Mymensingh Higher Secondary 2019#7
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
$1000!$ is divided by $4^a$,What is the highest value of $a$?
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: Mymensingh Higher Secondary 2019#7
Answer:$497$
Solution:$4^a=2^{2a}$
Using Legendre's theorem
$2a=\lfloor \dfrac {1000}{2^1}\rfloor+\lfloor \dfrac {1000}{2^2}\rfloor+\lfloor \dfrac {1000}
{2^3}\rfloor+\lfloor \dfrac {1000}{2^4}\rfloor+\lfloor \dfrac {1000}{2^5}\rfloor+\lfloor \dfrac
{1000}{2^6}\rfloor+\lfloor \dfrac {1000}{2^7}\rfloor+\lfloor \dfrac {1000}{2^8}\rfloor+\lfloor
\dfrac {1000}{2^9}\rfloor+\lfloor \dfrac {1000}{2^{10}}\rfloor=994\Rightarrow a=497$
Solution:$4^a=2^{2a}$
Using Legendre's theorem
$2a=\lfloor \dfrac {1000}{2^1}\rfloor+\lfloor \dfrac {1000}{2^2}\rfloor+\lfloor \dfrac {1000}
{2^3}\rfloor+\lfloor \dfrac {1000}{2^4}\rfloor+\lfloor \dfrac {1000}{2^5}\rfloor+\lfloor \dfrac
{1000}{2^6}\rfloor+\lfloor \dfrac {1000}{2^7}\rfloor+\lfloor \dfrac {1000}{2^8}\rfloor+\lfloor
\dfrac {1000}{2^9}\rfloor+\lfloor \dfrac {1000}{2^{10}}\rfloor=994\Rightarrow a=497$