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:

1. Trailing zeros: Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

Back to the original question: If m is the product of all integers from 1 to 40, inclusive, what is the greatest integer p for which 10^p is a factor of m? (A) 7 (B) 8 (C) 9 (D) 10 (E) 11

Given: \(m=40!\). So we should find the # of trailing zeros in 40!, as it'll be the greatest value of p for which 40!/10^p will be an integer.

40! has \(\frac{40}{5}+\frac{40}{5^2}=8+1=9\) trailing zeros, which means that 40! ends with 9 zeros so p=9 is the greatest integer for which 10^p (10^9) is a factor of 40!.

To find what k is, find how many times you can make 10 with the numbers in 1-40.

First, find the 10's as they are going to be most restrictive. Only 10, 20, 30 and 40 have 10 as a factor. 4 total.

Next find 5's. 5, 15, 25 and 35 each have a 5, with 25 having 2. 5 total.

You dont need to find the twos as you know there are more than 5's.

Add the two you get 9.

I would suggest that you refrain from actually counting the 5s/10s. Not only can it be tedious with bigger numbers, but it is also prone to errors. With smaller numbers, I don't think it will matter very much what you do. For some theory on these kind of questions, check out this link: [url] if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html#p834162[/url]

Once you understand such problems, you wouldn't care how big or small the numbers are... You would be able to solve such questions easily. _________________

As for your question, forget 40!. Let's look at 10!

What is the maximum value of p such that 10^p is a factor of 10! To make a 10, you need a 2 and a 5. In 10!, will you have more 2s or more 5s?

1*2*3*4*5*6*7*8*9*10 - Every alternate number has at least one 2 in it. Every fifth number has at least one 5 in it. You certainly have more 2s than 5s. If you have eight 2s and two 5s, how many 10s can you make? You certainly need a 5 to make a 10. Since you have only two 5s, you can make only two 10s. It doesn't matter how many 2s are extra. You cannot make a 10 with only 2s. _________________

The first four terms give rise to four 0's, the next three to three 0's and the last one to two 0's and the total is 9.

Note: You may also multiply 5 with 4 and/or 6 and/or 8 as they contain a 2. Similarly for 15, 25 and 35. But the number of 0's is restricted by the number of 5's. _________________

1. Trailing zeros: Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

Back to the original question: If m is the product of all integers from 1 to 40, inclusive, what is the greatest integer p for which 10^p is a factor of m? (A) 7 (B) 8 (C) 9 (D) 10 (E) 11

Given: \(m=40!\). So we should find the # of trailing zeros in 40!, as it'll be the greatest value of p for which 40!/10^p will be an integer.

40! has \(\frac{40}{5}+\frac{40}{5^2}=8+1=9\) trailing zeros, which means that 40! ends with 9 zeros so p=9 is the greatest integer for which 10^p (10^9) is a factor of 40!.

It explains this method in detail. Also, you divide by 5 because to make a 10, you need a 2 and a 5. The number of 5s will be fewer than the number of 2s hence all you need to do is find the number of 5s you have. You can make that many 10s. _________________

Re: If m is the product of all integers from 1 to 40, inclusive [#permalink]

Show Tags

14 Nov 2014, 08:23

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. _________________

The only time you can lose is when you give up. Try hard and you will suceed. Thanks = Kudos. Kudos are appreciated

https://gmatclub.com/forum/rules-for-posting-in-verbal-gmat-forum-134642.html When you post a question Pls. Provide its source & TAG your questions Avoid posting from unreliable sources.

My posts https://gmatclub.com/forum/beauty-of-coordinate-geometry-213760.html#p1649924 https://gmatclub.com/forum/calling-all-march-april-gmat-takers-who-want-to-cross-213154.html https://gmatclub.com/forum/possessive-pronouns-200496.html https://gmatclub.com/forum/double-negatives-206717.html

Re: If m is the product of all integers from 1 to 40, inclusive [#permalink]

Show Tags

28 Jun 2016, 04:35

1

This post was BOOKMARKED

I know this is an old post but Bunuel I understand your answer and I did remember to use the formula too however I didnt know what to do with 40/25. To take that as a 2 or a 1. Is it that we have to take an INT value or do we round it up? Just stuck there

Thanks Parth

Bunuel wrote:

MRHDK1 wrote:

Hey guys,

I need some help cracking this question, as I simply don't understand the explanation;

If m is the product of all integers from 1 to 40, inclusive, what is the greatest integer p for which 10^p is a factor of m?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11

1. Trailing zeros: Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

Back to the original question: If m is the product of all integers from 1 to 40, inclusive, what is the greatest integer p for which 10^p is a factor of m? (A) 7 (B) 8 (C) 9 (D) 10 (E) 11

Given: \(m=40!\). So we should find the # of trailing zeros in 40!, as it'll be the greatest value of p for which 40!/10^p will be an integer.

40! has \(\frac{40}{5}+\frac{40}{5^2}=8+1=9\) trailing zeros, which means that 40! ends with 9 zeros so p=9 is the greatest integer for which 10^p (10^9) is a factor of 40!.

Re: If m is the product of all integers from 1 to 40, inclusive [#permalink]

Show Tags

28 Jun 2016, 05:15

Expert's post

ppb1487 wrote:

I know this is an old post but Bunuel I understand your answer and I did remember to use the formula too however I didnt know what to do with 40/25. To take that as a 2 or a 1. Is it that we have to take an INT value or do we round it up? Just stuck there

Thanks Parth

You should take only the quotient into the account, thus 40/25 = 1.

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

They say you get better at doing something by doing it. then doing it again ... and again ... and again, and you keep doing it until one day you look...