Is integer n odd

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Is integer n odd

by aatech » Sat May 31, 2008 11:08 am
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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by netigen » Sat May 31, 2008 1:16 pm
Assuming some factors:

n = 1 x n
2n = 1 x 2n x 2 x n

(B) hold true only if the factors are as shown above

which means n = prime

note n=2 does not hold good for B hence n cannot be 2

so B is sufficient to ans the question

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by chidcguy » Sat May 31, 2008 1:54 pm
Does n need to be prime?

Take 9 and 18 as n and 2n, which satisfy the condition in (B)

9 has factors 1, 3, 9 and 18 has 1,2,3,6,9,18

How ever I agree that I arrived at B by brute force of taking a bunch of sets of n,2n

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by netigen » Sat May 31, 2008 2:23 pm
you are right n need not be prime.

I should have cross checked my assumption with A which tells me that n can not be prime :) until and unless n=3

I think the conclusion is that n=odd (not just prime)

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by aatech » Sat May 31, 2008 4:40 pm
OA is B.. Thanks guys

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by aj5105 » Fri Jan 16, 2009 9:16 am
anybody more on this problem,please?

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by Ian Stewart » Fri Jan 16, 2009 10:29 am
Statement 1 is irrelevant.

Statement 2 is trickier than most GMAT DS statements. You might persuade yourself it's sufficient by picking a few numbers, but if you want to be sure of your answer, it's probably easiest to see why Statement 2 is sufficient by looking at numerical examples, and seeing how we can list all of the divisors of a number. Let's first take an odd number, say n = 45 = 3^2 * 5. n has six divisors, all of course odd:

1, 3, 5, (3^2), (3*5), (3^2 * 5) = 1, 3, 5, 9, 15, 45

Now, 2n will have all of those odd divisors, but will have just as many even divisors: you find the even divisors of 2n by doubling all of the odd divisors:

1, 3, 5, (3^2), (3*5), (3^2 * 5) = 1, 3, 5, 9, 15, 45
2*1, 2*3, 2*5, 2*(3^2), 2*(3*5), 2*(3^2 * 5) = 2, 6, 10, 18, 30, 90

are all of the divisors of 90. The same will be true of any odd n: if n is odd, 2n has twice as many divisors as n.

Instead start with an even number, say n = 54 = 2*(3^3). This number has four odd divisors, and four even divisors:

1, 3, 3^2, 3^3 = 1, 3, 9, 27
2, 2*3, 2*(3^2), 2*(3^3) = 2, 6, 18, 54

When we look at 2n = 108 = (2^2)*(3^3) , we'll have all of these divisors, but also all the divisors we get by doubling the divisors in the second row:

1, 3, 3^2, 3^3 = 1, 3, 9, 27
2, 2*3, 2*(3^2), 2*(3^3) = 2, 6, 18, 54
2^2, (2^2)*3, (2^2)*(3^2), (2^2)*(3^3) = 4, 12, 36, 108

We don't get twice as many, of course -- we get 50% more divisors. From this one example, hopefully it's clear why, for any even number n, 2n will never have twice as many divisors as n; 2n will always have less than twice as many if n is even.

Of course, if you understand why we find every divisor by doing as we did above, you'll likely know how, from the exponents in a prime factorization, to calculate the number of divisors of an integer. You can use that as well to answer the question.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Re: Is integer n odd

by logitech » Fri Jan 16, 2009 10:37 am
Is the integer n odd

1) n is divisible by 3

3- odd
6- even

INSUF

2) 2n is divisible by twice as many positive integers as n[/quote]

2x3 = 6

ODD + Twice as many (1,2,3,6,) vs ( 1,3)

2x2 = 4

EVEN 2 vs 2

SUF

One more with an ODD

2x9 = 18

ODD

1,2,3,6,9,18 VS 1,3,9 ( Twice as many )
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by goelmohit2002 » Sun Aug 23, 2009 11:24 pm
is there a algebraic way to solve this problem rather than relying on number picking ?

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by navalpike » Mon Aug 24, 2009 10:58 am
In my opinion, this is really not a problem that lends to algebra. And if it did, Ian would have showed us.

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by drabblejhu » Sun Sep 27, 2009 10:47 am
I relied on picking numbers based on Ron Purewal's advice that odd/evens follow predictable patterns. At the same time, thanks for the more conceptual explanation! Helps.

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by arora007 » Thu Aug 12, 2010 8:03 am
"When odd number n is doubled, 2n has twice as many factors as n."

an excellent takeaway!!
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