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: Product of sequence [#permalink]
17 Sep 2010, 02:29

15

This post received KUDOS

Expert's post

13

This post was BOOKMARKED

rafi wrote:

If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which \(\frac{N}{10^m}\) is an integer?

a. 3 b. 6 c. 7 d. 8 e. 10

How do you solve these sort of questions quickly Thanks

We should determine # of trailing zeros of N=3*6*9*12*15*...*99 (a sequence of 0's of a number, after which no other digits follow).

Since there are at least as many factors 2 in N as factors of 5, then we should count the number of factors of 5 in N and this will be equivalent to the number of factors 10, each of which gives one more trailing zero.

Factors of 5 in N: once in 15; once in 30; once in 45; once in 60; twice in 75 (5*5*3); once in 90;

1+1+1+1+2+1=7 --> N has 7 trailing zeros, so greatest integer \(m\) for which \(\frac{N}{10^m}\) is an integer is 7.

Re: Product of sequence [#permalink]
20 Oct 2010, 03:14

Expert's post

nonameee wrote:

Bunuel, is it necessary to count the number of trailing zeros? I have solved the problem by counting the number of 5's in N.

It's basically the same. Since there are at least as many factors 2 as factors of 5 in N, then finding the number of factors of 5 in N would be equivalent to the number of factors 10, each of which gives one more trailing zero. _________________

Re: Product of sequence [#permalink]
15 Nov 2012, 11:10

Bunuel wrote:

It's basically the same. Since there are at least as many factors 2 as factors of 5 in N, then finding the number of factors of 5 in N would be equivalent to the number of factors 10, each of which gives one more trailing zero.

How did you know that 2 factors and 5 factors in N are same?

Re: Product of sequence [#permalink]
16 Nov 2012, 02:32

Expert's post

Amateur wrote:

Bunuel wrote:

It's basically the same. Since there are at least as many factors 2 as factors of 5 in N, then finding the number of factors of 5 in N would be equivalent to the number of factors 10, each of which gives one more trailing zero.

How did you know that 2 factors and 5 factors in N are same?

No, that's not what I'm saying (see the red part). The power of 2 in N is at least as high as the power of 5 in N.

We are told that N=3*6*9*12*15*18*21*...*90*93*96*99 --> as you can observe, the power of 2 in N will be higher than the power of 5 (there are more even numbers than multiples of 5).

If N is the product of all multiples of 3 between 1 and 100 [#permalink]
16 Nov 2012, 06:58

3

This post received KUDOS

I did it in a different way..... since it is multiplication of all 3 multiples.... 3*6*9*..... *99=(3^33)(1*2*3*4*5*......33)=(3^33)*33! (3 power 33 because a 3 can be extracted from each number inside) (3^33) doesn't have any multiples between 1-9 which can contribute a 0..... so number of trailing 0's should be number of trailing 0's of 33! which is 7. So C is the answer... we don't need to count 5's and 2's and complicate things in this case! Let me know if you think this approach of mine has loop holes.

Last edited by Amateur on 23 Jan 2015, 08:04, edited 2 times in total.

Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
16 Nov 2012, 07:47

i got answer as '6' bunuel am i missing something? I may be completely wrong. Below is my approach :-

N= 3*6*9*12*.......*99

Then total multiple of 3 will be 33

then every alternate number have factor of 2 in it so total factor of 2 will be 6 Since highest number which is multiple of 2 is 96 which have total 6 factors of 2 _________________

Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
04 Jun 2013, 04:48

5

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

rafi wrote:

If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which \(\frac{N}{10^m}\) is an integer?

A. 3 B. 6 C. 7 D. 8 E. 10

How do you solve these sort of questions quickly Thanks

Responding to a pm:

First, check out this post. It is an application of a concept that discusses the maximum power of a number in a factorial. This post discusses how and why we find the maximum power. http://www.veritasprep.com/blog/2011/06 ... actorials/

Once you are done, note that this question can be easily broken down into the factorial form.

Re: If N is the product of all multiples of 3 between 1 and 100 [#permalink]
17 Aug 2013, 11:37

rafi wrote:

If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which \(\frac{N}{10^m}\) is an integer?

A. 3 B. 6 C. 7 D. 8 E. 10

How do you solve these sort of questions quickly Thanks

stunning math.

my solution: 3.6.9.12..............................99 = 3^33 . (1.2.3.4.5...............................33) factors that can bring zeros are = 2,5,10,12,15,20,22,25,30,32 (All from 2 , 0 and 5 ) But only 25 can produce two 5.

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...