Factorials
Number of trailing zeros in n! :
Number of trailing zeros in n! = no. of times n! is divided by 10
= Highest power of 10 which divides n!
= no. of 5’s in n!
\(= \left( {\frac{n}{5}} \right) + \left( {\frac{n}{{25}}} \right) + \left( {\frac{n}{{125}}} \right) + ........\)
In doing this we consider the quotient as whole number if we get a decimal.
We use another simple method to count no. of trailing zeros i.e.,
Divide the number ‘n’ in n! by 5 and its subsequent quotients by 5 as long as the quotient is non-zero (each time ignore any non-zero remainder). Finally, add up all these non zero quotients and that will be the no. of zeros in n!
Ex : No. of trailing zeros at the end of 100!.
=\(= \frac{{100}}{5} + \frac{{100}}{{25}} + \frac{{100}}{{125}}.......\)
\(\begin{gathered}
= 20 + 4 + 0.8 \hfill \\
= 24 \hfill \\
\end{gathered} \) (Only whole numbers to be considered)
No. of trailing zeros are 20 + 4= 24