700-800 Level PS Question

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davidlee05: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Wed Nov 07, 2012 8:23 am

by davidlee05 » Sat Mar 16, 2013 9:56 pm

How many factors does 36^2 have?

A. 2
B. 8
C. 24
D. 25
E. 26

OA: [spoiler]D[/spoiler]

Can someone please explain the OA? Thanks.

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Source: Beat The GMAT — Problem Solving | Sat Mar 16, 2013 9:56 pm

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Anju@Gurome: GMAT Instructor; Posts: 511; Joined: Wed Aug 11, 2010 9:47 amFollowed by: 86 members; Location: Delhi, India; Thanked: 344 times

by Anju@Gurome » Sat Mar 16, 2013 9:59 pm

davidlee05 wrote:How many factors does 36^2 have?

For any real number N, if the prime factorization of N is (p^a)*(q^b)*(r^c)*..., then the number of factors of N is given by (a + 1)(b + 1)(c + 1)...

Here, 36^2 = 2^4 * 3^4

So, number of factors of 36^2 = (4 + 1)(4 + 1) = 25

The correct answer is D.

Last edited by Anju@Gurome on Sat Mar 16, 2013 10:05 pm, edited 1 time in total.

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davidlee05: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Wed Nov 07, 2012 8:23 am

by davidlee05 » Sat Mar 16, 2013 10:03 pm

Can you please expand on your method?

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Anju@Gurome: GMAT Instructor; Posts: 511; Joined: Wed Aug 11, 2010 9:47 amFollowed by: 86 members; Location: Delhi, India; Thanked: 344 times

by Anju@Gurome » Sat Mar 16, 2013 10:11 pm

davidlee05 wrote:Can you please expand on your method?

Please check my previous post.
I've edited the reply with the explanation.

To explain more, let us assume for some real number N the prime factorization of N is (p^a)*(q^b)*(r^c)*...

Now number of factors of N = (Number of ways to select the prime factor p)*(Number of ways to select the prime factor q)*(Number of ways to select the prime factor r)*... = (a + 1)(b + 1)(c + 1)...

Now, for each prime factor we may or may not select that factor. When none of the prime factors are selected we get the factor 1. Hence, for the prime factor p, number of ways to select p to include it in a factor of N is exponent of p in the prime factorization of N, i.e. a and number of ways to not select p is 1. That's why we are adding 1 to each of the exponents of the prime factors of N.

Hope that helps.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Mar 17, 2013 3:17 am

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