Question

Number of zeros in the expansion of $100!$ is

• A. $24$
• B. $25$
• C. $27$
• D. $20$

Answer 1

The correct option is A $24$

Let $T(n,p)$ denote the highest power of a prime $p$ in $n!$.
To find number of zeros in $n!$, it is sufficient to find the maximum power of $10$ contained in $n!$. Since $10=2\times 5$ and both $2$ and $5$ are its unique factors, required answer is $min\left(T\right(n,2),T(n,5\left)\right)$ . It is easy to see that abundance of $2$ in more than $5$ in $n!$ $($every alternate number has at least one $2$ contributed to the factorial while one out of five contribute a $5).$
Answer = $T(n,5)$.
We first evaluate $T(n,p)$. For this, simply count the number of multiples of $p$(contribute a power of p) plus multiples of $p^2$ (contribute two powers of p but one accounted in previous term) plus multiples of $p^3$(contribute three powers, one accounted in each of previous two terms) and so on.
Therefore, $T(n,p)={\sum }_{i:p^i<n}\left[\frac{n}{p^i}\right]$
Answer = $T(100,5)=\left[\frac{100}{5}\right]+\left[\right[\frac{100}{25}]=24$