How many odd, positive divisors does 540 have? : GMAT Problem Solving (PS)
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How many odd, positive divisors does 540 have?

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How many odd, positive divisors does 540 have? [#permalink]

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New post 10 Dec 2010, 10:15
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How many odd, positive divisors does 540 have?

A. 6
B. 8
C. 12
D. 15
E. 24
[Reveal] Spoiler: OA

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Last edited by Bunuel on 21 Mar 2012, 03:45, edited 1 time in total.
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Re: positive divisors [#permalink]

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shrive555 wrote:
How many odd, positive divisors does 540 have?

6
8
12
15
24


Make a prime factorization of a number: 540=2^2*3^3*5 --> get rid of powers of 2 as they give even factors --> you'll have 3^3*5 which has (3+1)(1+1)=8 factors.

Another example: 60=2^2*3*5 it has (2+1)(1+1)(1+1)=12 factors out of which (1+1)(1+1)=4 are odd: 1, 3, 5 and 15 the same # of odd factors as 60/2^2=15 has.

Answer: B.

MUST KNOW FOR GMAT:

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
For more on number properties check: math-number-theory-88376.html
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Re: positive divisors [#permalink]

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New post 10 Dec 2010, 11:08
Thanks for the explanation
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Re: positive divisors [#permalink]

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New post 10 Dec 2010, 11:32
Quote:
MUST KNOW FOR GMAT:

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.


i remembered that, but just caught in the term " odd factors " which is cleared now. All right, this was an odd factor case in which we don't have to consider Power of 2. what if we are asked about only even +ve factors. would then we be considering only power of 2 ?
for example: 540 = 2^2*3^3*5 ----
=>(2+1) = 3
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Re: positive divisors [#permalink]

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New post 10 Dec 2010, 11:40
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shrive555 wrote:
Quote:
MUST KNOW FOR GMAT:

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.


i remembered that, but just caught in the term " odd factors " which is cleared now. All right, this was an odd factor case in which we don't have to consider Power of 2. what if we are asked about only even +ve factors. would then we be considering only power of 2 ?
for example: 540 = 2^2*3^3*5 ----
=>(2+1) = 3


No. One way will be to count # of all factors and subtract # of odd factors.

540=2^2*3^3*5 has (2+1)(3+1)(1+1)=24 factors out of which 8 are odd, so 24-8=16 are even.
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Re: positive divisors [#permalink]

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New post 10 Dec 2010, 11:46
Bunuel wrote:
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No. One way will be to count # of all factors and subtract # of odd factors.

540=2^2*3^3*5 has (2+1)(1+1)(1+1)=24 factors out of which 8 are odd, so 24-8=16 are even.


:-D I'll stick to this one way method, i can sense the other way will be unmanageable for me

Thanks a lot B
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Re: How many odd, positive divisors does 540 have? [#permalink]

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New post 31 Oct 2016, 06:42
shrive555 wrote:
How many odd, positive divisors does 540 have?

A. 6
B. 8
C. 12
D. 15
E. 24


\(540 = 2^2 x 3^3 x 5^1\)

Odd positive Divisors = ( 3 + 1 ) ( 1 + 1 ) => 8

Hence answer will be (B) 8

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Re: How many odd, positive divisors does 540 have?   [#permalink] 31 Oct 2016, 06:42
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