Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Mar 2015, 18:19

### 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

# A positive odd number n has exactly 8 factors. How many

Author Message
TAGS:
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1268
Followers: 23

Kudos [?]: 148 [0], given: 0

A positive odd number n has exactly 8 factors. How many [#permalink]  19 Jul 2006, 05:13
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
A positive odd number n has exactly 8 factors. How many prime factors does it have?

(1) n is less than 150
(2) n is greater than 100
VP
Joined: 25 Nov 2004
Posts: 1495
Followers: 6

Kudos [?]: 38 [0], given: 0

Re: DS: 8 factors [#permalink]  19 Jul 2006, 06:06
kevincan wrote:
A positive odd number n has exactly 8 factors. How many prime factors does it have?

(1) n is less than 150
(2) n is greater than 100

guess E. there are two numbers that satisfy the above properties: 105 and 135.

factors:
105= 1, 3, 5, 7, 15, 21, 35, 105
135 = 1, 3, 5, 9, 15, 27, 45, 135.

prime factors:
105: 3, 5, 7
135: 3, 5
Senior Manager
Joined: 29 Jun 2005
Posts: 403
Followers: 1

Kudos [?]: 15 [0], given: 0

this is where one should make an educated guess.

Or there is any short way?
SVP
Joined: 24 Sep 2005
Posts: 1895
Followers: 11

Kudos [?]: 133 [0], given: 0

Re: DS: 8 factors [#permalink]  20 Jul 2006, 09:32
kevincan wrote:
A positive odd number n has exactly 8 factors. How many prime factors does it have?

(1) n is less than 150
(2) n is greater than 100

Agree with MA ...and how to find those numbers not by blind guessing
Employ the formula of number of factors:
Assume that n is constituted by prime factors a,b,c,etc ... the number of factors of n is the product of those (factors' powers + 1).
Observe that 8= 1*8 or 2*4 or 2*2*2 , in which 1*8 is impractical coz 1 can't be (factor's power+1) --> we have two cases: 2*4 or 2*2*2. That is to say, n can have 2 prime factors or 3 prime factors. Since n is an odd number, n can't have factor of 2. Now, look at statement 1 and 2, we can estimate the biggest prime factor that n can have is 7 . Thus we have two cases: n has prime factor 3, 5 and 7 .
The case with 3 prime factors fits the case of 2*2*2 --> each prime factor has power of 1 --> n= 3*5*7= 105
The case with 2 prime factors fits the case of 2*4 --> one prime factor has power of 1, the other 3 ..try 3^3 *5 , we have n= 135

==> we can point out at least 2 cases of n which lie in the open range of either statement 1 or 2 or both statements.

--> E.

It's just my thought of how to look at this problem from another angle but it's not really advisable in real test
Intern
Joined: 17 Jul 2006
Posts: 26
Followers: 0

Kudos [?]: 2 [0], given: 0

To elaborate on the previous post, the problem is testing a very important concept which is to to find out the # of factors of a positive integer.

This concept is very often tested in GMAT in different ways. Given below is the concept explained in more details and some sample problems.

Concept:

A positive integer n which has prime factorization a^x * b^y * c^z has (x+1)*(y+1)*(z+1) factors. (note a, b and c are distinct prime factors of n).

eg1: 36 = 2^2 * 3^2

The # of positive factors of 36 = (2+1) * (2+1) = 9

Note that the # of factors include 1 and 36 as well.

eg2: 360 = 2^3 * 3^2 * 5

# of factors = (3+1) * (2+1) * (1+1) = 4 * 3 * 2 = 24

explanation: to obtain a factor of 360, from the above list, I can pick a 2 or a 3 or a 5 or any combination of 2,3,5. (think of this as a probability or a combination problem).

The # of 2's I can select are (one 2 or two 2's or three 2's or zero 2's)
The # of 3's I can select are (one 3 or two 3's or zero 3's)
The # of 5's I can select are (one 5 or zero 5's).

Note that select zero 2's, zero3's and zero 5's gives you the trivial case of 1 as a factor.

The total # of combinations = 4*3*2=24

Applications:

Application 1:

What are the Number of positive factors for 1800?

1800 = 18 * 100 = 2^3 * 3^2 * 5^2

# of factors = 4 * 3 * 3 = 36

Application 2:

How many positive integers k are there such that 100k is a factor of (2^2)(3) (5^3)?

100 = 4 *25 = 2^2 * 5^2

Given 100 k is a factor of (2^2) (3) (5^3)

Which means, 2^2 * 5^2 * k is a factor of (2^2) (3) (5^3)

You see that k can be equal to 1, 3, or 5 or 15. (note that positive integer means 1 also).

So, there are four integers 1, 3, 5 and 15 that would satisfy the given condition.

In other words, the problem translates to finding the # of factors for 15.

15 = 3 * 5

and hence, the # of factors = (1+1) * (1+1) = 2 * 2 = 4.

3. Approximately what percent of the positive factors of 1800 are less than 30?

(A) 22% (B) 28% (C) 32% (D) 38% (E) 42%

First we need to determine the # of factors for 1800.

In order to find out the # of factors, express 1800 = 18*100 = 2^3 * 3^2 * 5^2

The # of factors = (3+1) * (2+1) * (2+1) = 4 * 3 * 3 = 36

The factors that are less than 30 are 1,2,3,4,5,6,8,9,10,12,15,18,20,24,25 (note that 1 is included but not 30).

% = 15/36 * 100 = 42%

4. What can you tell about an integer n that has only (and only) 3 factors.

a. Note that 3 itself is a prime number with only 1 and 3 as the factors.
So, the number n has the prime factorization of n = p^2 where p is a prime number.

b. From the above result, one can see that n is a perfect sqaure.

for example, n = 4, 9, 25, 49, 121 etc

c. Note that 4 satisfies the above condition and hence, in general we cannot conclude that n is odd. However, if it is given that n>4, we safely conclude that n is odd.

d. Similarly, if a positive integer n has only 5 factors, what can you tell about n.

n can be written as p^4 where p is prime. Hence, you can conclude that sqrt(n) = p^2 is a perfect sqaure.

-mathguru
_________________
SVP
Joined: 24 Sep 2005
Posts: 1895
Followers: 11

Kudos [?]: 133 [0], given: 0

Thank you for elaborating

How stunned I am to know that you were an Olympiad winner in maths . I'm sure GMATclubbers will benefit a lot if you often chime in so please be around
CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 18

Kudos [?]: 110 [0], given: 0

Dilshod wrote:
this is where one should make an educated guess.

Or there is any short way?

Yes there is a short way.

E

Total factors = 8

8 = 8*1 or 4*2 or 2*2*2
Case1 when we take 8 = 8*1
Case2 when we take 8 = 4*2
Case3 when we take 8 = 2*2*2
Total factors of a number x^a * y^b * z^c are (a+1) (b+1) (c+1)
St1:
Case1: n = x^7 where x is a prime number. Taking least odd prime n goes beyod 150. Reject
Case2: n = x^3 * y^1. Only values of x and y are x = 3 and y =5 for n to be less than 150. Then n = 135
Case3: n = x^1 * y^1 * z^1. Only values of x,y and z are x = 3 y = 5 and z = 7 for n to be less than 150. So n = 105
So n may be 105 or 135: INSUFF

St2: Clearly INSUFF. in all three cases above x,y and z could be any prime number greater than 2.

Combined:
Still n could be 105 or 135: SUFF
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Intern
Joined: 17 Jul 2006
Posts: 26
Followers: 0

Kudos [?]: 2 [0], given: 0

Dahiya - well done and thanks for clarifying the application of the concept explained above. Precisely. Only three possibilities if a number has only 8 factors - p^7 or p^3 q or p q r where p,q,r are prime factors.

-mathguru
_________________
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1268
Followers: 23

Kudos [?]: 148 [0], given: 0

Great work on the part of all three of you!
Similar topics Replies Last post
Similar
Topics:
3 The positive integer 200 has how many factors? 2 17 Sep 2012, 05:24
8 How many different prime numbers are factors of the positive 11 08 Oct 2010, 16:50
If integer n has exactly 3 positive factors, including 1 and 4 30 Jun 2008, 12:37
How many different prime numbers are factors of the positive 4 22 Oct 2006, 22:37
11 How many positive three digit integers have exactly 7 factor 19 07 Jul 2006, 07:25
Display posts from previous: Sort by