How many trailing zeroes (zeroes at the end of the number) does 60! have?

Explanatory Answer :

We will write the given factorial in the form as 2×5=10 Now, we will consider the factorial in terms of the power of 2.

So, we get 2^5 = 32; 2^6 = 64;

Thus, we get the value of 2^5 is lesser than the given factorial 60! whereas the value of 2^6 is greater than the given factorial 60! .

⇒ So, we get 2^5>2^6

Now, we will consider the factorial in terms of the power of 5.

So, we get 5^2 = 25; 5^3 = 125;

Thus, we get the value of 5^2 is lesser than the given factorial 60! whereas the value of 5^3 is greater than the given factorial 60!

⇒ So, we get 5^2>5^3

Now, we will compare the least powers from 2 and 5.

⇒5^2>2^5 , and we consider the integer with the least power, so we get 5^2 as an integer with the least power.

The number of trailing zeros in the decimal representation of n!

, the factorial of a non negative integer can be determined by this formula: n/5+n/5^2+n/5^3+.....+n/5^k where k must be chosen such that 5^k+1>n

So, we have the value of k

as 2 and thus we get 5^3=125>60

So, writing the powers of 5 in the given factorial 60! , so we get

⇒ Powers of 5

in 60! =[60/5]+[60/5^2]

By simplification, we get

⇒ Powers of 5 in 60!

=[60/5]+[60/2^5]

By dividing the number 5 , we get

⇒ Powers of 5

in 60! =[12]+[2.4]

Since the powers are greatest integer functions, we get

⇒ Powers of 5 in 60!

=12+2=14

Thus the number of zeros in the given factorial 60! is 14 .

so , A is the correct option