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

It is currently 17 Nov 2019, 15:20

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 trailing Zeroes does 49! + 50! have?

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

Hide Tags

Find Similar Topics 
Intern
Intern
avatar
B
Joined: 04 Jan 2017
Posts: 7
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 23 Mar 2017, 06:08
Hi bunuel,
What if the question was 48!+49!. Please correct me if i'm wrong with my calculation;
Factorization leads to 48!(1+49) => 48!*50 =>No of trailing zeroes in 48! is 9
Since 50=2*2*5; one more trailing zero will be added since we have either a 2 or a 5
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59095
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 23 Mar 2017, 06:21
wwambugu14 wrote:
Hi bunuel,
What if the question was 48!+49!. Please correct me if i'm wrong with my calculation;
Factorization leads to 48!(1+49) => 48!*50 =>No of trailing zeroes in 48! is 9
Since 50=2*2*5; one more trailing zero will be added since we have either a 2 or a 5


48!*50

48! will have 48/5 + 48/25 = 9 + 1 = 10 zeros.

50 = 10*5. 10 there will add 1 more zero. So, will 5 because 48! definitely has an extra 2 for that 5 to make one more trailing zero. Therefore, 48!*50 will have 10 + 1 + 1 = 12 trailing zeros.
_________________
Director
Director
User avatar
P
Joined: 04 Sep 2015
Posts: 662
Location: India
WE: Information Technology (Computer Software)
Premium Member Reviews Badge
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 26 Mar 2017, 01:06
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

49!+50!=49!(1+50)==49!(51)

now for zero to be the end there has to be a 5 therefore 49/5+9/5== 9+1=10 B
Director
Director
avatar
G
Joined: 02 Sep 2016
Posts: 643
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 03 Sep 2017, 03:47
Bunuel

We calculate trailing zeroes using 5 and not 2. In this, do we assume that the number of two's will be at least equal to the 5's?
_________________
Help me make my explanation better by providing a logical feedback.

If you liked the post, HIT KUDOS !!

Don't quit.............Do it.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59095
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 03 Sep 2017, 04:38
Senior Manager
Senior Manager
User avatar
G
Status: love the club...
Joined: 24 Mar 2015
Posts: 265
How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 16 Sep 2017, 10:21
chetan2u wrote:
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22


Self made



hi

great ....

during multiplication, zeros are added ...
during addition, the fewer zeros of the two or (more??) are counted....
is there any rule, however, for subtraction ...?

thanks in advance ..
Senior PS Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 3308
Location: India
GPA: 3.12
How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 16 Sep 2017, 13:41
chetan2u wrote:
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

Self made

Rule : n! is divided by prime number m to leave trailing zeroes.
The number of trailing zeroes is the sum of \(\frac{n!}{(m)}, \frac{n!}{(m^2)}, till \frac{n!}{(m^x)}\) such that \(m^x < n\)

\(49! + 50! = 49! + 50*49! = 49!(1+50) = 49!*51\)

Therefore, 49! has \((\frac{49}{5})9+(\frac{49}{5^2})1 = 10\) zeroes (Option B)

_________________
You've got what it takes, but it will take everything you've got
VP
VP
User avatar
D
Joined: 09 Mar 2016
Posts: 1229
How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 07 Aug 2018, 11:09
Bunuel wrote:
chetan2u wrote:
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22


Self made


Factor out 49! from the expression: \(49! + 50!=49!(1+50)=49!*51\).

51 won't contribute to the number of zeros at the end of the number, therefore all zeros will come from 49!.

Trailing zeros in 49!: \(\frac{49}{5}+\frac{49}{5^2}=9+1=10\)

Answer: B.

THEORY:

Trailing zeros:

Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.

For example 125,000 has 3 trailing zeros;

The number of trailing zeros 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 \(5^{(k+1)} \gt n\)

It's easier if we consider an example:

How many zeros are in the end (after which no other digits follow) of 32!?

\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\). Notice that the last denominator (\(5^2\)) must be less than 32. Also notice that we take into account only the quotient of the division, that is \(\frac{32}{5}=6\).

So there are 7 zeros in the end of 32!.

Another example, how many trailing zeros does 125! have?

\(\frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31\),

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.



Bunuel why 51 won't contribute to the number of zeros at the end of the number ? :? i dont get


\(\frac{49}{5} = 9\)

\(\frac{51}{5^2}\) = 2

\(9+2 =11\) :?
VP
VP
User avatar
D
Joined: 09 Mar 2016
Posts: 1229
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 08 Aug 2018, 01:18
CounterSniper many thanks for great explanation :) i have one question if after factoring out we have 49!(1+50) = 49! *51

why Bunuel uses 49! here twice ---> Trailing zeros in 49!: \(\frac{49}{5}+\frac{49}{5^2}=9+1=10\) we have only one 49! left after factoring out :?

have a good day :)
Director
Director
User avatar
G
Joined: 20 Feb 2015
Posts: 737
Concentration: Strategy, General Management
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 08 Aug 2018, 01:29
Is there a glitch ?
How come my post went to page 1 ?
VP
VP
User avatar
D
Joined: 09 Mar 2016
Posts: 1229
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 08 Aug 2018, 01:34
CounterSniper wrote:
Is there a glitch ?
How come my post went to page 1 ?



:lol: i marked it as the best community reply :cool: :-)
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13602
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

Show Tags

New post 24 Oct 2019, 03:31
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.
_________________
GMAT Club Bot
Re: How many trailing Zeroes does 49! + 50! have?   [#permalink] 24 Oct 2019, 03:31

Go to page   Previous    1   2   [ 32 posts ] 

Display posts from previous: Sort by

How many trailing Zeroes does 49! + 50! have?

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





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