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

 It is currently 21 Oct 2018, 19:20

### 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

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

# How many terminating zeroes does 200! have?

Author Message
TAGS:

### Hide Tags

Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 478
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
How many terminating zeroes does 200! have?  [#permalink]

### Show Tags

26 Dec 2011, 11:31
4
00:00

Difficulty:

15% (low)

Question Stats:

71% (00:55) correct 29% (01:46) wrong based on 128 sessions

### HideShow timer Statistics

How many terminating zeroes does 200! have?

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

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Intern
Joined: 29 Jun 2011
Posts: 5
Re: How many terminating zeroes does 200! have?  [#permalink]

### Show Tags

26 Dec 2011, 21:45
2
1
number property states

n!/p + n!/p^2 + n!/p^3 +......

where p is the prime number

to find restricting zeros we need to find number of 5's or 2's in 200! as (10=5*2)

200!/2=100 2's
200/2^2=50 2's
200/2^3=25 2's
200!/2^4 =12 2's
200/2^5=6 2's
200/2^6=3 2's
200/2^7=1 2's
total there are

100+50+25+12+6+3+1= 197 2's

number of 5's in 200! are
40+8+1=49

5 being less in number becomes the restriction for number of 10's in the equation.
hence the answer will be C.
Manager
Joined: 26 Apr 2011
Posts: 213
Re: How many terminating zeroes does 200! have?  [#permalink]

### Show Tags

26 Dec 2011, 23:03
hyena1986 wrote:
number property states

n!/p + n!/p^2 + n!/p^3 +......

where p is the prime number

to find restricting zeros we need to find number of 5's or 2's in 200! as (10=5*2)

200!/2=100 2's
200/2^2=50 2's
200/2^3=25 2's
200!/2^4 =12 2's
200/2^5=6 2's
200/2^6=3 2's
200/2^7=1 2's
total there are

100+50+25+12+6+3+1= 197 2's

number of 5's in 200! are
40+8+1=49

5 being less in number becomes the restriction for number of 10's in the equation.
hence the answer will be C.

nice explained .
could you please clarify whether zeros coming from 10,20......200 also included in this calculation as you have given above?
Senior SC Moderator
Joined: 22 May 2016
Posts: 2038
How many terminating zeroes does 200! have?  [#permalink]

### Show Tags

10 Aug 2017, 07:41
enigma123 wrote:
How many terminating zeroes does 200! have?

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

Factors of 10 create the zeros. In order to count how many tens and thus how many zeros, use the method described by hyena1986 - with one minor adjustment.

There will be many more powers of 2 than there are powers of 5. Every 5 will have a matching 2 (to get 5*2=10), but not every 2 will have a matching 5.

So consider only powers of 5. Divide the factorial number (w/o !) by separate and increasing powers of 5.

Don't worry about remainders (we just need how many powers of $$5^{n}$$ exist in the factorial, not how many "fit" perfectly).

$$\frac{200}{5^{1}}$$ = 40

$$\frac{200}{5^{2}}$$ = $$\frac{200}{25}$$ = 8

$$\frac{200}{5^{3}}$$ = $$\frac{200}{125}$$ = 1

$$5^4$$ is 600. Too large.

Add the results: 49 + 8 + 1 = 49.

_________________

___________________________________________________________________
For what are we born if not to aid one another?
-- Ernest Hemingway

Manager
Joined: 27 Jan 2016
Posts: 142
Schools: ISB '18
GMAT 1: 700 Q50 V34
Re: How many terminating zeroes does 200! have?  [#permalink]

### Show Tags

10 Aug 2017, 07:50
Counting the number of 5's gives us the number of trailing zeroes

There are 40 multiples of 5 within 1 to 200

Every multiple of 25 contributes, an additional 5. Since there are 8 multiples of 25, we have 8 additional 5's

Also 125, a multiple of 25, contains one more 5.

So 40+8+1 = 49
Manager
Joined: 27 Jan 2016
Posts: 142
Schools: ISB '18
GMAT 1: 700 Q50 V34
Re: How many terminating zeroes does 200! have?  [#permalink]

### Show Tags

10 Aug 2017, 07:52
hyena1986 wrote:
number property states

n!/p + n!/p^2 + n!/p^3 +......

where p is the prime number

to find restricting zeros we need to find number of 5's or 2's in 200! as (10=5*2)

200!/2=100 2's
200/2^2=50 2's
200/2^3=25 2's
200!/2^4 =12 2's
200/2^5=6 2's
200/2^6=3 2's
200/2^7=1 2's
total there are

100+50+25+12+6+3+1= 197 2's

number of 5's in 200! are
40+8+1=49

5 being less in number becomes the restriction for number of 10's in the equation.
hence the answer will be C.

In a given range, the number of 5's is always less than the number of 2's. So its always sufficient to count the number of 5's in these kind of problems.
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3907
Location: United States (CA)
Re: How many terminating zeroes does 200! have?  [#permalink]

### Show Tags

15 Aug 2017, 10:49
enigma123 wrote:
How many terminating zeroes does 200! have?

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

Recall that each 5-and-2 pair (which makes 10 when multiplied) in the factorization of a number results in one trailing zero. Thus, we need to determine the number of 5-and-2 pairs within the prime factorization of that number.

Since we know there are fewer 5s than 2s in 200!, 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!

Since there are 49 factors of 5 within 200!, there are 49 5-and-2 pairs and thus 49 trailing zeros.

_________________

Scott Woodbury-Stewart
Founder and CEO

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

Director
Joined: 13 Mar 2017
Posts: 622
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: How many terminating zeroes does 200! have?  [#permalink]

### Show Tags

04 Oct 2017, 05:56
1
enigma123 wrote:
How many terminating zeroes does 200! have?

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

No. of zeroes at the end of 200! = [200/5]+[200/25]+[200/125] = 40 + 8 + 1 = 49
where [x] denotes greatest integer function less than equal to x

_________________

CAT 2017 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.

MBA Social Network : WebMaggu

Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)

What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".

Re: How many terminating zeroes does 200! have? &nbs [#permalink] 04 Oct 2017, 05:56
Display posts from previous: Sort by