14.0 Exponent of a Prime Number in $n!$

Let, $$ \Rightarrow n! = {p_1}^{{\alpha _1}}.{p_2}^{{\alpha _2}}.{p_3}^{{\alpha _3}}.{p_4}^{{\alpha _4}}....$$

Then ${\alpha _k}$ is given as, $$ \Rightarrow {\alpha _k} = \left[ {\frac{n}{{{p_k}}}} \right] + \left[ {\frac{n}{{{p_k}^2}}} \right] + \left[ {\frac{n}{{{p_k}^3}}} \right] + ..... + \left[ {\frac{n}{{{p_k}^m}}} \right]$$

where ${p^m} \le n \le {p^{m + 1}}$.

Here $\left[ x \right]$ is the greatest integer less than or equal to $x$.

Question 28. Find the number of zeros in $55!$.

Solution: The product of $5$ and $2$ gives $10$, i.e. the trailing zeros. The maximum number of such pair in a factorial gives the total number of zeros.

Let us consider, $$ \Rightarrow 55! = {2^{{\alpha _1}}}{.5^{{\alpha _2}}}.{p_1}^{{\alpha _3}}.{p_2}^{{\alpha _4}}....$$

Now, the number of $2$'s, i.e. the exponent of $2$ in the above equation is,

$$ \Rightarrow {\alpha _1} = \left[ {\frac{{55}}{{{2^1}}}} \right] + \left[ {\frac{{55}}{{{2^2}}}} \right] + \left[ {\frac{{55}}{{{2^3}}}} \right] + \left[ {\frac{{55}}{{{2^4}}}} \right] + \left[ {\frac{{55}}{{{2^5}}}} \right] + \left[ {\frac{{55}}{{{2^6}}}} \right]$$

$$ \Rightarrow {\alpha _1} = \left[ {\frac{{55}}{2}} \right] + \left[ {\frac{{55}}{4}} \right] + \left[ {\frac{{55}}{8}} \right] + \left[ {\frac{{55}}{{16}}} \right] + \left[ {\frac{{55}}{{32}}} \right] + \left[ {\frac{{55}}{{64}}} \right]$$

$$ \Rightarrow {\alpha _1} = \left[ {27.5} \right] + \left[ {13.75} \right] + \left[ {6.875} \right] + \left[ {3.4375} \right] + \left[ {1.7187} \right] + \left[ {0.8593} \right]$$

$$\begin{equation} \begin{aligned} \Rightarrow {\alpha _1} = 27 + 13 + 6 + 3 + 1 + 0 \\ \Rightarrow {\alpha _1} = 50 \\\end{aligned} \end{equation} $$

Thus the number of $2$s are $50$.

Now, the number of $5$'s, i.e. the exponent of $5$ in the above equation is

$$\begin{equation} \begin{aligned} \Rightarrow {\alpha _2} = \left[ {\frac{{55}}{{{5^1}}}} \right] + \left[ {\frac{{55}}{{{5^2}}}} \right] + \left[ {\frac{{55}}{{{5^3}}}} \right] \\ \Rightarrow {\alpha _2} = \left[ {\frac{{55}}{5}} \right] + \left[ {\frac{{55}}{{25}}} \right] + \left[ {\frac{{55}}{{125}}} \right] \\ \Rightarrow {\alpha _2} = 11 + 2 + 0 \\ \Rightarrow {\alpha _2} = 13 \\\end{aligned} \end{equation} $$

Thus the maximum number of pairs of $2$ and $5$ are $13$.

Thus there are $13$ zeros at the end of $55!$