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:

Re: M01-Q35 Please help with this question. thanks [#permalink]
04 Apr 2009, 20:09

4

This post received KUDOS

Prime factorize 90 90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

Re: M01-Q35 Please help with this question. thanks [#permalink]
30 Jan 2010, 12:43

cramya wrote:

Prime factorize 90 90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

I don't have 12 in my answer choices and i answered incorrectly Although i understood the question perfectly

If @x is the number of distinct positive divisors of x, what is the value of @@90 ?

Re: M01-Q35 Please help with this question. thanks [#permalink]
22 Sep 2010, 05:13

cramya wrote:

Prime factorize 90 90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(

Re: M01-Q35 Please help with this question. thanks [#permalink]
22 Sep 2010, 05:18

cnrnld wrote:

cramya wrote:

Prime factorize 90 90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(

Re: M01-Q35 Please help with this question. thanks [#permalink]
22 Sep 2010, 07:43

shrouded1 wrote:

cnrnld wrote:

cramya wrote:

Prime factorize 90 90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(

Re: M01-Q35 Please help with this question. thanks [#permalink]
22 Sep 2010, 07:57

4

This post received KUDOS

Expert's post

cnrnld wrote:

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(

MUST KNOW FOR GMAT:

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 \(@x\) is the number of distinct positive divisors of \(x\) , what is the value of \(@@90\)? (A) 3 (B) 4 (C) 5 (D) 6 (E) 7

The question defines \(@x\) as the number of distinct positive divisors of \(x\). Say \(@6=4\), as 6 have 4 distinct positive divisors: 1, 2, 3, 6.

Question: \(@@90=?\)

\(90=2*3^2*5\), which means that the number of factors of 90 is: \((1+1)(2+1)(1+1)=12\). So \(@90=12\) --> \(@@90=@12\) --> \(12=2^2*3\), so the number of factors of 12 is: \((2+1)(1+1)=6\).

Re: M01-Q35 Please help with this question. thanks [#permalink]
26 Sep 2011, 08:45

To get number of positive factors of any number follow below 4 steps:

1. The prime factorization raised to the power: 90= (3^2)(2^1)(5^1)(1^0) 2. The powers: 2, 1, 1, 0 3. Add one to the powers: 3,2,2,1 4. Multiply the results: (3)(2)(2)(1)=12

and repeat the steps for 12. thats it you are done! _________________

Re: M01-Q35 Please help with this question. thanks [#permalink]
26 Sep 2012, 05:57

Use the Fundamental Counting Theory...

Prime factorization of 90= 2*3^2*5 Add 1 to each exponent and multiply= (2)(3)(2)=12 distinct positive divisors of 90 Prime factorizatino of 12=3*2^2 (2)(3)=6 distinct factors of 12

Answer is D

gmatclubot

Re: M01-Q35 Please help with this question. thanks
[#permalink]
26 Sep 2012, 05:57