How many factors does 36^2 have? : GMAT Problem Solving (PS)
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# How many factors does 36^2 have?

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Current Student
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How many factors does 36^2 have? [#permalink]

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13 Nov 2006, 06:58
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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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13 Nov 2006, 07:15
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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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08 Dec 2006, 08:14
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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08 Dec 2006, 10:46
36^2 = 2^4 * 3^4

number of factors = ( 4+1) * (4+1) = 25
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08 Dec 2006, 11:22
AK, can you please explain the logic you used. I dont get it.
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08 Dec 2006, 11:46
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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08 Dec 2006, 13:20
Ak I did not know this. Thank you! Did you mean:

N = a ^ p * b ^ q * c ^ r
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08 Dec 2006, 14:26
AK 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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10 May 2008, 18:05
How many factors does 36^2 have?
2
8
24
25
26
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10 May 2008, 18:39
That is correct, but how did you get 25?
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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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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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26 Jul 2008, 09:25
How many factors does 36^2 have?
2
8
24
25
26
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26 Jul 2008, 10:07
36^2 = 6*6*6*6 = 2^4*3^4
Total factors = (4+1)*(4+1) = 5*5 = 25

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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

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

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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]

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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]

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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]

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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

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