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Is integer n odd

This topic has 1 expert reply and 10 member replies
aatech Master | Next Rank: 500 Posts Default Avatar
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Is integer n odd

Post 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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goelmohit2002 Legendary Member Default Avatar
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Post 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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navalpike Master | Next Rank: 500 Posts Default Avatar
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Post 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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arora007 Community Manager
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Post 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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drabblejhu MBA Student
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Post 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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logitech Legendary Member
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Post 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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Ian Stewart GMAT Instructor
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Post 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.

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aj5105 Legendary Member Default Avatar
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Post Fri Jan 16, 2009 9:16 am
anybody more on this problem,please?

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aatech Master | Next Rank: 500 Posts Default Avatar
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Post Sat May 31, 2008 4:40 pm
OA is B.. Thanks guys

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netigen Legendary Member Default Avatar
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Post 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 Smile until and unless n=3

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

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chidcguy Master | Next Rank: 500 Posts Default Avatar
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Post 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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netigen Legendary Member Default Avatar
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Post 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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