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:

A number N when expressed as product of prime factors gives. [#permalink]

Show Tags

06 Oct 2010, 05:07

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

43% (02:42) correct
57% (01:07) wrong based on 33 sessions

HideShow timer Statistics

A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pairwise co-prime.

A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pairwise co-prime.

A. 1027 B. 3125 C. 243 D. 729 E. None of these

Question is quite ambiguous but here is what I think:

So we are looking for all cases of N = x * y * z like:

In this case factors x * y * z would be co-prime (wont share any common factor but 1) as each prime will be only in one factor.

As there are total of 5 primes in N then there are following cases possible:

1. {1}*{1}*{factor with all five primes} - 1 (1*1*N);

2. {1}*{factor with one prime}*{factor with four primes} - \(C^1_5*C^4_4=5\) (\(C^1_5\) - choosing which prime will be in one-prime factor, the rest primes go to the third factor);

3. {1}*{factor with two prime}*{factor with three primes} - \(C^2_5*C^3_3=10\) (\(C^2_5\) - choosing which 2 primes will be in two-prime factor, the rest primes go to the third factor);

4. {factor with one prime}*{factor with one prime}*{factor with three primes} - \(C^3_5=10\) (\(C^3_5\) - choosing which 3 primes will be in three-prime factor, from the 2 primes left one will go to the first factor and another to the second);

5. {factor with one prime}*{factor with two primes}*{factor with two primes} - \(C^1_5*\frac{C^2_4}{2}=15\) (\(C^1_5\) - choosing which 1 primes will be in one-prime factor, 4 primes left can be split among two factors (into two groups) in \(\frac{C^2_4}{2}\) ways, dividing by 2 as the order of the factors (groups) does not matter);

Total: \(1+5+10+10+15=41\).

No such answer among answer choices.

So I guess it's meant that the order of the factors is important, for example N= (2^5) * (3^2) * (5^4 * 7 * 11^3) is different from N= (3^2) * (2^5) * (5^4 * 7 * 11^3), then case 1 can be arranged in 3 ways and cases 2, 3, 4, and 5 in 3! ways: \(3*1+3!(5+10+10+15)=243\).

So answer: C.

But if the order of the factors is important then the problem can be solved easier: each of the five primes can be part of the first, the second or the third factor so each prime has 3 options. Total ways to distribute 5 primes would be \(3*3*3*3*3=3^5=243\).

Answer: C.

Anyway not a GMAT question, so I wouldn't worry about it at all.
_________________

Re: A number N when expressed as product of prime factors gives. [#permalink]

Show Tags

06 Oct 2010, 14:42

pzazz12 wrote:

A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pairwise co-prime.

A.1027 B.3125 C.243 D.729 E.None of these

The trick is treat \(2^5, 3^2, 5^4, 7, 11^3\) as 5 objects which you need to place in 3 groups. Order being irrelevant as that cannot matter

So the answer would be : \(\frac{5!}{0!0!5!} + \frac{5!}{1!0!4!} + \frac{5!}{1!1!3!} + \frac{5!}{2!0!3!} + \frac{5!}{2!1!2!}\) \(= 1 + 5 + 20 + 10 + 30 =66\)

Now its been too long since I learnt P&C, and this is a hard question ... so I am not 100% confident, but I would guess (e)

What I am 100% confident about is that this is beyond a GMAT difficultly level
_________________

A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pairwise co-prime.

A.1027 B.3125 C.243 D.729 E.None of these

The trick is treat \(2^5, 3^2, 5^4, 7, 11^3\) as 5 objects which you need to place in 3 groups. Order being irrelevant as that cannot matter

So the answer would be : \(\frac{5!}{0!0!5!} + \frac{5!}{1!0!4!} + \frac{5!}{1!1!3!} + \frac{5!}{2!0!3!} + \frac{5!}{2!1!2!}\) \(= 1 + 5 + 20 + 10 + 30 =66\)

Now its been too long since I learnt P&C, and this is a hard question ... so I am not 100% confident, but I would guess (e)

What I am 100% confident about is that this is beyond a GMAT difficultly level

I think 66 contains duplication.

For example if we manually count case 1-1-3 we will get: {2}-{3}-{5, 7, 11} {2}-{5}-{3, 7, 11} {2}-{7}-{3, 5, 11} {2}-{11}-{3, 5, 7} {5}-{3}-{2, 7, 11} {7}-{3}-{2, 5, 11} {11}-{3}-{2, 5, 7} {5}-{7}-{2, 3, 11} {5}-{11}-{2, 3, 7} {7}-{11}-{2, 3, 5}

Total of 10 cases (when order is not important) but as per your formula it's \(\frac{5!}{1!1!3!}=20\). The same with 1-2-2, it's also twice as much. If you subtract this duplications 10 and 15 then you'll get 41 as in my calculations.
_________________

Re: A number N when expressed as product of prime factors gives. [#permalink]

Show Tags

06 Oct 2010, 15:17

Good spot ! Agreed.

I can intuitively tell why the duplication exists. Its sort of because the two "equal" groups in 3-1-1 and 2-2-1 are somehow treated as "distint" in these formulae. I agree with the 41. Surely order can't metter though
_________________

Re: A number N when expressed as product of prime factors gives. [#permalink]

Show Tags

08 Oct 2010, 03:29

I've seen this question being posted on beatthegmat.com and according to experts there, it is beyond GMAT. How can one solve this question within 2 minutes? or even 3 mins I think. "pairwise co-prime" statement is ambiguous.
_________________

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

A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of threee factors such that the factors are pairwise co-prime. 1027 3125 243 729
_________________

Re: A number N when expressed as product of prime factors gives. [#permalink]

Show Tags

22 Nov 2016, 08:46

pzazz12 wrote:

A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pairwise co-prime.

A.1027 B.3125 C.243 D.729 E.None of these

“Pairwise coprime” means that if we choose any 2 numbers from a given set of numbers they will be co-prime to each other.

This question is about splitting into groups*.

We have 5 possible cases including those which have 1 as a factor:

1*1*(group of 5 primes) 1*(group of 1 prime)(group of 4 primes) 1*(group of 2 primes) (group of 3 primes) (group of 1 prime)*(group of 1 prime)*(group of 3 primes) (group of 1prime)*(group of 2 primes)*(group of 2 primes)

* - in each group primes are different and choosing 1 as a factor means choosing 0 from a group of a given 5 different prime factors.

Values of given answer options suggest that order of factors matters (we won’t be able to achieve such big numbers otherwise).

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...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...