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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] New post 23 Oct 2006, 09:58
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Is the integer N odd?
(1) N is divisible by 3
(2)2N is divisible by twice as many positive integer as N.
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 [#permalink] New post 23 Oct 2006, 10:08
Is the integer N odd?
(1) N is divisible by 3
(2)2N is divisible by twice as many positive integer as N.

from one n could be 6 or 9 ( even or odd)...not suff

from two

assume n = 3 devisible by 3,1 , 2n = 6 devisble by 6,3,2,1

assume n = 5 devisble by 5,2 2n = 10 devisble by 10,5,2,1

assume n = 9 devisble by 3 , 9 ,1 while 18 is devisble by 2,9,3,2,6,1

suff and n is odd

My answer is B
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Re: DS - good concept [#permalink] New post 23 Oct 2006, 21:13
I think B too.

1 is not sufficient. N can be either odd or even.

For 2: All the prime factors of an odd number N will be odd. The prime factors to 2N will be all the factors of N as well as the factors of N multiplied by 2.

For example: 21 = 1,3,7,21
42 = 1,3,7,21,
2,6,14,42 (above factors multiplied by 2)

For an even number N, 2N can't have all distinct prime factors by multiplying N's factors by 2
10 - 1,2,5,10
20 - 1,2,5,10
4, ,20
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 [#permalink] New post 24 Oct 2006, 07:48
yezz wrote:
Is the integer N odd?
(1) N is divisible by 3
(2)2N is divisible by twice as many positive integer as N.

from one n could be 6 or 9 ( even or odd)...not suff

from two

assume n = 3 devisible by 3,1 , 2n = 6 devisble by 6,3,2,1

assume n = 5 devisble by 5,2 2n = 10 devisble by 10,5,2,1

assume n = 9 devisble by 3 , 9 ,1 while 18 is devisble by 2,9,3,2,6,1

suff and n is odd

My answer is B

IMHO E it is
yezz and what if n=4?
4 is devisible by 4,2,1 and 8 is divisible by 8, 4, 2,1 n is even
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 [#permalink] New post 24 Oct 2006, 07:58
Yurik79, I did not understand how your example satisfies the 2nd option.
8 is divisible by 4 positive numbers and 4 is divisible by 3 numbers. 8 should have 6 positive factors to meet the requirement stated in 2. Am I missing something?
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 [#permalink] New post 24 Oct 2006, 08:14
I tried solving it in another way:

Let O be any odd number. O is prime factorized into a, b and c.

O = a^m x b^n x c^l
Therefore number of factors of O = (m+1)(n+1)(l+1)
2xO = 2^1 x a^m x b^n x c^l
Therefore number of factors of 2O = (1+1)(m+1)(n+1)(l+1); this is twice the number of factors of O.

Now let E be any even number.
E= 2^m x d^n x y^l
Therefore number of factors of E = (m+1)(n+1)(l+1)
// At the minimum m+1 = 2
2xE = 2^(m+1) x d^n x y^l
Therefore number of factors of 2E = (m+2)(n+1)(l+1)

The number of prime factors of 2E will be twice of those of E only if (m+2) = 2(m+1)
This is possible only if m=0. Since E is an even number, m cannot be 0.
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 [#permalink] New post 24 Oct 2006, 08:15
agreed on (B). Statement 1 will always result in just one more postive factor.
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 [#permalink] New post 24 Oct 2006, 08:36
mst wrote:
Yurik79, I did not understand how your example satisfies the 2nd option.
8 is divisible by 4 positive numbers and 4 is divisible by 3 numbers. 8 should have 6 positive factors to meet the requirement stated in 2. Am I missing something?

:oops: my fault))you are absolutely correct thanks for pointing my mistake
yezz I am sorry agree B
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  [#permalink] 24 Oct 2006, 08:36
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