Question

What is the highest power of \[5\] that divides $x = 100!$.

Answer 1

Hint: By the definition of factorial, factorial of a number is the product of all natural numbers from one to that number. So to check the highest power of a certain number in a factorial we have to check the factor in every terms of the product.

Formula used:
For any natural number $n,$ $n!$ (factorial of $n$) is given by $n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1$
For any $a,x,y$, we have ${a^x} \times {a^y} = {a^{x + y}}$

Complete step-by-step answer:
Here we are asked to find the highest power of $5$ that divides $100!$.
Given $x = 100!$
For any natural number $n,$ $n!$ (factorial of $n$) is given by $n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1$
$ \Rightarrow x = 100 \times 99 \times 98 \times ... \times 3 \times 2 \times 1 - - - (i)$
To find the highest power of $5$ that divides $x$, we have to find how many times $5$ is multiplied in the above product.
For that first we can list the multiples of $5$ in the representation $(i)$.
We can see there are $20$ multiples of $5$ which are $5,10,15,20,...,95,20$.
Let $A = 5 \times 10 \times 15 \times ... \times 95 \times 100$
So, we can write $(i)$ as $x = A \times B$ where $B$ is the product of the remaining terms which are not included in $A$.
Since $B$ does not contain any multiples of $5$, we can leave $B$ and focus on $A$.
We can rewrite $A$ as
$A = 5 \times 1 \times 5 \times 2 \times 5 \times 3... \times 5 \times 19 \times 5 \times 20$
Taking all $5$ together we have,
$ \Rightarrow A = 5 \times 5 \times ... \times 5(20times) \times 1 \times 2 \times ... \times 19 \times 20$
$ \Rightarrow A = {5^{20}} \times 1 \times 2 \times ... \times 19 \times 20$
Now we can see that the expression $1 \times 2 \times ... \times 19 \times 20$ contains $4$ multiples of $5$ which are $5,10,15\& 20$.
This gives $A = {5^{20}} \times 5 \times 10 \times 15 \times 20 \times C$, where $C$ is the product of remaining terms.
$ \Rightarrow A = {5^{20}} \times 5 \times 1 \times 5 \times 2 \times 5 \times 3 \times 5 \times 4 \times C$
$ \Rightarrow A = {5^{20}} \times 5 \times 5 \times 5 \times 5 \times 1 \times 2 \times 3 \times 4 \times C$
Simplifying we get,
$ \Rightarrow A = {5^{20}} \times {5^4} \times 1 \times 2 \times 3 \times 4 \times C$
Returning to the equation $x = A \times B$ we get,
$ \Rightarrow x = {5^{20}} \times {5^4} \times 1 \times 2 \times 3 \times 4 \times C \times B$
$ \Rightarrow x = {5^{24}} \times D$ ( since ${a^x} \times {a^y} = {a^{x + y}}$ and $D = 1 \times 2 \times 3 \times 4 \times C \times B$)
Since $D$ contains no multiple of $5$, we have the highest power of $5$ in $x$ is $24$.
$\therefore $ The answer is $24$.

Note: We have another simple method to solve this question.
For some natural number $n$ and prime number $p$, highest power of $p$ dividing $n!$ is given by $\dfrac{n}{p} + \dfrac{n}{{{p^2}}} + \dfrac{n}{{{p^3}}} + ...$
Here, $n = 100,p = 5$
So highest power of $5$ dividing $100!$ is $\dfrac{{100}}{5} + \dfrac{{100}}{{{5^2}}} + \dfrac{{100}}{{{5^3}}} + ... = \dfrac{{100}}{5} + \dfrac{{100}}{{25}} + \dfrac{{100}}{{125}} + ... = 20 + 4 + 0 = 24$