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stonecold
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If integer N has p factors; how many factors will 2N have ? Updated on: 07 Aug 2016, 21:55
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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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

58% (01:20) correct 42% (00:59) wrong based on 451 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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E
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manhasnoname
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If integer N has p factors; how many factors will 2N have ? 07 Aug 2016, 21:48
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stonecold
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
Show more

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?
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stonecold
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If integer N has p factors; how many factors will 2N have ? 07 Aug 2016, 21:56
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chnarjun
stonecold
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?
Show more

Hi Its indeed E
Thanks
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If integer N has p factors; how many factors will 2N have ? 15 Dec 2016, 03:48
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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.

Answer E.
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If integer N has p factors; how many factors will 2N have ? 22 Dec 2025, 11:33
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Is 2 already a factor of N? We don't know, so we don't know if 2N means introducing a new factor to N or not. No hard algebra needed.

For example:
If N=3*5 has 4 factors, and
2N=2*3*5 has 8 factors.

If N were 2*5 has 4 factors, and
2N=(2^2)*5 has 3*2=6 factors.

In essence increasing the exponent of 2 and adding a (1+1), representing the new "2" to N's factors, to the factor number formula returns different results, so without more information on N, the number of factors of 2N cannot be determined.
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