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Q.
The number of zeros at the end of $100!$ is
Permutations and Combinations
Solution:
In terms of prime factors $100 !$ can be written as $2^{a} \cdot 3^{b} \cdot 5^{c} \cdot 7^{d}\cdots$
Now, $E_{2}(100 !)$
$=\left[\frac{100}{2}\right]+\left[\frac{100}{2^{2}}\right]+\left[\frac{100}{2^{3}}\right]+\left[\frac{100}{2^{4}}\right]+\left[\frac{100}{2^{5}}\right]+\left[\frac{100}{2^{6}}\right]$
$=50+25+12+6+3+1=97$
and, $E_{5}(100 !)=\left[\frac{100}{5}\right]+\left[\frac{100}{5^{2}}\right]$
$=20+4=24$
$100 ! =2^{97} \cdot 3^{b} \cdot 5^{24} \cdot 7^{d} \ldots $
$=2^{73} \cdot 3^{b} \cdot(2 \times 5)^{24} \cdot 7^{d} \ldots $
$=2^{73} \cdot 3^{b} \cdot(10)^{24} \cdot 7^{d} \ldots$
Hence, number of zeros at the end of $100 !$ is $24 .$