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# If integer N has p factors; how many factors will 2N have ?

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Current Student
Joined: 12 Aug 2015
Posts: 2562
Schools: Boston U '20 (M)
GRE 1: Q169 V154
If integer N has p factors; how many factors will 2N have ?  [#permalink]

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Updated on: 07 Aug 2016, 21:55
2
00:00

Difficulty:

35% (medium)

Question Stats:

58% (01:12) correct 42% (00:59) wrong based on 92 sessions

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If integer N has p factors; how many factors will 2N have ?

A) p
B) 2p
C) p + 1
D) 2p + 1
E) Cannot be determined

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Originally posted by stonecold on 07 Aug 2016, 21:06.
Last edited by stonecold on 07 Aug 2016, 21:55, edited 2 times in total.
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Joined: 21 Apr 2016
Posts: 163
Re: If integer N has p factors; how many factors will 2N have ?  [#permalink]

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07 Aug 2016, 21:48
stonecold wrote:
if integer N has p factors ; how many factors will 2N have ?
A) p
B) 2p
C) P+1
D) 2p+1
E) Cannot be determined

How is the OA C? shouldn't it be E

Case 1:
Let N = 6 => 2 x 3 (prime factorization) => p = 4;

2N = 12 => 2^2 x 3 (prime factorization) => no. of factors = 6; (p+2)

Case 2:
Let N = 9 => 3^2 (prime factorization) => p = 3;

2N = 18 => 3^2 x 2 (prime factorization) => no of factors = 6; (2p)

For different N different results are observed. So, shouldn't E be the right answer?
Current Student
Joined: 12 Aug 2015
Posts: 2562
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Re: If integer N has p factors; how many factors will 2N have ?  [#permalink]

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07 Aug 2016, 21:56
chnarjun wrote:
stonecold wrote:
if integer N has p factors ; how many factors will 2N have ?
A) p
B) 2p
C) P+1
D) 2p+1
E) Cannot be determined

How is the OA C? shouldn't it be E

Case 1:
Let N = 6 => 2 x 3 (prime factorization) => p = 4;

2N = 12 => 2^2 x 3 (prime factorization) => no. of factors = 6; (p+2)

Case 2:
Let N = 9 => 3^2 (prime factorization) => p = 3;

2N = 18 => 3^2 x 2 (prime factorization) => no of factors = 6; (2p)

For different N different results are observed. So, shouldn't E be the right answer?

Hi Its indeed E
Thanks
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Joined: 13 Oct 2016
Posts: 359
GPA: 3.98
Re: If integer N has p factors; how many factors will 2N have ?  [#permalink]

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15 Dec 2016, 03:48
1
Our number is $$N=2^a*q^b*r^c*s^d …$$

Number of factors of N: $$(a + 1)*(b + 1)*(c + 1)*(d + 1) … = p$$

$$2N = 2*2^a*q^b*r^c*s^d … = 2^{(a+1)} *q^b*r^c*s^d …$$

Total # of factors of $$2N = (a + 2)*(b + 1)*(c + 1)*(d + 1) …$$

For simplicity sake we denote $$(b + 1)*(c + 1)*(d + 1) … = X$$

$$(a + 2)*X = (a + 1 + 1)*X = (a + 1)*X + X = p + (b + 1)*(c + 1)*(d + 1) …$$

In order to find our total # of factors of 2N we need to know number of odd factors of N and this is not given in the question.

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Re: If integer N has p factors; how many factors will 2N have ?  [#permalink]

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04 Feb 2019, 19:33
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Re: If integer N has p factors; how many factors will 2N have ?   [#permalink] 04 Feb 2019, 19:33
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