The number of zeros at the end of 100! is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To find the number of zeros at the end of $100!$, we need to calculate the exponent of the highest power of $5$ that divides $100!$, denoted as $E

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(D) To find the number of zeros at the end of $100!$, we need to calculate the exponent of the highest power of $5$ that divides $100!$, denoted as $E_5(100!)$.Using Legendre's Formula: $E_p(n!) = \sum_{k=1}^{\infty} \left[ \frac{n}{p^k} \right]$.For $n = 100$ and $p = 5$:$E_5(100!) = \left[ \frac{100}{5} \right] + \left[ \frac{100}{25} \right] + \left[ \frac{100}{125} \right]$$E_5(100!) = 20 + 4 + 0 = 24$.Therefore, there are $24$ zeros at the end of $100!$.

The number of zeros at the end of $100!$ is

Similar Questions

Let $N = n(n+1)(n+2)(n+3)$ where $n$ is a natural number,and $d$ is the number of divisors of $N$. Which of the following is true?

The largest value of $n$, for which $40^n$ divides $60!$, is

The coefficient of $x^3 y^4 z^5$ in the expansion of $(x y+y z+x z)^6$ is

The sum of all positive divisors of $960$ is

The sum of the divisors of $9000$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module