I am getting 14.

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.

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

shortcut :

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.]

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.

anshumishra wrote:

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

shortcut :

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.]

Perfectly said !

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...

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?

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

60/5 = 12
60/25 = 2
60/125 = 0

thus 12+2 = 14.

=(60/5)+(60/25) = 14

the prime building block of 10 is 2 and 5..we have 2 in abundance,so count the 5..
60/5=12
then 25 contain an extra 5,50 contain an extra 5(5*2*5)..so 2 5's
total 12+2=14

To find the number of p (prime number) in a certain n!, you add up the quotients of n/p, n/p^2, n/p^3.... so on.

60/5=12
60/25=2

12+2=14

Answer is D

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

(D) 14

[60/5]=12

[60/(5^2)]=2

[60/(5^3)]=0

Required number = 12+2

= 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.