraic way to solve this. I mean, I really hate having to just pick numbers, but is that the only way to do this? _________________ https://www.beatthegmat.com/the-retake-osirus-blog-t51414.html Brandon Dorsey GMAT Instructor Veritas Prep Buy any Veritas Prep book(s) and receive access to 5 Practice Cats for free! Learn More. Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now! • 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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