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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] New post 10 Dec 2010, 10:15
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A
B
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E

Difficulty:

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Question Stats:

58% (01:47) correct 42% (01:02) wrong based on 158 sessions
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] New post 10 Dec 2010, 11:00
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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.

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] 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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

I'm the Dumbest of All !!

Senior Manager
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Status: Do and Die!!
Joined: 15 Sep 2010
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Kudos [?]: 74 [0], given: 193

Re: positive divisors [#permalink] New post 10 Dec 2010, 11:46
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.


:-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] New post 28 Nov 2013, 10:00
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Re: How many odd, positive divisors does 540 have?   [#permalink] 28 Nov 2013, 10:00
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