Is the integer n odd
1) n is divisible by 3
2) 2n is divisible by twice as many positive integers as n
Is integer n odd
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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
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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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
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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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)
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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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.
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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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 )
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 )
LGTCH
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- drabblejhu
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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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"When odd number n is doubled, 2n has twice as many factors as n."
an excellent takeaway!!
an excellent takeaway!!
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