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Re: Help: Factors problem !! [#permalink]
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.

Re: Help: Factors problem !! [#permalink]
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.

Re: Help: Factors problem !! [#permalink]
14 Oct 2010, 13:58

1

This post received KUDOS

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) _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

Re: Help: Factors problem !! [#permalink]
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.

Re: How many factors does 36^2 have? [#permalink]
25 Oct 2014, 05:23

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