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How many factors does 36^2 have? [#permalink]
 13 Nov 2006, 07:58
Question Stats:
77% (01:18) correct
22% (00:34) wrong based on 18 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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Senior Manager
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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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Senior Manager
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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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Senior Manager
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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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Senior Manager
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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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Director
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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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Senior Manager
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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, 19:05
How many factors does 36^2 have?
2
8
24
25
26
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Director
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Re: ManhattanGMAT Factors [#permalink]
10 May 2008, 19:39
That is correct, but how did you get 25?
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Re: ManhattanGMAT Factors [#permalink]
10 May 2008, 19:42
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, 07:04
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, 10:25
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, 11:07
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, 23:28
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, 05:22
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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, 07:56
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, 08:27
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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, 08:28, edited 1 time in total.
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Re: Forgot an important shortcut, pls help! [#permalink]
28 Aug 2008, 08:27
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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, 08:40
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, 08:40
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