Question

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

• A. $36$
• B. $18$
• C. $24$
• D. None of these

Answer 1

The correct option is C $24$

In terms of prime factors $100!$ can be written as $2^a.3^b.5^c.7^d....$

Now we use the formula to determine the factorial number $100!$ and that is given by

$E_2(100!)=\frac{100}{2}+\frac{100}{2^2}+\frac{100}{2^3}+\frac{100}{2^4}+\frac{100}{2^5}+\frac{100}{2^6}=50+25+12+6+3+1=97$

And

$E_5(100!)=\frac{100}{5}+\frac{100}{5^2}=20+4=24$

Thus $(100!)=2^{97}.3^b.5^{24}.7^d....=2^{73}.3^b.{(2\times 5)}^{24}.7^d...=2^{73}.3^b.{\left(10\right)}^{24}.7^d..$

So the number of zeros at the end of $100!$ are $24$.