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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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10 Dec 2010, 11:15
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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, 04:45, edited 1 time in total.
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10 Dec 2010, 12:00
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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.

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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10 Dec 2010, 12:08
Thanks for the explanation
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10 Dec 2010, 12:32
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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10 Dec 2010, 12:40
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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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10 Dec 2010, 12:46
Bunuel wrote:
Quote:

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.

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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28 Nov 2013, 11:00
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Re: How many odd, positive divisors does 540 have? [#permalink]

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26 Sep 2015, 01:40
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Re: How many odd, positive divisors does 540 have? [#permalink]

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31 Oct 2016, 06:11
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Re: How many odd, positive divisors does 540 have? [#permalink]

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31 Oct 2016, 07:42
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, 07:42
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