Question

how many trailing zeros will be there after the rightmost non-zero digit in the value of 25!

Answer 1

$\underline{\huge{\textbf{Concept\: Behind:}}}$

$\maltese \: \: \sf\textsf{Factorial of a Number 'n' is the product of} \\ \sf \textsf{all the numbers from n all the way down to 1.}\\ \quad \bigstar \quad \displaystyle\boxed{\large{\mathbf{n! = n \times (n-1) \times (n-2) \times \cdots \times 3 \times 2\times 1}}}$

$\textbf{\textdagger}\: \: \sf\textsf{There is a zero (or zeroes) in the End of Product(\textbf{EOP})}\\\sf\textsf{only when the No. to be multiplied consists a power of 10.}\\\sf\textsf{And the No. of zeroes in EOP depends on the power of 10.}$

$\maltese \: \: \sf\textbf{Finding the trailing zeroes in a factorial of} \\ \sf\textbf{of a number is tantamount to computing the highest} \\ \sf \textbf{power of 10 that can divide the number.}$

$\textbf{\textdagger} \: \: \textit{But It is also to be noted that},\\\sf\textsf{As 10 = 2 \times 5 , A zero will be at EOP,} \\\sf\textsf{if a 2 is multiplied with a 5.}$

$\textbf{\textdagger} \: \: \sf\textsf{Any Even Number when Multiplied with }\\\sf\textsf{a power of 5, it results 0 at the EOP.}$

$\begin{tabular}{|c|l|}\cline{1-2}\bf Number & \bf Value of it's Factorial\\ \cline{1-2}1&1\\2&2\\3&6\\4&24\\5&120\\6&720\\7&5040\\...&......0\\\cline{1-2}\end{tabular}$

$\maltese \: \: \sf\textbf{The trailing zeroes depends on} \\ \sf \textbf{powers of 2 or Powers of 5, whichever is less.}$

$\maltese\: \: \sf\textsf{No. of Zeroes in EOP of}\\ \quad \bigstar \quad\displaystyle\boxed{ \textsf{ 2^{m} \times 5^{n} = } \: \begin{cases}m&\textsf{if m < n}\\n&\textsf{if n < m}\end{cases}}$

$\maltese \: \: \sf\textsf{In any factorial, there will be} \\ \qquad \displaystyle\boxed{\sf \textbf{more powers of 2 than the powers of 5.}}$

$\textbf{\textdagger} \: \: \sf\textsf{To find the highest power of 10 that divides a number,}\\\sf\textsf{10 is expressed as product of prime Factors.}\\\sf\textsf{10 = 2 \times 5}$

$\textbf{\textdagger}\: \: \sf\textsf{Here as the powers are same,}\\\sf\textsf{highest number i.e., 5 is considered}\\\sf\textsf{for dividing the number successively.}\\\sf\textsf{to find the highest power of 5 that divides the number}$

$\textbf{\textdagger}\: \: \sf\textsf{Only the quotient has to be taken into the account}\\\sf\textsf{(ignoring remainders) while dividing the number successively,}\\\sf\textsf{till the division with 5 can no longer be done.}$

$\textbf{\textdagger}\: \: \sf\textsf{Adding all quotients will give the highest power of 5 that divides the number.}$

$\textbf{\textdagger}\: \: \sf\textsf{This Power of 5 is the \textbf{Required No. of trailing zeroes}}$



$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\begin{array}{r|l}5 & 25 \\ \cline{1-2} 5 & \bf 5\\ \cline{1-2} & \bf 1\end{array}\begin{array}{cc}&\\\left \big \}\: \: \mathsf{5+1} =\: \bf 6 \: \: \begin{cases}\bf Five\: \sf 5's &\rm 5, 10, 15, 20, 25\\\bf One\: \sf 5 &25\end{cases} \atop \right.&\end{array}$



$\underline{\underline{\large{\textsf{Explanation of the concept:}}}}$

$\begin{array}{r|l}5 & 25 \\ \cline{1-2} & \bf 5\\\end{array}$

$\textbf{\textdagger}\: \:\sf\textsf{When 25 is divided by 5, a quotient of 5 is obtained,} \\ \sf\textsf{which means that there exists 5 Multiples of 5,}\\\sf\textsf{each of them having at least one 5 in it.}$

$\left \big \{\ \atop } \right. \right. \begin{array}{r|l}5 & 25 \\ \cline{1-2} 5 & \bf 5\\ \cline{1-2} & \bf 1\\\end{array}$

$\textbf{\textdagger}\: \: \sf\textsf{When 5 is divided again by 5,}\\\sf\textsf{Which means 25 is divided by multiple of 25,}\\\sf\textsf{(5 \times 5 = 25, here two 5's are the two divisors of successive division)}\\\sf\textsf{a quotient of 1 is obtained,}\\ \sf \textsf{which means there exists 1 multiple of 25.}$

$\textbf{\textdagger}\: \: \sf\textsf{25 is square of 5. Hence, it has two 5's in it.}\\\sf\textsf{One got counted as multiple of 5}\\\sf\textsf{and the \textbf{other} is getting counted as Multiple of 25.}\\\sf\textbf{So, the quotient 1 is obtained.}$

$\sf\textsf{The successive division cannot be continued further.}$

$\sf\textsf{The Two quotients obtained are 5 and 1}$

$\textbf{\textdagger}\: \: \sf\textsf{The highest power of 5 that can divide 25 = 5+1 = 6.}$

$\sf\textsf{So, There are 6 5's in 25!}$

$\sf\underline{\large{\textsf{(Multiples of 5 in 25! = 5, 10, 15, 20, 25)}}}$

$\sf\textsf{These 6 5's get multiplied with 6 powers of 2}\\\sf\textsf{Resulting 6 Zeroes at EOP}$



$\blacksquare \: \: \sf\textsf{The No. of trailing zeros after} \\ \sf \textsf{the rightmost non-zero digit in the value of 25! is } \: \: \boxed{\utilde{\Large{\textbf{6}}}}$