# Find the no. of trailing zeroes of $20190!!$

Find the no. of trailing zeroes of $$20190!!$$

What I did was I found the no. of trailing zeroes of $$20190!$$ and divided it by $$2$$ and got $$2522$$, but the correct answer was $$2521$$. What's the correct solution here?

The answer by liaombro already mentioned this, but did not go into details, so I would like to explain it a little bit here.

Here, $$n!!$$ is taken to denote the double factorial of $$n$$.

By definition, \begin{align*} 20190!! &= (20190) (20188) (20186) \dotsm (2) \\ &= 2(10095) \cdot 2(10094) \cdot 2(10093) \dotsm 2(1) \\ &= 2^{10095} \cdot (10095)(10094)(10093) \dotsm (1) \\ &= 2^{10095} \cdot 10095!. \end{align*}

Now you should be able to figure out the number of trailing zeros of $$10095$$ by yourself. After that, note that $$10095!$$ contains strictly less factors of $$5$$ than $$2$$. Therefore, multiplying $$10095!$$ by $$2^{10095}$$ does not introduce new trailing zeros.

As per the other answers, $$20190!!=2^{10095}\cdot 10095!$$

Now, according to Legendre's Theorem $$\nu_p(n!)=\frac{n-s_p(n)}{p-1}$$ With $$p=2$$ $$\nu_2(10095!)=\frac{10095-s_2(10011101101111)}{2-1}=10095-10=10085$$ With $$p=5$$ $$\nu_5(10095!)=\frac{10095-s_5(310340)}{5-1}=\frac{10095-11}{4}=2521$$ and finally $$20190!!=2^{10095+10085}\cdot 5^{\color{red}{2521}} \cdot Q$$ where $$Q$$ is not contributing to the trailing zero's, since it doesn't contain any $$2$$ or $$5$$ in its factorization.

Hint: $$20190!! = 2^{10095} \cdot 10095!$$

• (+1) for the hint. Commented May 12, 2019 at 10:14