It is currently 19 Oct 2017, 09:51

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# How many odd, positive divisors does 540 have?

Author Message
TAGS:

### Hide Tags

Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 313

Kudos [?]: 592 [0], given: 193

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

### Show Tags

10 Dec 2010, 11:15
9
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

62% (00:55) correct 38% (01:03) wrong based on 486 sessions

### HideShow timer Statistics

How many odd, positive divisors does 540 have?

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

_________________

I'm the Dumbest of All !!

Last edited by Bunuel on 21 Mar 2012, 04:45, edited 1 time in total.
Edited the question

Kudos [?]: 592 [0], given: 193

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128872 [7], given: 12183

### Show Tags

10 Dec 2010, 12:00
7
KUDOS
Expert's post
14
This post was
BOOKMARKED
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
_________________

Kudos [?]: 128872 [7], given: 12183

Senior Manager
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 390

Kudos [?]: 47 [0], given: 46

Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)

### Show Tags

10 Dec 2010, 12:08
Thanks for the explanation
_________________

Kudos [?]: 47 [0], given: 46

Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 313

Kudos [?]: 592 [0], given: 193

### Show Tags

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
_________________

I'm the Dumbest of All !!

Kudos [?]: 592 [0], given: 193

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128872 [1], given: 12183

### Show Tags

10 Dec 2010, 12:40
1
KUDOS
Expert's post
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.
_________________

Kudos [?]: 128872 [1], given: 12183

Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 313

Kudos [?]: 592 [0], given: 193

### Show Tags

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
_________________

I'm the Dumbest of All !!

Kudos [?]: 592 [0], given: 193

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16674

Kudos [?]: 273 [0], given: 0

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

### Show Tags

28 Nov 2013, 11:00
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16674

Kudos [?]: 273 [0], given: 0

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

### Show Tags

26 Sep 2015, 01:40
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16674

Kudos [?]: 273 [0], given: 0

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

### Show Tags

31 Oct 2016, 06:11
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Math Forum Moderator
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3003

Kudos [?]: 1087 [0], given: 325

Location: India
GPA: 3.5
Re: How many odd, positive divisors does 540 have? [#permalink]

### Show Tags

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

_________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )

Kudos [?]: 1087 [0], given: 325

Re: How many odd, positive divisors does 540 have?   [#permalink] 31 Oct 2016, 07:42
Display posts from previous: Sort by