How many zeroes does $2012!$ end with?
My idea is:
402 zeroes come from $2\times 5$, 80 from $2\times 25$, 16 from $2\times 125$ and 3 from $2\times 625$
How can we "show" that this is true?
Answer
The correct answer is 501.
In order to find the number of zeros is same as finding the number of factors of powers of $5$. There are more factors of powers of $2$ than the factors of powers of $5$.
For instance $10! = 3628800 = \hspace{3pt}2^8 \hspace{3pt} 3^4 \hspace{3pt}5^2\hspace{3pt} 7$
$$\left \lfloor \frac{n}{p} \right \rfloor +\left \lfloor \frac{n}{p^2} \right \rfloor +\left \lfloor \frac{n}{p^3}\right \rfloor + \cdots \left \lfloor \frac{n}{p^{k-1}} \right \rfloor$$
where $\left \lfloor \frac{n}{p^k} \right \rfloor=0$.
In this case $k=5$ because $\left \lfloor \frac{2012}{5^5} \right \rfloor=0$
$$\left \lfloor \frac{2012}{5} \right \rfloor =402, \hspace{6pt} \left \lfloor \frac{2012}{5^2} \right \rfloor = \left \lfloor \frac{402}{5} \right \rfloor =80$$
$$\left \lfloor \frac{2012}{5^3} \right \rfloor = \left \lfloor \frac{80}{5} \right \rfloor =16, \hspace{6pt} \left \lfloor \frac{2012}{5^4} \right \rfloor = \left \lfloor \frac{16}{5} \right \rfloor =3$$
$$\left \lfloor \frac{2012}{5} \right \rfloor+ \left \lfloor \frac{2012}{5^2}\right \rfloor +\left \lfloor \frac{2012}{5^3}\right \rfloor +\left \lfloor \frac{2012}{5^4} \right \rfloor = 402+80+16+3=501$$
No comments:
Post a Comment