Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Can I ask Bunuel or some other math guru to look at this problem?

I initially thought that 28 is the correct answer. But after some deliberation I don't think it's correct. E.g., 2^2 doesn't have 4 divisors, but only 3: 1, 2, 4.

The method of finding the number of factors (also called divisors) is based on the concept of finding the basic factors (which make up all other factors) and then combining them in different ways to make as many different factors as possible.

Let us say the question asks us to find the number of factors of 72. We know that 72 = 8x9 = 2^3 x 3^2 - This is called prime factorization. We have essentially brought down 72 to its basic factors. We find that 72 has three 2s and two 3s. We can combine them in various ways e.g. I could take one 2 and two 3s and make 2x3x3 = 18. Similarly, I could take three 2s and no 3 to make 2x2x2 = 8 Since we have three 2s, we can choose a 2 in four ways (take no 2, take one 2, take two 2s or take three 2s) Since we have two 3s, we can choose a 3 in three ways (take no 3, take one 3 or take two 3s) Every time we make a different choice, we get a different factor of 72. Since we can choose 2s in 4 ways and 3s in 3 ways, together we can choose them in 4x3 = 12 ways. Therefore, 72 will have 12 factors.

This is true for any positive integer N. If N = 36 x 48 = 2^2 x 3^2 x 2^4 x 3 = 2^6 x 3^3. To make factors of N, we have seven ways to choose a 2 and four ways to choose a 3. Therefore, we can make 7 x 4 = 28 combinations of these prime factors to give 28 factors of 36x48. Note: I could write 36 x 48 as 72 x 24 or 64 x 27 or 8 x 8 x 27 or many other ways. The answer doesn't change because it is still the same number N.

Generalizing, if N = p^a x q^b x r^c ... where p, q, r are all distinct prime numbers, the total number of factors of N (including 1 and N ) is (a + 1)(b + 1) (c + 1)... Remember, the '+1' is because of an option of dropping that particular prime number from our factor.

On that note, if I tell you that a positive integer N has total 7 factors, what can you say about N?

Can I ask Bunuel or some other math guru to look at this problem?

I initially thought that 28 is the correct answer. But after some deliberation I don't think it's correct. E.g., 2^2 doesn't have 4 divisors, but only 3: 1, 2, 4.

Thank you.

It is asking about divisors/factors (all) not distinct (unique) divisors/factors. HTH
_________________

"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

Can I ask Bunuel or some other math guru to look at this problem?

I initially thought that 28 is the correct answer. But after some deliberation I don't think it's correct. E.g., 2^2 doesn't have 4 divisors, but only 3: 1, 2, 4.

Thank you.

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: If p and q are prime numbers, how many divisors does the product \(p^3*q^6\) have?

According to above the number of distinct factors of \(p^3*q^6\) would be \((3+1)(6+1)=28\).

Answer: D.

As for your doubt: 2^2=4 thus 4 must have (2+1)=3 factors: 1, 2, and 4.

To determine the number of positive factors of any integer:

1) Prime-factorize the integer 2) Add 1 to each exponent 3) Multiply

A question in OG11 (I think) asked for the number of positive factors of 441.

Since 441 = 3^2 * 7^2, we get (2+1)(2+1) = 9 factors.

Here's the reasoning. To determine how many factors can be created from 3^2 * 7^2, we need to determine the number of choices we have of each prime factor:

For 3, we can use 3^0, 3^1, or 3^2, giving us 3 choices. For 7, we can use 7^0, 7^1, or 7^2, giving us 3 choices.

Multiplying, we get 3*3 = 9 possible factors.
_________________

GMAT Tutor and Instructor GMATGuruNY@gmail.com New York, NY

Got another one that I'm stumped on. Thanks for the help in advance!

If P and Q are prime numbers, how many divisors does the product of (P^3)(Q^6) have?

A 9 B 12 C 18 D 28 E 36

When you need to find the number of divisors of a number, you use this approach: Break down the number into its prime factors. e.g. \(N = a^p*b^q*c^r\)... where a, b and c are all distinct prime factors of N. p, q and r are the powers of the prime factors in N Total number of divisors of N = (p+1)(q+1)(r+1)...

e.g. Total number of factors of 36 (\(= 2^2*3^2\)) is (2+1)(2+1) = 9

In this question, you need to find the total number of divisors of \((P^3)(Q^6)\) where P and Q are prime. Total number of factors = (3+1)(6+1) = 28

Note: They should have mentioned that P and Q are distinct prime numbers. If P and Q are not distinct e.g. \(3^3*3^6 = 3^9\) and its total number of divisors is (9+1) = 10. But from the options, it is obvious they intend you to take them as distinct. Still, erroneous question.
_________________

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

If p and q are prime numbers, how many divisors does the product p^3 * q^6 have?

A) 9 B) 12 C) 18 D) 28 E) 36

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:

According to the above, \(p^3*q^6\) will have \((3+1)(6+1)=28\) different positive factors.

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...