BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
Trailiiiiiing zer0s
Source: Beat The GMAT — Problem Solving |
-
Brian@VeritasPrep
- GMAT Instructor
- Posts: 1031
- Location: Malibu, CA
- Thanked: 716 times
- GMAT Score:750
Good question - I've seen a few variations on this one, and the biggest "catch" is what makes the answer here 14.
They're asking how many 0s this number has at the end, or in other words how many times this number can be divided by 10. If you prime-factor out 10, that means that we need to know how many pairings of 2*5 this number has. (For example, 2*15 is 30, which is divisible by 10 once because 30 is 2*3*5, and the 2 and 5 form a 10)
Well, there will be tons of 2s - every second number has at least one factor of 2. But there will be fewer 5s - just consider 5!: 1*2*3*2*2*5 ---> There are 3 2s and one 5. So we really need to know how many 5s this number has.
We can find multiples of 5 at:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 ----> there are 12 multiples of 5 in this set
HOWEVER - keep in mind that both 25 and 50 are divisible by 5 twice. 25 = 5*5 and 50 = 2*5*5. So each of those numbers provides and extra factor of 5, so we're up to 14 5s, and enough 2s to pair with each, so this number will have 14 trailing zeroes.
They're asking how many 0s this number has at the end, or in other words how many times this number can be divided by 10. If you prime-factor out 10, that means that we need to know how many pairings of 2*5 this number has. (For example, 2*15 is 30, which is divisible by 10 once because 30 is 2*3*5, and the 2 and 5 form a 10)
Well, there will be tons of 2s - every second number has at least one factor of 2. But there will be fewer 5s - just consider 5!: 1*2*3*2*2*5 ---> There are 3 2s and one 5. So we really need to know how many 5s this number has.
We can find multiples of 5 at:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 ----> there are 12 multiples of 5 in this set
HOWEVER - keep in mind that both 25 and 50 are divisible by 5 twice. 25 = 5*5 and 50 = 2*5*5. So each of those numbers provides and extra factor of 5, so we're up to 14 5s, and enough 2s to pair with each, so this number will have 14 trailing zeroes.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
GMAT Instructor
Chief Academic Officer
Veritas Prep
Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
-
anshumishra
- Legendary Member
- Posts: 543
- Thanked: 147 times
shortcut :Ramit88 wrote:how many trailing zeroes does 60! have?
a 12
b 9
c 15
d 14
e 13
ANS A
can someone explain this concept in depth
The number of trailing zeros in n! can be determined with this formula:
n/5 + n/5^2+n/5^3+...+n/5^k ; where 5^(k+1) > n
So, number of trailing zeros in 60! (since, 5^3 > 60) = 60/5 + 60/5^2 = 12+2 = 14.
Hope that helps !
[Note : why does this formula work ? The formula actually counts the number of factors of 5 in n!, but since there are at least as many factors of 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.]
Thanks
Anshu
(Every mistake is a lesson learned )
Anshu
(Every mistake is a lesson learned )
I approached the problem as below.little timeconsuming may be...
60! = 1.2.3.4.5...........58.59.60
Which of these factors will produce trailing zeros.
One zero - (2,5),(10),(12,15),(20),(30),(32,35),(40),(42,45),(52,55),(60),
Two zeros - (50,6),(8,25)
The count is 14. Reason i approach the problem like this: a number or pair of number once used cannot be used further to produce a trailing zero. So when a factor has been used , it cannot be used any further to do calculation.
60! = 1.2.3.4.5...........58.59.60
Which of these factors will produce trailing zeros.
One zero - (2,5),(10),(12,15),(20),(30),(32,35),(40),(42,45),(52,55),(60),
Two zeros - (50,6),(8,25)
The count is 14. Reason i approach the problem like this: a number or pair of number once used cannot be used further to produce a trailing zero. So when a factor has been used , it cannot be used any further to do calculation.
-
Abhishek009
- Master | Next Rank: 500 Posts
- Posts: 359
- Location: Kolkata, India
- Thanked: 50 times
Perfectly said !anshumishra wrote:shortcut :Ramit88 wrote:how many trailing zeroes does 60! have?
a 12
b 9
c 15
d 14
e 13
ANS A
can someone explain this concept in depth
The number of trailing zeros in n! can be determined with this formula:
n/5 + n/5^2+n/5^3+...+n/5^k ; where 5^(k+1) > n
So, number of trailing zeros in 60! (since, 5^3 > 60) = 60/5 + 60/5^2 = 12+2 = 14.
Hope that helps !
[Note : why does this formula work ? The formula actually counts the number of factors of 5 in n!, but since there are at least as many factors of 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.]
If U want to reduce the calculations then proceed as follows
60 / 5 = 12
12 /5 = 2
Ignore the decimal part and add those , it gives you 14.
Again it can be done by :
60/5 + 60 / 25
=>12 + 2 = 14
Both the methods are same...
Abhishek
WOW never seen a problem like this before! I very much doubt I would have been able to do this in under 2 mins in the exam, but perhaps I will now having read this post.
What level difficulty would this be?
What level difficulty would this be?
total no of factor of 5 = 60 /5 = 12
total no of factor of 5^2 = 60/25 = 2
total no of factor of 5^3 = 60 /125 = 0
So total no of trailing zeros = 12 + 2 = 14
total no of factor of 5^2 = 60/25 = 2
total no of factor of 5^3 = 60 /125 = 0
So total no of trailing zeros = 12 + 2 = 14
60/5 = 12
60/25 = 2
60/125 = 0
thus 12+2 = 14.
60/25 = 2
60/125 = 0
thus 12+2 = 14.
For Understanding Sustainability,Green Businesses and Social Entrepreneurship visit -https://aamthoughts.blocked/
(Featured Best Green Site Worldwide-https://bloggers.com/green/popular/page2)
(Featured Best Green Site Worldwide-https://bloggers.com/green/popular/page2)
-
ronnie1985
- Legendary Member
- Posts: 626
- Location: Ahmedabad
- Thanked: 31 times
answer is D.
The number of trailing zeroes depends on the highest factor of 5 in that factorial.
so the sequence for highest factor(hf) of 5 between the range of factorials is as follows:'
from
0 to 5!= highest factor of 5= 1
5! to 10! hf(5)=2
10! to 15! =3
15! to 20! =4
20! to 25!=6 (here in 25 we have 5*5 so highest factor raises to 6 instead of 5)
25! to 30! =7
30! to 35! =8
35! to 40! =9
40! to 45!=10
45! to 50!=12(here in 50 we have 5*5*2 so highest factor raises to 12 instead of 11)
50! to 55!=13
55! to 60!=14
The number of trailing zeroes depends on the highest factor of 5 in that factorial.
so the sequence for highest factor(hf) of 5 between the range of factorials is as follows:'
from
0 to 5!= highest factor of 5= 1
5! to 10! hf(5)=2
10! to 15! =3
15! to 20! =4
20! to 25!=6 (here in 25 we have 5*5 so highest factor raises to 6 instead of 5)
25! to 30! =7
30! to 35! =8
35! to 40! =9
40! to 45!=10
45! to 50!=12(here in 50 we have 5*5*2 so highest factor raises to 12 instead of 11)
50! to 55!=13
55! to 60!=14
Simplest way is
60/5 = 12
12/5 = 2
12+2 = 14 zeroes.
Let me know if you want to know the logic behind this.
60/5 = 12
12/5 = 2
12+2 = 14 zeroes.
Let me know if you want to know the logic behind this.
Best,
Bharat
If you find my post helpful, please click on the 'Thank' icon.
Bharat
If you find my post helpful, please click on the 'Thank' icon.
