GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Aug 2019, 13:50

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

How many terminating zeroes does 121! have?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Intern
Intern
User avatar
B
Joined: 07 Dec 2016
Posts: 38
Reviews Badge
How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 14 Apr 2017, 03:59
2
8
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

71% (00:51) correct 29% (01:10) wrong based on 137 sessions

HideShow timer Statistics

How many terminating zeroes does 121! have?

A) 21
B) 24
C) 28
D) 35
E) 41

_________________
Cheers!
If u like my post..... payback in Kudos!! :beer
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 57025
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 14 Apr 2017, 04:06
2
2
amathews wrote:
How many terminating zeroes does 121! have?

A) 21
B) 24
C) 28
D) 35
E) 41


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

Fro example, 125000 has 3 trailing zeros (125000);

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:

According to above 121! has \(\frac{121}{5}+\frac{121}{25}=24+4=28\) trailing zeros.

Answer: C.

For more check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.



Hope this helps.
_________________
Retired Moderator
avatar
P
Joined: 04 Aug 2016
Posts: 481
Location: India
Concentration: Leadership, Strategy
GPA: 4
WE: Engineering (Telecommunications)
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 14 Apr 2017, 08:00
is it really a 700 level qs?

24 + 4 = 28

C
Board of Directors
User avatar
D
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4568
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)
GMAT ToolKit User
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 14 Apr 2017, 08:02
amathews wrote:
How many terminating zeroes does 121! have?

A) 21
B) 24
C) 28
D) 35
E) 41


\(\frac{121}{5} = 24\)
\(\frac{24}{5} = 4\)

So, Total no of trailing zeroes is 28 , answer will be (C) 28
_________________
Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 9535
Location: Pune, India
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 22 Aug 2018, 00:52
2
amathews wrote:
How many terminating zeroes does 121! have?

A) 21
B) 24
C) 28
D) 35
E) 41


Responding to a pm:

Terminating 0s are formed by creation of 10 ( = 2*5)
121! will certainly have more 2s than 5s so 5 will be our limiting factor. We need to find the number of 5s in 121! and that will give us the number of 10s (and hence the number so terminating 0s)

121/5 = 24
24/5 = 4

There will be 24+4 = 28 5s in 121! and hence 28 terminating zeroes.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
IIMA, IIMC School Moderator
User avatar
V
Joined: 04 Sep 2016
Posts: 1365
Location: India
WE: Engineering (Other)
CAT Tests
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 23 Aug 2018, 00:19
Thanks VeritasKarishma for your two cents.

Quote:
121! will certainly have more 2s than 5s so 5 will be our limiting factor.


To confirm your approach for highlighted text, you did not perform 121! as 121 * 120 * ...1
Instead you found a pattern that just as 10! has more even numbers from 1 to 10
than multiples of 5 from 1 to 10, similarly 121! will have follow same pattern, correct?
_________________
It's the journey that brings us happiness not the destination.

Feeling stressed, you are not alone!!
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 9535
Location: Pune, India
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 23 Aug 2018, 08:59
adkikani wrote:
Thanks VeritasKarishma for your two cents.

Quote:
121! will certainly have more 2s than 5s so 5 will be our limiting factor.


To confirm your approach for highlighted text, you did not perform 121! as 121 * 120 * ...1
Instead you found a pattern that just as 10! has more even numbers from 1 to 10
than multiples of 5 from 1 to 10, similarly 121! will have follow same pattern, correct?


Yes. It is true for all numbers. Every second number has a 2 while only every fifth number has a 5.
Check this post for more:
https://www.veritasprep.com/blog/2011/0 ... actorials/
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Target Test Prep Representative
User avatar
D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 7367
Location: United States (CA)
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 26 Aug 2018, 19:14
1
amathews wrote:
How many terminating zeroes does 121! have?

A) 21
B) 24
C) 28
D) 35
E) 41


To determine the number of trailing zeros in a number, we need to determine the number of factors of 10, which is equal to the number of 5- and-2 pairs within the prime factorization of that number.

Since we know there are fewer 5s in 121! than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 121!, we can use the following shortcut in which we divide 121 by 5, then divide the quotient of 121/5 by 5 and continue this process until we no longer get a nonzero quotient.

121/5 = 24 (we can ignore the remainder)

24/5 = 4 (we can ignore the remainder)

Since 4/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 121!.

Thus, there are 24 +4 = 28 factors of 5 within 120! and so there are 28 5-and-2 pairs, creating 28 trailing zeros.

Answer: C
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern
Intern
avatar
B
Joined: 25 Feb 2016
Posts: 11
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 27 Aug 2018, 05:09
Bunuel Please explain how n5+n52+n53+...+n5kn5+n52+n53+...+n5k , where k must be chosen such that 5^(k+1)>n

I don't get this part "where k must be chosen such that 5^(k+1)>n"
VP
VP
User avatar
D
Status: Learning stage
Joined: 01 Oct 2017
Posts: 1031
WE: Supply Chain Management (Energy and Utilities)
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 27 Aug 2018, 06:32
1
onyx12102 wrote:
Bunuel Please explain how n5+n52+n53+...+n5kn5+n52+n53+...+n5k , where k must be chosen such that 5^(k+1)>n

I don't get this part "where k must be chosen such that 5^(k+1)>n"


Hi onyx12102,

IMO, we will carry out the operations(successive division & addition) as long as \({Denominator}\leq{Numerator}\) Or,\(5^k\leq{n}\)

Here n=121, So, \(5^k\) is less than 121 when k=2 (\(5^3=125\), DISCARD)

So, \(\frac{121}{5}+\frac{121}{5^2}\)
_________________
Regards,

PKN

Rise above the storm, you will find the sunshine
Manager
Manager
avatar
B
Joined: 20 Apr 2019
Posts: 81
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 10 Aug 2019, 21:48
VeritasKarishma wrote:
amathews wrote:
How many terminating zeroes does 121! have?

A) 21
B) 24
C) 28
D) 35
E) 41


Responding to a pm:

Terminating 0s are formed by creation of 10 ( = 2*5)
121! will certainly have more 2s than 5s so 5 will be our limiting factor. We need to find the number of 5s in 121! and that will give us the number of 10s (and hence the number so terminating 0s)

121/5 = 24
24/5 = 4

There will be 24+4 = 28 5s in 121! and hence 28 terminating zeroes.

Why do have to divide the 24 by 5 again?
Intern
Intern
avatar
B
Joined: 05 Jul 2019
Posts: 23
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 10 Aug 2019, 22:32
to calculating terminating zeroes, divide the number by powers for 5...

121/5 + 121/5^2 = 121/5 + 121/25 = 24 + 4 =28
GMAT Club Bot
Re: How many terminating zeroes does 121! have?   [#permalink] 10 Aug 2019, 22:32
Display posts from previous: Sort by

How many terminating zeroes does 121! have?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne