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

It is currently 24 May 2019, 03:11

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 200! have?

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

Hide Tags

 
Intern
Intern
avatar
Joined: 11 Dec 2003
Posts: 49
Location: IN
How many terminating zeroes does 200! have?  [#permalink]

Show Tags

New post 15 Apr 2005, 06:42
3
37
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

64% (00:56) correct 36% (01:40) wrong based on 511 sessions

HideShow timer Statistics

How many terminating zeroes does 200! have?

(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
Most Helpful Community Reply
Director
Director
avatar
Joined: 29 Nov 2012
Posts: 740
Re: How many terminating zeroes does 200! have?  [#permalink]

Show Tags

New post 03 Oct 2013, 05:11
10
6
To find number of terminating zero's count the number of 5's

\(\frac{200}{5}\) = 40

\(\frac{200}{25}\) = 8

\(\frac{200}{125}\) = 1


If you add them its 49 zero's

Explanation of this concept on this thread in detail...

question-on-terminating-zero-s-154440.html
General Discussion
Director
Director
avatar
Joined: 13 Nov 2003
Posts: 755
Location: BULGARIA
  [#permalink]

Show Tags

New post 15 Apr 2005, 07:39
1
You have 40 multiples of 5, 8 of 25 and 1 of 125. This will give 49 zeros
Intern
Intern
avatar
Joined: 11 Dec 2003
Posts: 49
Location: IN
  [#permalink]

Show Tags

New post 15 Apr 2005, 08:40
You have the correct answer.

Please explain how did you go for 5, 25, 125 what is the reasoning for your approach to this soln.
Director
Director
avatar
Joined: 13 Nov 2003
Posts: 755
Location: BULGARIA
  [#permalink]

Show Tags

New post 16 Apr 2005, 00:15
4
in order to determine the zeros you need to know how many 2 and 5 200! contains. Apparently, 2 are much more than 5 since every second number is a multiple of 2. That is why when you find the multiples of 5 you will get the number of zeros. Now every 5-th number is a 5 multiple, where every 25-th has two fives, and 125 has 3 fives. So you have 40 times 5, 8 times 5x5 and 1 time 5x5x5. But note that 25 and 125 already contain one five from the fives multiples, and you will have 40+8+1 times 5 which is 49.
Hope that helps
Intern
Intern
avatar
Joined: 23 Dec 2014
Posts: 17
Re: How many terminating zeroes does 200! have?  [#permalink]

Show Tags

New post 25 Feb 2015, 08:28
1
To solve this question, it would be good to know the formular: n!/5^x and in this example: 200/5=40 200/25=8 and 200/125=1 and then the sum is 49, Hence C
Manager
Manager
User avatar
Joined: 27 Jun 2014
Posts: 67
Location: New Zealand
Concentration: Strategy, General Management
GMAT 1: 710 Q43 V45
GRE 1: Q161 V163

GRE 2: Q159 V166
GPA: 3.6
WE: Editorial and Writing (Computer Software)
How many terminating zeroes does 200! have?  [#permalink]

Show Tags

New post 25 Feb 2015, 12:36
2
4
TS wrote:
How many terminating zeroes does 200! have?

(A) 40
(B) 48
(C) 49
(D) 55
(E) 64


Nice question. Answer is C.

To answer a question about trailing zeros in a factorial, we must realize that the trailing zeros depend on the number of 10s (i.e. 10s or its multiples) in the factorial. 10=5*2, so we are looking for 5s and 2s. By virtue of being the more frequent number, we don't need to count the number of 2s, because there will always be equal or more 2s than 5s in a factorial. So, essentially we are looking for 5s and its exponents.

Start with finding 5s. 200/5= 40. So we have 40 fives (and 40 twos) which means we'll have 40 zeros. If the answer was a decimal, such as 201/5, round down.

Now look for two 5s, that is 25. So, 200/25= 8, so we have eight more zeros.

Lastly, 5^3= 125, and we have one such number in 200!, which gives us one more trailing zero. Therefore, we have 40+8+1=49 trailing zeros.

Hope this helps.
_________________
"Hardwork is the easiest way to success." - Aviram

One more shot at the GMAT...aiming for a more balanced score.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55272
Re: How many terminating zeroes does 200! have?  [#permalink]

Show Tags

New post 25 Feb 2015, 12:47
EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14198
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: How many terminating zeroes does 200! have?  [#permalink]

Show Tags

New post 25 Feb 2015, 16:45
1
2
Hi All,

It looks like everyone who posted in this thread understands the correct answer. For anyone who doesn't quite get why the math works the way that it does, here are some examples to prove the point.

200! = (200)(199)(198)(197)....(2)(1) so we know that it's a gigantic number. The reason why it will end in a "string" of 0s is because of all of the multiples of 5 involved.

When multiplying integers, there are two ways to get a number that ends in a 0:
1) multiply a multiple of 5 by an even number
2) multiply a multiple of 10 by an integer.

(5)(2) = 10 so we get one 0 for every multiple of 5

However, 25 = (5)(5). It has TWO 5s in it, so there will be two 0s.

eg (25)(4) = 100

With 125 = (5)(5)(5), we have THREE 5s, so there will be three 0s.

eg (125)(8) = 1,000

So when we divide 200 by 5.....200/5 = 40 multiples of 5. SOME of those multiples of 5 are multiples of 25 (or 125) though. Each of those special cases has to be accounted for.

25, 50, 75, 100, 125, 150, 175 and 200 are all multiples of 25, so they each include one extra 0 (and 125 includes two extra 0s).

So we have 40 + 8 + 1 = 49 zeroes.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
SVP
SVP
avatar
B
Joined: 06 Nov 2014
Posts: 1877
Re: How many terminating zeroes does 200! have?  [#permalink]

Show Tags

New post 28 Feb 2015, 06:46
1
1
TS wrote:
How many terminating zeroes does 200! have?

(A) 40
(B) 48
(C) 49
(D) 55
(E) 64


Multiplication of 2s and 5s gives a 0.
We need equal number of 2s and 5s.
Since there are less powers of 5 than 2, we'll consider powers of 5 and multiply with equal number of 2s.
5, 25, 125 are the only three powers of 5 below 200.
Now, 200/5 = 40. Hence there are 40 multiples of 5 which when multiplied by multiples of 2 will contribute a 0.
200/25 = 8. Hence there are 8 multiples of 25 which when multiplied by multiples of 2 will contribute a 0.
200/125 = 1 (1 is the quotient). Hence there is 1 multiple of 125 which when multiplied by multiples of 2 will contribute a 0.

Hence there are 40+8+1 = 49 terminating zeros.
Hence option C.

--
Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: http://www.optimus-prep.com/gmat-on-demand-course
Current Student
User avatar
B
Joined: 20 Jan 2017
Posts: 58
Location: United States (NY)
Schools: CBS '20 (A)
GMAT 1: 750 Q48 V44
GMAT 2: 610 Q34 V41
GPA: 3.92
Reviews Badge
Re: How many terminating zeroes does 200! have?  [#permalink]

Show Tags

New post 25 Jan 2017, 10:11
1) To paraphrase this question, we need to find the number of prime factor pairs 2*5 in the product 200!
2) In a range of consecutive integers, there will always be more prime factors 2, than prime factors 5, so we only need to calculate the number of prime factors of 5.
3) 200/5=40; 200/25=8; 200/125=1
4) 40+8+1=49

Posted from my mobile device
Target Test Prep Representative
User avatar
G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2823
Re: How many terminating zeroes does 200! have?  [#permalink]

Show Tags

New post 08 Mar 2018, 17:44
TS wrote:
How many terminating zeroes does 200! have?

(A) 40
(B) 48
(C) 49
(D) 55
(E) 64


To determine the number of trailing zeros in a number, we need to determine the number of 5-and-2 pairs within the prime factorization of that number. Note that each 5-and-2 pair creates a “10,” which contributes one trailing zero.

Since we know there are fewer 5s in 200! than there are 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 200!, we can use the following shortcut in which we divide 200 by 5, then divide the quotient of 200/5 by 5 and continue this process until we no longer get a nonzero quotient.

200/5 = 40

40/5 = 8

8/5 = 1 (we can ignore the remainder)

Since 1/5 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 200!.

Thus, there are 40 + 8 + 1 = 49 factors of 5 within 200!. This means that there are 49 5-and-2 pairs in 200!; therefore, there are 49 terminating zeros in 200!.

Answer: C
_________________

Jeffrey Miller

Head of GMAT Instruction

Jeff@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
Joined: 18 Nov 2014
Posts: 13
Re: How many terminating zeroes does 200! have?  [#permalink]

Show Tags

New post 23 Mar 2019, 05:06
Number of trailing zero's are only possible due to the number 0f pairs of (2*5)
Thus we must count the number of 2's and 5's in 200! But as the number of 5's is less in 200!, it makes sense to count the total number of 5's in it.

Let's count the number of 5's exactly once in each sequence=> Equal to the number of terms in a sequence

1) 5, 10,15,20,25.......200 We can use the direct formula to count no. of terms : [(Last-First)/Difference] +1
=[(200-5)/5]+1
=39+1
=40

2) 25,50, 75,100......200
=[(200-25)/5] + 1
=7+1
=8

3) 125
=1

Thus the total number of 5's in all the above :
40+8+1=49

Thus the total number of trailing zeroes is 49
GMAT Club Bot
Re: How many terminating zeroes does 200! have?   [#permalink] 23 Mar 2019, 05:06
Display posts from previous: Sort by

How many terminating zeroes does 200! 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®.