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How many factors does 36^2 have?

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How many factors does 36^2 have? [#permalink] New post 15 Aug 2010, 02:47
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Question Stats:

81% (01:22) correct 19% (00:47) wrong based on 26 sessions
How many factors does 36^2 have?

A. 2
B. 8
C. 24
D. 25
E. 26
[Reveal] Spoiler: OA
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Re: Help: Factors problem !! [#permalink] New post 15 Aug 2010, 03:04
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praveengmat wrote:
How many factors does 36^2 have?
A 2
B 8
C 24
D 25
E 26
Please help as to how to solve this problem with 1 minute !!


Finding the Number of Factors of an Integer:

First make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2

Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors.

Back to the original question:

How many factors does 36^2 have?

36^2=(2^2*3^2)^2=2^4*3^4 --> # of factors (4+1)*(4+1)=25.

Answer: D.

Or another way: 36^2 is a perfect square, # of factors of perfect square is always odd (as perfect square has even powers of its primes and when adding 1 to each and multiplying them as in above formula you'll get the multiplication of odd numbers which is odd). Only odd answer in answer choices is 25.

Hope it helps.
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Re: Help: Factors problem !! [#permalink] New post 15 Aug 2010, 03:11
Bunuel wrote:
praveengmat wrote:
How many factors does 36^2 have?
A 2
B 8
C 24
D 25
E 26
Please help as to how to solve this problem with 1 minute !!


Finding the Number of Factors of an Integer:

First make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2

Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors.

Back to the original question:

How many factors does 36^2 have?

36^2=(2^2*3^2)^2=2^4*3^4 --> # of factors (4+1)*(4+1)=25.

Answer: D.

Or another way: 36^2 is a perfect square, # of factors of perfect square is always odd (as perfect square has even powers of its primes and when adding 1 to each and multiplying them as in above formula you'll get the multiplication of odd numbers which is odd). Only odd answer in answer choices is 25.

Hope it helps.



Thanks a ton !!.. loved the approach !
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Re: Help: Factors problem !! [#permalink] New post 14 Oct 2010, 13:58
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Factors of a perfect square can be derived by using prime factorization and then using the formula to find perfect square's factors.

In this case (36)^2= (2^2*3^2)^2=2^4*3^4 or (36)^2=(6^2)^2=(6)^4=(2*3)^4=2^4*3^4

And now you can use the formula explained above by Bunuel to determine the answer, which is (4+1)*(4+1)=5*5=25=Odd(Trick is there must be odd number of factors of a perfect square and only 25 is odd in answer choices, so it can be solved within 30 seconds or less :) )

Please! go through the GMAT Math Book by GMAT CLUB (written by bunuel & walker), all of these tips & tricks are written there. (even I have compiled them in one .pdf file and is shared here on Math forum)
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Number of factors [#permalink] New post 03 Feb 2011, 08:31
How many factors does 36^2 have?
A. 2
B. 8
C. 24
D. 25
E. 26
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Re: Number of factors [#permalink] New post 03 Feb 2011, 08:40
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Re: Help: Factors problem !! [#permalink] New post 29 Aug 2011, 09:57
This method is worth bookmarking. Appreciate it
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Re: Help: Factors problem !! [#permalink] New post 01 Sep 2011, 19:17
The Easy Answer! (Applicable only in case of perfect square numbers)
A perfect square always have a odd number of factors.
36^2 is a perfect square.
Given the answer choices, the only odd number of factor is 25.

So, The definite answer is D.
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Re: Help: Factors problem !! [#permalink] New post 02 Sep 2011, 08:08
36^2 = 2^4 3^4

total factors = (4+1)(4+1) = 25.

Answer is D.
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Re: Help: Factors problem !! [#permalink] New post 03 Sep 2011, 10:04
The odd number of factors for perfect squares solves this in no time. Nice trick.
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Re: Help: Factors problem !!   [#permalink] 03 Sep 2011, 10:04
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