Mymensingh Higher Secondary 2019#7
Posted: Fri Feb 22, 2019 5:12 pm
$1000!$ is divided by $4^a$,What is the highest value of $a$?
Answer:$497$
Solution:$4^a=2^{2a}$
Using Legendre's theorem
$2a=\lfloor \dfrac {1000}{2^1}\rfloor+\lfloor \dfrac {1000}{2^2}\rfloor+\lfloor \dfrac {1000}
{2^3}\rfloor+\lfloor \dfrac {1000}{2^4}\rfloor+\lfloor \dfrac {1000}{2^5}\rfloor+\lfloor \dfrac
{1000}{2^6}\rfloor+\lfloor \dfrac {1000}{2^7}\rfloor+\lfloor \dfrac {1000}{2^8}\rfloor+\lfloor
\dfrac {1000}{2^9}\rfloor+\lfloor \dfrac {1000}{2^{10}}\rfloor=994\Rightarrow a=497$