Last visit was: 24 Apr 2026, 08:26 It is currently 24 Apr 2026, 08:26
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 12 Mar 2026
Posts: 1,841
Own Kudos:
8,511
 [22]
Given Kudos: 707
Location: India
Posts: 1,841
Kudos: 8,511
 [22]
Find the number of trailing zeros in (23!+24!+26!). 18 Nov 2019, 03:48
2
Kudos
Add Kudos
20
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

29% (01:50) correct 71% (01:26) wrong based on 197 sessions

History

Date
Time
Result
Not Attempted Yet
Find the number of trailing zeros in (23!+24!+26!).

A. 2
B. 4
C. 5
D. 8
E. 10
ShowHide Answer
Official Answer
E
Most Helpful Reply
User avatar
pk123
User avatar
Joined: 16 Sep 2011
Last visit: 26 Apr 2020
Posts: 104
Own Kudos:
122
 [8]
Given Kudos: 158
Products:
My Reviews
Posts: 104
Kudos: 122
 [8]
Find the number of trailing zeros in (23!+24!+26!). 18 Nov 2019, 04:09
6
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
nick1816
Find the number of trailing zeros in (23!+24!+26!).

A. 2
B. 4
C. 5
D. 8
E. 10
Show more

(23!+24!+26!)
=23!*(1+24+26*25*24)
=23!*(25+26*25*24)
=23!*25*(1+26*24)
=23!*25*(1+(25+1)*(25-1))
=23!*25*(1+25*25-1)
=23!*25*625
=23!*25^3
no of zeros in 23!= [23/5]+[23/25]=4
no of zeros in 25^3 or 5^6 = 6
so total no of zeros =4+6=10
General Discussion
User avatar
uchihaitachi
User avatar
Joined: 20 Aug 2017
Last visit: 06 Jul 2024
Posts: 89
Own Kudos:
241
 [2]
Given Kudos: 174
Posts: 89
Kudos: 241
 [2]
Find the number of trailing zeros in (23!+24!+26!). 20 Nov 2019, 02:09
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
nick1816
Find the number of trailing zeros in (23!+24!+26!).

A. 2
B. 4
C. 5
D. 8
E. 10
Show more

To find the number of zeros, we need to check for the powers of 5 as it will be the limiting factor.
23!(1+24+26*25*24)
23!(25 + 26*25*24)
23!*25(1+26*24)
23!*25*625.
Now 23! has 4 powers of 5 and 625*25 has 6 powers of 5.
Total = 4+6 = 10(5^10).

OA - E
User avatar
dips1122
User avatar
Joined: 12 Oct 2019
Last visit: 24 Feb 2022
Posts: 72
Own Kudos:
49
 [1]
Given Kudos: 92
Location: India
Concentration: Marketing, General Management
Schools: ISB'22 (A) Tuck '23 (A$) Ross '23 (WL) IIMA PGPX'22 (WA) Fuqua '23 (WL) Darden '23 (WD) Cambridge '22 (D)
GMAT 1: 720 Q48 V41
GMAT 2: 730 Q50 V39
GMAT 3: 760 Q50 V44
GPA: 4
WE:Information Technology (Computer Software)
Products:
My Reviews
Schools: ISB'22 (A) Tuck '23 (A$) Ross '23 (WL) IIMA PGPX'22 (WA) Fuqua '23 (WL) Darden '23 (WD) Cambridge '22 (D)
GMAT 3: 760 Q50 V44
Posts: 72
Kudos: 49
 [1]
Find the number of trailing zeros in (23!+24!+26!). 20 Nov 2019, 06:27
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
pk123
nick1816
Find the number of trailing zeros in (23!+24!+26!).

A. 2
B. 4
C. 5
D. 8
E. 10

(23!+24!+26!)
=23!*(1+24+26*25*24)
=23!*(25+26*25*24)
=23!*25*(1+26*24)
=23!*25*(1+(25+1)*(25-1))
=23!*25*(1+25*25-1)
=23!*25*625
=23!*25^3
no of zeros in 23!= [23/5]+[23/25]=4
no of zeros in 25^3 or 5^6 = 6
so total no of zeros =4+6=10
Show more

Hi Could you plz explain how "no of zeroes in 5^ 6 = 6??" Isn't 5^6 = 15625?
Also what method did u use for zeroes in 23!?
User avatar
pk123
User avatar
Joined: 16 Sep 2011
Last visit: 26 Apr 2020
Posts: 104
Own Kudos:
122
 [1]
Given Kudos: 158
Products:
My Reviews
Posts: 104
Kudos: 122
 [1]
Find the number of trailing zeros in (23!+24!+26!). 20 Nov 2019, 08:48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
dips1122
pk123
nick1816
Find the number of trailing zeros in (23!+24!+26!).

A. 2
B. 4
C. 5
D. 8
E. 10

(23!+24!+26!)
=23!*(1+24+26*25*24)
=23!*(25+26*25*24)
=23!*25*(1+26*24)
=23!*25*(1+(25+1)*(25-1))
=23!*25*(1+25*25-1)
=23!*25*625
=23!*25^3
no of zeros in 23!= [23/5]+[23/25]=4
no of zeros in 25^3 or 5^6 = 6
so total no of zeros =4+6=10

Hi Could you plz explain how "no of zeroes in 5^ 6 = 6??" Isn't 5^6 = 15625?
Also what method did u use for zeroes in 23!?
Show more

To calculate the number of zeros we need to see how we could get 10s which is possible only when we have combination of 2s and 5s as prime factors
So for 23!= 23*22*21*20*19*18*17*........3*2*1

Think as when you multiply each item how many zeros you will get depends on how many 5's and 2's are there in the multiplication....and how would you find how many 2's and 5's are can be found from [ number/5] + [ number /5*5] +[number /5*5*5]+ so on

Why do we need to do this...think of 5!= 5*4*3*2*1=120...how many zeros are there ...1 right but 2s are plenty ....total 3 2's.
so any consecutive multiplication 2's outnumber 5's

coming to 23!, number of zeros can be found by number of 5 it has so [23/5] +[23/5*5]+[23/5*5*5] =4 +0 +0 =4

Number of 5 in 5^6 means how many times 5 appear in 5^6 , which is 6

Hope this answers your question
User avatar
gvij2017
User avatar
Joined: 09 Aug 2017
Last visit: 18 Jun 2024
Posts: 663
Own Kudos:
508
Given Kudos: 778
Posts: 663
Kudos: 508
Find the number of trailing zeros in (23!+24!+26!). 23 Nov 2019, 22:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Though I understood my mistake, I want your help to solve such question.

23! = a0000
24!= b0000
25!= c000000

Now, while summing up all these, I understood my mistake that a+b can lead to another zero.

So to get the number of zero of an expression, get the expression in multiplication/division form only. No addition or subtraction.
User avatar
pk123
User avatar
Joined: 16 Sep 2011
Last visit: 26 Apr 2020
Posts: 104
Own Kudos:
122
Given Kudos: 158
Products:
My Reviews
Posts: 104
Kudos: 122
Find the number of trailing zeros in (23!+24!+26!). 23 Nov 2019, 23:15
Kudos
Add Kudos
Bookmarks
Bookmark this Post
gvij2017
Though I understood my mistake, I want your help to solve such question.

23! = a0000
24!= b0000
25!= c000000

Now, while summing up all these, I understood my mistake that a+b can lead to another zero.

So to get the number of zero of an expression, get the expression in multiplication/division form only. No addition or subtraction.
Show more

Yes absolutely...to get number of zeros, better solve the factorial and bring it to one expression...and in this example we can see 24! and 25! both have 23! in its expanded form ...so anywhere we come through similar expression first simplify and get to mostly simpler version that will help minimize errors
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,973
Own Kudos:
1,117
Posts: 38,973
Kudos: 1,117
Find the number of trailing zeros in (23!+24!+26!). 23 Jan 2024, 00:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109814 posts
Krunaal
Tuck School Moderator
853 posts