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# Is the integer n odd? (1) n is divisible by 3. (2) 2n is

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Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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05 Jan 2010, 22:35
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Is the integer n odd?

(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n
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Math Expert
Joined: 02 Sep 2009
Posts: 47923
Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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06 Jan 2010, 04:56
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Is the integer n odd ?

(1) n is divisible by 3. Clearly not sufficient. Consider n = 3 and n = 6.

(2) 2n is divisible by twice as many positive integers as n.

TIP:
When odd number n is doubled, 2n has twice as many factors as n.
That's because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2.

Sufficient.

Answer: B.

For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html
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Re: Is N ODD  [#permalink]

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14 Mar 2010, 03:28
Bunuel wrote:
Is the integer n odd
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

(1) 3 or 6. Clearly not sufficient.

(2) TIP:
When odd number n is doubled, 2n has twice as many factors as n.
Thats because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2.

Sufficient.

Answer: B.

P.S. You can attach the screenshot of a question directly to the post so that everyone will see it.

Nice explanation. Didnt know abt this rule.
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Re: Is the integer n odd  [#permalink]

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10 Dec 2010, 06:42
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ajit257 wrote:
Is the integer n odd?
(1) n is divisible by 3.
(2) 2n is divisible by twice as many positive integers as n.

What do you mean by the second statement.

Merging similar topics.

2n is divisible by twice as many positive integers as n, means that the # of factors of 2n is twice the # of factors of n.

For example: # of factors of 3 is two (1, and 3 itself) and the # of factors of 2*3=6 is four (1, 2, 3, and 6 itself), so the # of factors of 6=2n is twice the # of factors of 3=n.

Hope it's clear.
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Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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12 Dec 2010, 05:12
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ajit257 wrote:
Is the integer n odd?
(1) n is divisible by 3.
(2) 2n is divisible by twice as many positive integers as n.

What do you mean by the second statement.

If $$n = a^p*b^q*c^r...$$ where a, b, c are distinct prime numbers,
Total number of factors of n = (p+1)(q+1)(r+1)...

(Check out the post: http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-writing-factors-of-an-ugly-number/ for complete explanation of this formula)

Let us assume $$n = 3^2*5^3$$ i.e. odd
Total number of factors of n = (2+1)(3+1) = 12

$$2n = 2^1*3^2*5^3$$
Total number of factors of n = (1+1)(2+1)(3+1) = 24 (Twice of 12 obtained before because of additional 2)

Now assume the case where n is already even:

$$n = 2^2*3^2*5^3$$ i.e. even
Total number of factors of n = (2+1)(2+1)(3+1) = 36

$$2n = 2^3*3^2*5^3$$
Total number of factors of n = (3+1)(2+1)(3+1) = 48 (More than 36 but not double because 3 is replaced by 4)

This is true for any prime factor. If that prime factor, p, is not in n, then p*n will have double the total number of factors.
If p is already in n, the total number of factors will increase but will never double.

*Edited*
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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Senior Manager Status: Bring the Rain Joined: 17 Aug 2010 Posts: 358 Location: United States (MD) Concentration: Strategy, Marketing Schools: Michigan (Ross) - Class of 2014 GMAT 1: 730 Q49 V39 GPA: 3.13 WE: Corporate Finance (Aerospace and Defense) Re: Is N ODD [#permalink] ### Show Tags 13 Dec 2010, 06:36 Thanks for the tip. _________________ Senior Manager Joined: 13 Mar 2012 Posts: 289 Concentration: Operations, Strategy Re: Prep Question....is n odd? [#permalink] ### Show Tags 16 Apr 2012, 01:12 some2none wrote: Bunuel wrote: Is the integer n odd (When even number is doubled, 2n has 1.5 more factors as n.) Sufficient. I'm afraid the above statement is incorrect. Let's take an even number, 4. The number of factors = 3 (1, 2, 4) If the statement above is true, then the 2n (4 * 2 = 8) will have 1.5 more factors, that is 3 + 1.5 = 4.5 factors. Instead the number of factors for 8 = 4 (1, 2, 4, 8) I tried to find out certain relationship between a even number and its double: If an even number is E = 2^x * p^a* q^b* r^c........ 2E= 2^(x+1) * p^a* q^b* r^c........ The number of factors that 2E has more than E is (a+1)*(b+1)* (c+1)*......... PS: request Bunuel to check whether the observation i derived is correct. thanx _________________ Practice Practice and practice...!! If my reply /analysis is helpful-->please press KUDOS If there's a loophole in my analysis--> suggest measures to make it airtight. Intern Joined: 26 Jul 2010 Posts: 24 Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is [#permalink] ### Show Tags 04 May 2012, 11:40 Quote: (2) TIP: When odd number n is doubled, 2n has twice as many factors as n. Thats because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2. Bunuel, could you please explain the above TIP with an example. If n=2 and n=15? Thanks. Math Expert Joined: 02 Sep 2009 Posts: 47923 Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is [#permalink] ### Show Tags 04 May 2012, 11:56 pgmat wrote: Quote: (2) TIP: When odd number n is doubled, 2n has twice as many factors as n. Thats because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2. Bunuel, could you please explain the above TIP with an example. If n=2 and n=15? Thanks. 15 has 4 factors: 1, 3, 5, and 15; 15*2=30 has 4*2=8 factors: 1, 1*2=2, 3, 3*2=6, 5, 5*2=10, 15, and 15*2=30. _________________ Senior Manager Joined: 13 Aug 2012 Posts: 441 Concentration: Marketing, Finance GPA: 3.23 Is the integer n odd? (1) n is divisible by 3. (2) 2n is [#permalink] ### Show Tags 23 Jan 2013, 00:24 (1) $$n = 3*I$$ If I is even then n is EVEN... If I is odd then n is ODD... INSUFFICIENT! 2) Let f(n) be the number of factors of n... Let f(2n) be the number of factors of 2n... Now let f(n) be some X ==> f(n) = R The statement says that f(2n) = 2*f(n) = 2*R... When n is EVEN: $$2n = 2^(p+1) * {n/2^{p}}$$where p is the number factor of 2 in n... $$f(2n) = f(2^{p+1}) * f( {n/2^{p}} ) = (p+2) * R/{p+1}$$ let p=1: f(2n) = 3 * R/2 let p=2: f(2n) = 4 * R/3 let p=3: f(2n) = 4 * R/4 When n is ODD: 2n = 2^1 * n f(2n) = 2 * f(n) = 2*R SUFFICIENT that n is ODD! Answer: B _________________ Impossible is nothing to God. Senior Manager Status: Gonna rock this time!!! Joined: 22 Jul 2012 Posts: 477 Location: India GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29 WE: Information Technology (Computer Software) Re: Prep Question....is n odd? [#permalink] ### Show Tags 24 Jan 2013, 23:53 Bunuel wrote: Is the integer n odd (1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n (1) 3 or 6. Clearly not sufficient. (2) TIP: When odd number n is doubled, 2n has twice as many factors as n. Thats because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2. Sufficient. Answer: B. For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html P.S. You can attach the screenshot of a question directly to the post so that everyone will see it. So The following doesn't happen for even numbers? When odd number n is doubled, 2n has twice as many factors as n. Thats because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2. _________________ hope is a good thing, maybe the best of things. And no good thing ever dies. Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595 My GMAT Journey : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992 Senior Manager Joined: 27 Jun 2012 Posts: 390 Concentration: Strategy, Finance Schools: Haas EWMBA '17 Re: Prep Question....is n odd? [#permalink] ### Show Tags 25 Jan 2013, 00:48 2 1 Sachin9 wrote: Bunuel wrote: Is the integer n odd (1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n (1) 3 or 6. Clearly not sufficient. (2) TIP: When odd number n is doubled, 2n has twice as many factors as n. Thats because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2. Sufficient. Answer: B. For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html P.S. You can attach the screenshot of a question directly to the post so that everyone will see it. So The following doesn't happen for even numbers? When odd number n is doubled, 2n has twice as many factors as n. Thats because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2. Yes it doesn't follow for even numbers. We can prove it using the formula: First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers. The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself. ODD Number: Finding the number of all factors of 45: $$45= 3^2*5^1$$ Total number of factors of 45 including 1 and 45 itself is $$((2+1)*(1+1)=3*2=6$$ factors. Finding the number of all factors of 90: $$90=2^1*3^2*5^1$$ Total number of factors of 90 including 1 and 90 itself is $$(1+1)*(2+1)*(1+1)=2*3*2=12$$ factors. Note that, the odd numbers have only odd prime factors. If you double it then you introduce factor of 2^1 in the prime factorization and hence you end up multiplying by (1+1) when finding total number of factors, which therefore gets doubled. EVEN Number: However, the same is not true for EVEN numbers. They already have prime factorization with 2^x (x>=1) and doubling that EVEN number only increments the exponent of factor 2, but not necessarily doubles the number of factors. Finding the number of all factors of 12: $$12= 2^2*3^1$$ Total number of factors of 12 including 1 and 12 itself is $$((2+1)*(1+1)=3*2=6$$ factors. Finding the number of all factors of 24: $$24= 2^3*3^1$$ Total number of factors of 24 including 1 and 24 itself is $$((3+1)*(1+1)=4*2=8$$ factors. _________________ Thanks, Prashant Ponde Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7 Reading Comprehension notes: Click here VOTE GMAT Practice Tests: Vote Here PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here Looking to finance your tuition: Click here Manager Status: *Lost and found* Joined: 25 Feb 2013 Posts: 122 Location: India Concentration: General Management, Technology GMAT 1: 640 Q42 V37 GPA: 3.5 WE: Web Development (Computer Software) Re: Number Properties [#permalink] ### Show Tags Updated on: 13 May 2013, 10:51 coolpintu wrote: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n I feel [B] is the answer. Please find below my explanation! Statement 1. It states that n is divisible by 3. Doesn't really prove conclusively that n is odd. n could be divisible by 6. Hence not sufficient. Statement 2. It states that 2n has twice the factors as n. Now, assuming n as odd, 2n would have the below factors {1,n,2n,2...} and n would have {1,n} as its factors. The extra factors appear due to the fact that 2 was multiplied. In case n is not a prime and has odd factors, like 3 and 7 in it, the value will double since 2 is multiplies to individual odd factors leading to double the factors again. For example if n = 3, 2n = 6, the factors double. It doesn't happen in case of n = 4. Once can similarly test the logic for complex numbers like 35. Hence sufficient. Hope I my explanation is satisfactory and correct! Regards, Arpan _________________ Feed me some KUDOS! *always hungry* My Thread : Recommendation Letters Originally posted by arpanpatnaik on 13 May 2013, 10:17. Last edited by arpanpatnaik on 13 May 2013, 10:51, edited 1 time in total. Manager Joined: 11 Jun 2010 Posts: 74 Re: Number Properties [#permalink] ### Show Tags 13 May 2013, 10:34 St1: N is divisible by 3 N can be 6 (even) or 9 (odd) Hence Insufficient St2: 2n is divisible by twice as many divisors as N if N = 3, divisors = {1,3} that is # divisors = 2 2N = 6 = {1,2,3,6} and divisors = 4 Good solution. Lets check for an even number also. N = 6 {1,2,3,6} and #divisors = 4 2N = 12 = {1,2,3,4,6,12} and #divisors = 6 Not a good solution as #divisors for 2N is not twice #divisors for N Hence sufficient to say that if 2n is divisible by twice as many divisors as N, then N = ODD Ans B Intern Joined: 05 Mar 2013 Posts: 44 Location: India Concentration: Entrepreneurship, Marketing GMAT Date: 06-05-2013 GPA: 3.2 Re: Number Properties [#permalink] ### Show Tags 13 May 2013, 23:14 coolpintu wrote: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n AD/BCE statement 1:- n is divisible by three.If n is 6 it is even and divisible by three .If n is 9 it is divisible by three. So A alone is not sufficient. AD out statement 2:- number of factors for n = $$a^p * b^q *c^r$$ ....where a,b,c.... are prime numbers and p,q,r.... are positive integers is given by$$(p+1)*(q+1)*(r+1)$$.... Now 2n has twice as many positive integers as n. This is only possible if n is odd.Why??IF n is odd then n doesn't have a 2 .But for 2n n will have a 2 and we will get (1+1) in our formula which doubles the number of factors.But if n is even then we will already have 2 .Lets say the number if 2's is p.The for 2n we will have p+1 2's which doesn't double the number of factors. So n must be odd for this condition to be true.So statement 2 is sufficient. B is the answer _________________ "Kudos" will help me a lot!!!!!!Please donate some!!! Completed Official Quant Review OG - Quant In Progress Official Verbal Review OG 13th ed MGMAT IR AWA Structure Yet to do 100 700+ SC questions MR Verbal MR Quant Verbal is a ghost. Cant find head and tail of it. Manager Joined: 04 Oct 2013 Posts: 173 Concentration: Finance, Leadership GMAT 1: 590 Q40 V30 GMAT 2: 730 Q49 V40 WE: Project Management (Entertainment and Sports) Re: Number Properties [#permalink] ### Show Tags 07 Jan 2014, 02:11 1. is clearly insufficient. n could be zero for instance and 0 is both even and a multiple of 3. 2. I figured this out pretty quickly, I started off visualizing n as a prime number because prime numbers have 2 factors. Moreover 2 is a prime number itself and has two factors as well, then I tried to visualize the same thing assuming n to be an odd number. Generally n must not be an even number because any even number will cause our "twice as many positive integer" rule to fall. since we are talking about factors: squares of prime numbers have 3 divisors -the prime number itself, its square, and one. Sufficient. _________________ learn the rules of the game, then play better than anyone else. BSchool Forum Moderator Joined: 12 Aug 2015 Posts: 2649 GRE 1: Q169 V154 Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is [#permalink] ### Show Tags 14 Jan 2017, 06:32 Great Official Question. Here is what i did in this one => We are asked if n is odd or not. We are told that n is an integer. Statement 1-> n/3=integer. E.g=> n=3 or n=6 etc. Hence not sufficient. Statement 2-> If n is odd => The number of factors of 2n will always be 2 times the number of factors of n. This is because any odd number will just have odd divisors.It cannot have any even divisor as 2 is not its prime factor. After we multiple 2 to n => The numb of factors will get doubled as for every odd factor there would be an even factor. Hence n is odd Thus-> This statement is sufficient. Hence B. _________________ MBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs! STONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+) AVERAGE GRE Scores At The Top Business Schools! Intern Joined: 30 Nov 2017 Posts: 43 Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is [#permalink] ### Show Tags 21 May 2018, 07:09 VeritasPrepKarishma wrote: Now assume the case where n is already even: $$n = 2^2*3^2*5^3$$ i.e. odd Just so that my understanding is correct, you meant "i.e. even" and not "i.e.odd". Is my understanding correct? DS Forum Moderator Joined: 22 Aug 2013 Posts: 1340 Location: India Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is [#permalink] ### Show Tags 21 May 2018, 11:10 1 Nived wrote: VeritasPrepKarishma wrote: Now assume the case where n is already even: $$n = 2^2*3^2*5^3$$ i.e. odd Just so that my understanding is correct, you meant "i.e. even" and not "i.e.odd". Is my understanding correct? Hello I think that was a typo. You could actually ignore the phrase "i.e.odd".. and the explanation will make perfect sense. And yes your understanding is correct - the number is obviously even. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8185 Location: Pune, India Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is [#permalink] ### Show Tags 21 May 2018, 23:10 Nived wrote: VeritasPrepKarishma wrote: Now assume the case where n is already even: $$n = 2^2*3^2*5^3$$ i.e. odd Just so that my understanding is correct, you meant "i.e. even" and not "i.e.odd". Is my understanding correct? Yes, even. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >

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Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is &nbs [#permalink] 21 May 2018, 23:10

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