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:

Numbers divisible by 50 not divisible by 100 would be divisible by \(2^1\) and \(5^2\), but power of 2 cannot be more than 1 as anything including \(2^2\) and \(5^2\) would be divisible by 100. So only 1 power of 2 allowed(2^1) and 5 powers of 5(\(5^2\),\(5^3\),\(5^4\),\(5^5\),\(5^6\)) =1*6*5*9 (1 powers of 2)*(6 powers of 3)*(5 powers of 5)*(9 powers of 7) =270

If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

A. 240 B. 345 C. 270 D. 120 E. None of these

To be divisible by 50, the factor must have \(2*5^2\) and to be not divisible by 100, it must NOT have \(2^2*5^2\). Hence the only constraints are on the power of 2 (which must be 1) and the power of 5 (which must be greater than or equal to 2)

The other prime factors can appear in any way in the factor. So number of factors = 1*(6)*(5)*(9) = 270 1 - because 2 can have a power in only one way 6 - because 3 can have 6 different powers (0/1/2/3/4/5) 5 - because 5 can have 5 different powers (2/3/4/5/6) 9 - because 7 can have 9 different powers (0/1/2/3/4/5/6/7/8)

Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]

Show Tags

21 May 2013, 22:42

VeritasPrepKarishma wrote:

GMATtracted wrote:

If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

A. 240 B. 345 C. 270 D. 120 E. None of these

To be divisible by 50, the factor must have \(2*5^2\) and to be not divisible by 100, it must NOT have \(2^2*5^2\). Hence the only constraints are on the power of 2 (which must be 1) and the power of 5 (which must be greater than or equal to 2)

The other prime factors can appear in any way in the factor. So number of factors = 1*(6)*(5)*(9) = 270 1 - because 2 can have a power in only one way 6 - because 3 can have 6 different powers (0/1/2/3/4/5) 5 - because 5 can have 5 different powers (2/3/4/5/6) 9 - because 7 can have 9 different powers (0/1/2/3/4/5/6/7/8)

Why can't 5 have 6 different powers. Because, you add 1 to the powers of 3 and 7, which I understand. Also, one 2 is understandable because more than one 2 will lead to divisibility by 100. Thanks!

If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

A. 240 B. 345 C. 270 D. 120 E. None of these

To be divisible by 50, the factor must have \(2*5^2\) and to be not divisible by 100, it must NOT have \(2^2*5^2\). Hence the only constraints are on the power of 2 (which must be 1) and the power of 5 (which must be greater than or equal to 2)

The other prime factors can appear in any way in the factor. So number of factors = 1*(6)*(5)*(9) = 270 1 - because 2 can have a power in only one way 6 - because 3 can have 6 different powers (0/1/2/3/4/5) 5 - because 5 can have 5 different powers (2/3/4/5/6) 9 - because 7 can have 9 different powers (0/1/2/3/4/5/6/7/8)

Why can't 5 have 6 different powers. Because, you add 1 to the powers of 3 and 7, which I understand. Also, one 2 is understandable because more than one 2 will lead to divisibility by 100. Thanks!

\(50 = 2*5^2\)

The power of 5 must be at least 2. It can be 3, 4, 5 or 6 as well. Here there are 5 different ways in which you can give a power to 5. _________________

Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]

Show Tags

29 Jun 2014, 07:16

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. _________________

If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]

Show Tags

31 Aug 2015, 09:11

here, we have to catch all factors for 50.

100=2^2*5^2 50=2*5^2

Now, to have number divisible by 50 and not by 100 we have to make sure that 2 is taken only once. As, if we pick 2 two's then it will form 100 with no. of 5 selected and since min 5 we have to select is 2, we cannot have 2 more than one.

Number= 2^7*3^5*5^6*7^8

So, there are 6 ways of picking 3, 9 ways of picking 7, 5 ways of picking 5 and only one way of picking 2 and that is 2^1.

Why only 5 ways for picking 5 when we have 7 ways(0,1,2,3,4,5,6)?

Reason: we need 50 so we have to pick 5^2 and greater than it. Also, we can pick only 2^1 because picking 2^2 will form 100 with 5 already selected. For example, 5^2*2^2=100. So, we cannot afford it.

Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by [#permalink]

Show Tags

06 Apr 2016, 11:12

If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

50=2*5^2 100=2^2*5^2 N=(2*2^6)*(5^2*5^4)*(3^5)*(7^8) =(2*5^2)*(2^6*5^4*3^5*7^8) We dont need any other power of 2 except (2*5^2) no of factors (power of 5+1)*(power of 3+1)*(power of 7+1)= 5*6*9=270

Hence C

gmatclubot

Re: If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by
[#permalink]
06 Apr 2016, 11:12

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...