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: If X = the product of four distinct prime numbers, how many [#permalink]
06 Sep 2013, 13:42
3
This post received KUDOS
Expert's post
salsal wrote:
If X = the product of four distinct prime numbers, how many factors does X have besides 1 and itself?
(a) 10 (b) 11 (c) 12 (d) 13 (e) 14
My approach:
Let's take the four primes: 2, 3, 5 and 7 --> each of the 4 primes will be a factor of the product
All the possible combinations of the four primes will be factors (10 possible combinations): 2x3 2x5 2x7 3x5 3x7 5x7 2x3x5 2x3x7 2x5x7 3x5x7
Thus, in total there are 10 + 4 = 14 factors.
Is my reasoning correct? Is there a faster way to figure out the number of possible combinations than listing them?
No that;s not correct.
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 QUESTION:
We are told that x=abcd, where a, b, c, and d are four distinct prime numbers. According to the above the # of factors of x including 1 and x is (1+1)(1+1)(1+1)(1+1)=16. Excluding 1 and x the # of factors is 16-2=14.
Re: If X = the product of four distinct prime numbers, how many [#permalink]
06 Sep 2013, 13:45
salsal wrote:
If X = the product of four distinct prime numbers, how many factors does X have besides 1 and itself?
(a) 10 (b) 11 (c) 12 (d) 13 (e) 14
My approach:
Let's take the four primes: 2, 3, 5 and 7 --> each of the 4 primes will be a factor of the product
All the possible combinations of the four primes will be factors (10 possible combinations): 2x3 2x5 2x7 3x5 3x7 5x7 2x3x5 2x3x7 2x5x7 3x5x7
Thus, in total there are 10 + 4 = 14 factors.
Is my reasoning correct? Is there a faster way to figure out the number of possible combinations than listing them?
I believe the formula for the amount of factors in an number was to prime factorize it, and then add 1 to all the exponents and multiply them. Since it's only going to be 4 prime numbers all to the 1st power that are multiplied, each of the exponents is 1. So basically this number will have 16 factors, because (2^4) = 16. The question asks you to exclude 2 of the factors, so the answer is 16.
Re: If X = the product of four distinct prime numbers, how many [#permalink]
29 Nov 2013, 11:25
salsal wrote:
If X = the product of four distinct prime numbers, how many factors does X have besides 1 and itself?
(a) 10 (b) 11 (c) 12 (d) 13 (e) 14
My approach:
Let's take the four primes: 2, 3, 5 and 7 --> each of the 4 primes will be a factor of the product
All the possible combinations of the four primes will be factors (10 possible combinations): 2x3 2x5 2x7 3x5 3x7 5x7 2x3x5 2x3x7 2x5x7 3x5x7
Thus, in total there are 10 + 4 = 14 factors.
Is my reasoning correct? Is there a faster way to figure out the number of possible combinations than listing them?
I took the combinations approach as well. Instead of listing them out you I picked combinations of 3, 2 and 1. Combination of all 4 numbers is out because we are told that number and itself should be excluded.
Re: If X = the product of four distinct prime numbers, how many [#permalink]
26 Feb 2015, 06:03
Hello from the GMAT Club BumpBot!
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. _________________
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...