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

It is currently 15 Nov 2018, 11:02

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
Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
  • Free GMAT Strategy Webinar

     November 17, 2018

     November 17, 2018

     07:00 AM PST

     09:00 AM PST

    Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
  • GMATbuster's Weekly GMAT Quant Quiz # 9

     November 17, 2018

     November 17, 2018

     09:00 AM PST

     11:00 AM PST

    Join the Quiz Saturday November 17th, 9 AM PST. The Quiz will last approximately 2 hours. Make sure you are on time or you will be at a disadvantage.

How many terminating zeroes does 121! have?

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

Hide Tags

Intern
Intern
User avatar
B
Joined: 07 Dec 2016
Posts: 41
Reviews Badge
How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 14 Apr 2017, 02:59
2
6
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

72% (00:51) correct 28% (01:05) wrong based on 160 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: 50613
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 14 Apr 2017, 03:06
1
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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Retired Moderator
avatar
P
Joined: 04 Aug 2016
Posts: 508
Location: India
Concentration: Leadership, Strategy
GPA: 4
WE: Engineering (Telecommunications)
Premium Member
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 14 Apr 2017, 07:00
is it really a 700 level qs?

24 + 4 = 28

C
Board of Directors
User avatar
P
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4210
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)
GMAT ToolKit User Premium Member
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 14 Apr 2017, 07:02
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
P
Joined: 16 Oct 2010
Posts: 8531
Location: Pune, India
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 21 Aug 2018, 23:52
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 >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Study Buddy Forum Moderator
User avatar
D
Joined: 04 Sep 2016
Posts: 1249
Location: India
WE: Engineering (Other)
Premium Member CAT Tests
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 22 Aug 2018, 23:19
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.

Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8531
Location: Pune, India
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 23 Aug 2018, 07:59
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 >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Target Test Prep Representative
User avatar
P
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4170
Location: United States (CA)
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 26 Aug 2018, 18:14
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

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

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, 04:09
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"
Director
Director
User avatar
P
Status: Learning stage
Joined: 01 Oct 2017
Posts: 930
WE: Supply Chain Management (Energy and Utilities)
Premium Member
Re: How many terminating zeroes does 121! have?  [#permalink]

Show Tags

New post 27 Aug 2018, 05:32
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

GMAT Club Bot
Re: How many terminating zeroes does 121! have? &nbs [#permalink] 27 Aug 2018, 05:32
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  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.