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How many factors does 36^2 have? [#permalink ]
13 Nov 2006, 06:58
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Question Stats:

76% (01:24) correct

24% (00:29) wrong

based on 38 sessions
How many factors does 36^2 have?

A. 2

B. 8

C. 24

D. 25

E. 26

OPEN DISCUSSION OF THIS QUESTION IS HERE: how-many-factors-does-36-2-have-126422.html

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Learnt this somewhere in this forum.
36^2=2^4*3^4
Take the powers of each prime factor, add 1 and multiply.
5*5=25 factors

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How many factors does 36^2 have
A)2
B)8
C)24
D)25
E)26
Plz explain ur method. thanks!

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36^2 = 2^4 * 3^4
number of factors = ( 4+1) * (4+1) = 25

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AK, can you please explain the logic you used. I dont get it.

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ncprasad wrote:

AK, can you please explain the logic you used. I dont get it.

express number ( say N) as products of its prime factors ( say a,b,c)

N = a ^ p + b ^ q + c ^ r

Now number of factors of given number N = ( p +1) * (q+1)* (r+1)

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Ak I did not know this. Thank you! Did you mean:
N = a ^ p * b ^ q * c ^ r

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AK wrote:

ncprasad wrote:

AK, can you please explain the logic you used. I dont get it.

express number ( say N) as products of its prime factors ( say a,b,c)

N = a ^ p + b ^ q + c ^ r

Now number of factors of given number N = ( p +1) * (q+1)* (r+1)

Thanks for this formula. I assume u meant multiplication above rathar than addition.

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ManhattanGMAT Factors [#permalink ]
10 May 2008, 18:05

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

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Re: ManhattanGMAT Factors [#permalink ]
10 May 2008, 18:39

That is correct, but how did you get 25?

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Re: ManhattanGMAT Factors [#permalink ]
10 May 2008, 18:42

the short cut for this type of question is to factor the original number to prime numbers 36^2 = 2^4 x 3^4 the quick trick is to add 1 to each exponent 4+1 and 4 +1 then multiply them 5 x 5 = 25

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Re: ManhattanGMAT Factors [#permalink ]
11 May 2008, 06:04

very cool trick...and this always works to find the # of factors from the # of prime factors?

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PS: Factors (n3.29) [#permalink ]
26 Jul 2008, 09:25

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

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Re: PS: Factors (n3.29) [#permalink ]
26 Jul 2008, 10:07

36^2 = 6*6*6*6 = 2^4*3^4 Total factors = (4+1)*(4+1) = 5*5 = 25 Answer D.

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Re: PS: Factors (n3.29) [#permalink ]
26 Jul 2008, 22:28

abhijit_sen wrote:

36^2 = 6*6*6*6 = 2^4*3^4 Total factors = (4+1)*(4+1) = 5*5 = 25 Answer D.

You answer is correct. Could you explain in more details. I still not understand. Thanks.

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Re: PS: Factors (n3.29) [#permalink ]
28 Jul 2008, 04:22
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The general rule... For finding the number of factors for any given number first break it down into powers of prime factors. pf1^a, pf2^b,...... Now, the total number of factors = (power of pf1 + 1) * (power of pf2 + 1)*...... or (a+1)*(b+1)*.... eg. let us say u want to find the number of factors of 12. 12=2*2*3 = 2^2 * 3^1 pf1=2 pf2=3 a=2 b=1 total number of factors = (a+1)(b+1) = (2+1)*(1+1) = 6 1,2,3,4,6,12 are the 6 factors of 12.

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Forgot an important shortcut, pls help! [#permalink ]
28 Aug 2008, 06:56

How many factors does 36^2 have? OA is 25. 36^2 = 2^4 x 3^4 Instead of listing out the factors, there was a equation for shortcut to calcluate the total number of factors. But I forgot, do you guys know? Tks

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Re: Forgot an important shortcut, pls help! [#permalink ]
28 Aug 2008, 07:27
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judokan wrote:

How many factors does 36^2 have? OA is 25. 36^2 = 2^4 x 3^4 Instead of listing out the factors, there was a equation for shortcut to calcluate the total number of factors. But I forgot, do you guys know? Tks

(4+1)*(4+1).

it is the product of all (powers of prime factors + 1)

Last edited by

bhushangiri on 28 Aug 2008, 07:28, edited 1 time in total.

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Re: Forgot an important shortcut, pls help! [#permalink ]
28 Aug 2008, 07:27
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If a number can be represented as a^x*b^y*c^z, where a, b, and c are prime factors. Then total number of factors are (x+1)*(y+1)*(z+1).

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Re: Forgot an important shortcut, pls help! [#permalink ]
28 Aug 2008, 07:40

judokan wrote:

How many factors does 36^2 have? OA is 25. 36^2 = 2^4 x 3^4 Instead of listing out the factors, there was a equation for shortcut to calcluate the total number of factors. But I forgot, do you guys know? Tks

2^0 2^1 ... 2^n ---> (n+1) factors

3^0 3^1 ... 3^n ---> (n+1) factors

Total factors.. combination of both = (n+1)(n+1)

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Re: Forgot an important shortcut, pls help!
[#permalink ]
28 Aug 2008, 07:40