Mymensingh Higher Secondary 2019#7

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samiul_samin
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Joined: Sat Dec 09, 2017 1:32 pm

Mymensingh Higher Secondary 2019#7

Unread post by samiul_samin » Fri Feb 22, 2019 5:12 pm

$1000!$ is divided by $4^a$,What is the highest value of $a$?

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samiul_samin
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Re: Mymensingh Higher Secondary 2019#7

Unread post by samiul_samin » Sat Feb 23, 2019 12:33 am

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$

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