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 350,000 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: Math problem involving powers and smallest possible value [#permalink]
24 Sep 2009, 10:11

1

This post received KUDOS

The answer is D: 75.

This is a question relating to prime factorability (is that a word?).

Basically, for any number, X, to be a factor of another number, Y, the prime factors of X must be present in the prime factors of Y. For example, if you reduce the number 50 = (5^2)*(2^1), it is a factor of any number which has these prime factors when reduced.

So, for 5^2 and 3^3 to be factors of this number, the number, when reduced to prime factors, must contain 5^2 and 3^3.

\(n*2^5*6^2*7^3\)

\(= n*2^5*(3^2*2^2)*7^3\)

\(= n*2^7*3^2*7^3\)

As seen by the above equation, we need at least one more factor of 3, as well as two more factors of 5. As a result:

Re: Math problem involving powers and smallest possible value [#permalink]
24 Sep 2009, 10:22

1

This post received KUDOS

Think of it this way:

You have the equation 10x. What is the smallest number x can be for the equation to be a multiple of 3 (for 3 to be a factor of the equation)?

5*2*x. So for the equation to be divisible by 3 there must be a 3 in the equation. Therefore x = 3

5*2*3 is divisible by 3

Now you have the equation 15x and want to know what is the smallest number x for it to be divisible by 3^2 (or 9) 15x = 5*3*x

There is already one three in the equation. You will need at least one more 3 to factor out the 9. The smallest value of x is 3 for the number to be divisible by 9

Therefore 5*3*3 = 45

Prime factorization

Last edited by lagomez on 24 Sep 2009, 10:23, edited 1 time in total.

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I was checking my phone all day. I wasn’t sure when I would receive the admission decision from Tepper. I received an acceptance from Goizueta in the early morning...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...