Question

The number of zeros at the end of $60!$ is

Answer 1

Here, we will find the number of zeros at the end of a given factorial. We will rewrite the given factorial in the form of the powers of 2 and 5 and the power with the least integer is considered. We will use the number of trailing zeros formula to find the number of zeros at the end of a given factorial.

Formula Used:
The number of trailing zeros in the decimal representation of $n!$ , the factorial of a non negative integer can be determined by this formula: $\dfrac{n}{5} + \dfrac{n}{{{5^2}}} + \dfrac{n}{{{5^3}}} + ..... + \dfrac{n}{{{5^k}}}$ where $k$ must be chosen such that ${5^{k + 1}} > n$
Complete step-by-step answer:
We are given a factorial $60!$.
We will write the given factorial in the form as $2 \times 5 = 10$
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 $60!$.
$\Rightarrow$ 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 $60!$.
$\Rightarrow$ So, we get ${5^2} > {5^3}$
Now, we will compare the least powers from $2$ and $5.$
$\Rightarrow {5^2} > {2^5}$ 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: $\dfrac{n}{5} + \dfrac{n}{{{5^2}}} + \dfrac{n}{{{5^3}}} + ..... + \dfrac{n}{{{5^k}}}$ 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
$\Rightarrow$ Powers of $5$ in 60!$$$$ = \left[ {\dfrac{{60}}{5}} \right] + \left[ {\dfrac{{60}}{{{5^2}}}} \right]
By simplification, we get
$\Rightarrow$ Powers of $5$ in $60!$ $= \left[ {\dfrac{{60}}{5}} \right] + \left[ {\dfrac{{60}}{{25}}} \right]$
By dividing the number $5$, we get
$\Rightarrow$ Powers of $5$ in 60!$$$$ = \left[ {12} \right] + \left[ {2.4} \right]
Since the powers are greatest integer functions, we get
$\Rightarrow$ Powers of $5$ in 60!$$$$ = 12 + 2 = 14
Thus the number of zeros in the given factorial $60!$ is $14$.
Therefore, the number of zeros at the end of $60!$ is 14.

Note: We know that number of zeros at the end is similar to the number of trailing zeros. The function which is rounding off the real number down to the integer less than the number is known as the greatest integer function. We should also note that every multiple of 5 will add a zero to the factorial 60, so we have 12 zeros. A square of 5 has two 5 so we have one zero and also since 50 is a multiple of 25, so we have one additional zero, so we will add two more zeros and thus in total we have 14 numbers of zeros at the end of the factorial 60.