|
Author |
Message |
|
TAGS:
|
|
|
Senior Manager
Joined: 16 Jul 2008
Posts: 291
Schools: INSEAD Dec'10
Followers: 2
Kudos [?]:
14
[0], given: 4
|
How many factors does 36^2 have? 2 8 24 25 26 [#permalink]
 02 Sep 2008, 13:14
How many factors does 36^2 have? 2 8 24 25 26
_________________
http://applicant.wordpress.com/
|
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1837
Location: New York
Followers: 20
Kudos [?]:
296
[0], given: 5
|
Re: MGMAT - good one [#permalink]
02 Sep 2008, 13:18
Nerdboy wrote: How many factors does 36^2 have?
2
8
24
25
26 2^4*3^4 No of factors = (4+1)*(4+1)=25
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
|
|
|
|
|
|
Intern
Joined: 08 Jun 2008
Posts: 14
Location: United States (AL)
Concentration: Strategy, Finance
GMAT 1: 710 Q46 V42
GPA: 3.81
WE: Information Technology (Accounting)
Followers: 0
Kudos [?]:
11
[0], given: 2
|
Re: MGMAT - good one [#permalink]
02 Sep 2008, 14:50
@x2suresh: Can you explain this a little further? I'm not sure I understand how how you did this.
Also, thanks for all your posts on this site. They are very helpful!
|
|
|
|
|
|
VP
Joined: 05 Jul 2008
Posts: 1442
Followers: 28
Kudos [?]:
153
[0], given: 1
|
Re: MGMAT - good one [#permalink]
02 Sep 2008, 15:12
|
|
|
|
|
|
Manager
Joined: 11 Apr 2008
Posts: 202
Followers: 2
Kudos [?]:
13
[0], given: 1
|
Re: MGMAT - good one [#permalink]
02 Sep 2008, 15:15
36^2=36*36 =4*9*4*9 =16*81 =2^4*3^4 =(4+1)(4+1) =25 ( For 16 you can see there are 5 factors and similarly for 81 there are 5 factors) So the ans is the total number of combinations. Is it right suresh?
_________________
Nobody dies a virgin, life screws us all.
|
|
|
|
|
|
GMAT Instructor
Joined: 24 Jun 2008
Posts: 973
Location: Toronto
Followers: 174
Kudos [?]:
452
[0], given: 3
|
Re: MGMAT - good one [#permalink]
02 Sep 2008, 18:31
The answer choices make this easy: every positive perfect square has an odd number of divisors (because in a perfect square's prime factorization, all of the exponents must be even; when you add one to each and multiply, you will be multiplying only odd numbers, and the product will therefore be odd). So without doing any work, the answer must be 25.
_________________
Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.
Private GMAT Tutor based in Toronto
|
|
|
|
|
|
Current Student
Joined: 11 May 2008
Posts: 562
Followers: 5
Kudos [?]:
13
[0], given: 0
|
Re: MGMAT - good one [#permalink]
02 Sep 2008, 18:57
yes IAN... it is def an out of the box thinking...
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1837
Location: New York
Followers: 20
Kudos [?]:
296
[0], given: 5
|
Re: MGMAT - good one [#permalink]
02 Sep 2008, 20:02
samiam7 wrote: @x2suresh: Can you explain this a little further? I'm not sure I understand how how you did this.
Also, thanks for all your posts on this site. They are very helpful! In general here is the formula for finding factors 2^n * 3^m factors for 2^n 2^0, 2^1, 2^2 ... 2^n ---> (n+1) factors for 2^n factors for 3^m 3^0, 3^1, 3^2,....3^m ---->(m+1) factors for 3^m So No.of factors for 2^n * 3^m = combination of factors for 2^n *3^m = (n+1)(m+1) Is it clear.
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
Last edited by x2suresh on 02 Sep 2008, 20:07, edited 1 time in total.
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1837
Location: New York
Followers: 20
Kudos [?]:
296
[0], given: 5
|
Re: MGMAT - good one [#permalink]
02 Sep 2008, 20:02
subarao wrote: 36^2=36*36
=4*9*4*9
=16*81
=2^4*3^4
=(4+1)(4+1)
=25 ( For 16 you can see there are 5 factors and similarly for 81 there are 5 factors) So the ans is the total number of combinations.
Is it right suresh? you are right! buddy.
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
|
|
|
|
|
|
VP
Joined: 05 Jul 2008
Posts: 1442
Followers: 28
Kudos [?]:
153
[0], given: 1
|
Re: MGMAT - good one [#permalink]
02 Sep 2008, 20:33
x2suresh wrote: samiam7 wrote: @x2suresh: Can you explain this a little further? I'm not sure I understand how how you did this.
Also, thanks for all your posts on this site. They are very helpful! In general here is the formula for finding factors 2^n * 3^m factors for 2^n 2^0, 2^1, 2^2 ... 2^n ---> (n+1) factors for 2^n factors for 3^m 3^0, 3^1, 3^2,....3^m ---->(m+1) factors for 3^m So No.of factors for 2^n * 3^m = combination of factors for 2^n *3^m = (n+1)(m+1) Is it clear. Just want to add that it does not need to be in the form of 2^n 3^m It just needs to be expressed in the form of its prime factors raised to some power.
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1837
Location: New York
Followers: 20
Kudos [?]:
296
[0], given: 5
|
Re: MGMAT - good one [#permalink]
02 Sep 2008, 21:17
icandy wrote: x2suresh wrote: samiam7 wrote: @x2suresh: Can you explain this a little further? I'm not sure I understand how how you did this.
Also, thanks for all your posts on this site. They are very helpful! In general here is the formula for finding factors 2^n * 3^m factors for 2^n 2^0, 2^1, 2^2 ... 2^n ---> (n+1) factors for 2^n factors for 3^m 3^0, 3^1, 3^2,....3^m ---->(m+1) factors for 3^m So No.of factors for 2^n * 3^m = combination of factors for 2^n *3^m = (n+1)(m+1) Is it clear. Just want to add that it does not need to be in the form of 2^n 3^m It just needs to be expressed in the form of its prime factors raised to some power. thats correct!! It is just an example.
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
|
|
|
|
|
|
Senior Manager
Joined: 16 Jul 2008
Posts: 291
Schools: INSEAD Dec'10
Followers: 2
Kudos [?]:
14
[0], given: 4
|
Re: MGMAT - good one [#permalink]
03 Sep 2008, 00:38
25 is the answer, if anyone doubts it. I just love these types of questions - high difficulty level, but with a 'shortcut' to solve it in less than a minute. I got this question in the CAT this weekend, couldn't figure out how to solve it and guessed wrong. Saw the explanation after the test and it struck me as beautiful 
_________________
http://applicant.wordpress.com/
|
|
|
|
|
|
Intern
Joined: 08 Jun 2008
Posts: 14
Location: United States (AL)
Concentration: Strategy, Finance
GMAT 1: 710 Q46 V42
GPA: 3.81
WE: Information Technology (Accounting)
Followers: 0
Kudos [?]:
11
[0], given: 2
|
Re: MGMAT - good one [#permalink]
03 Sep 2008, 04:46
That makes sense now. Thanks guys!
|
|
|
|
|
|
Intern
Joined: 02 Sep 2008
Posts: 46
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: MGMAT - good one [#permalink]
03 Sep 2008, 11:31
Hi,
I have thought on it to learn the concept in depth.
we have only just two prime number (2 and 3)
for example, how many factors does 900?
900 = 2^2 * 3^2 * 5^2
so it is 2^n * 3^m * 5^k
= (n+1) (m+1) (k+1)
= 3 * 3 * 3
= 27.
so, no of factors for 900 is 27.
Is answer correct? or there is a different theory for more than 2 prime numbers.
Thanks
|
|
|
|
|
|
|
Re: MGMAT - good one
[#permalink]
03 Sep 2008, 11:31
|
|
|
|
|
|
|
|
|
|
|
close
