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How many factors does 36^2 have? 2 8 24 25 26

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How many factors does 36^2 have? 2 8 24 25 26 [#permalink] New post 02 Sep 2008, 12:14
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How many factors does 36^2 have?

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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 12:18
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
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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 13:50
@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!
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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 14:12
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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 14:15
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?
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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 17:31
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.
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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 17:57
yes IAN... it is def an out of the box thinking...
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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 19:02
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.
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Last edited by x2suresh on 02 Sep 2008, 19:07, edited 1 time in total.
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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 19:02
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.
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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 19:33
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.
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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 20:17
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.
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Re: MGMAT - good one [#permalink] New post 02 Sep 2008, 23:38
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 :)
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Re: MGMAT - good one [#permalink] New post 03 Sep 2008, 03:46
That makes sense now. Thanks guys!
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Re: MGMAT - good one [#permalink] New post 03 Sep 2008, 10:31
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, 10:31
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