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

 It is currently 19 Nov 2018, 16:10

### 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

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### How to QUICKLY Solve GMAT Questions - GMAT Club Chat

November 20, 2018

November 20, 2018

09:00 AM PST

10:00 AM PST

The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.
• ### The winning strategy for 700+ on the GMAT

November 20, 2018

November 20, 2018

06:00 PM EST

07:00 PM EST

What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.

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

Author Message
TAGS:

### Hide Tags

Director
Joined: 04 Sep 2015
Posts: 510
Location: India
WE: Information Technology (Computer Software)
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

26 Mar 2017, 00:06
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
Joined: 02 Sep 2016
Posts: 688
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

03 Sep 2017, 02:47
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
Joined: 02 Sep 2009
Posts: 50670
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

03 Sep 2017, 03:38
Shiv2016 wrote:
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?

Yes, since 2 < 5, then in n! there will be at leas as many 2's as 5's.
_________________
Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 267
How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

16 Sep 2017, 09:21
chetan2u wrote:
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

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 ...?

Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3307
Location: India
GPA: 3.12
How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

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

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
Joined: 09 Mar 2016
Posts: 1101
How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

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

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!: $$\frac{49}{5}+\frac{49}{5^2}=9+1=10$$

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

$$9+2 =11$$
Director
Joined: 20 Feb 2015
Posts: 796
Concentration: Strategy, General Management
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

07 Aug 2018, 23:51
1
dave13 wrote:
Bunuel wrote:
chetan2u wrote:
How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

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!: $$\frac{49}{5}+\frac{49}{5^2}=9+1=10$$

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

$$9+2 =11$$

The formula works only when we have a factorial !
when we have an integer , it has to be factorised.

consider 6!
6!= 6*5*4*3*2*1
no of zeroes = no of 5*2 pairs
we have only one in this case , so no of trailing zeroes = 1
using the formula
6/5=1

now consider 6
6=2*3
doesn't have a five in it , and therefore doesn't have a trailing zero

similarly 51 is an integer
factors of 51 are 1,3,17,51
no 5's or 2's , so no trailing zeroes
VP
Joined: 09 Mar 2016
Posts: 1101
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

08 Aug 2018, 00:18
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
Joined: 20 Feb 2015
Posts: 796
Concentration: Strategy, General Management
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

08 Aug 2018, 00:24
1
dave13 wrote:
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

Thats how the formula works .

you keep on dividing the numerator with increasing powers 5 .

you might find this useful !!

https://gmatclub.com/forum/gmat-math-book-87417.html
Director
Joined: 20 Feb 2015
Posts: 796
Concentration: Strategy, General Management
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

08 Aug 2018, 00:29
Is there a glitch ?
How come my post went to page 1 ?
VP
Joined: 09 Mar 2016
Posts: 1101
Re: How many trailing Zeroes does 49! + 50! have?  [#permalink]

### Show Tags

08 Aug 2018, 00:34
CounterSniper wrote:
Is there a glitch ?
How come my post went to page 1 ?

i marked it as the best community reply
Re: How many trailing Zeroes does 49! + 50! have? &nbs [#permalink] 08 Aug 2018, 00:34

Go to page   Previous    1   2   [ 31 posts ]

Display posts from previous: Sort by