Sylhet - 2014

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Mahfuz Sobhan
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Joined: Sat Feb 07, 2015 5:40 pm

Sylhet - 2014

Unread post by Mahfuz Sobhan » Tue Jul 07, 2015 2:45 am

$${x} = {m} \times ({m} + 1) \times ({m} + 2) \times ({m} + 3) \times ........ ....\times ({3m} - 1) \times {3m}$$. Here ${x}$ is divisible by ${3}^{k}$, if
m=1000 then find the maximum possible value of k.
Last edited by Mahfuz Sobhan on Wed Jul 08, 2015 2:59 pm, edited 6 times in total.

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seemanta001
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Location: Chittagong

Re: Sylhet - 2014

Unread post by seemanta001 » Tue Jul 07, 2015 9:09 pm

Since $x$ is divisible by $3k$, the maximum possible value of $3k$ will be $x$.
So, the maximum value if $k$ will be $x/3$.
We can also write it as $m^2(m+1)(m+2)(m+3)....(3m-1)$.
So, in this case the maximum value of $k$ will be $1000^2$x$1001$x$1002$......x$2999$.
"Sometimes it's the very people who no one imagines anything of who do the things no one can imagine"

Tahmid
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Joined: Wed Mar 20, 2013 10:50 pm

Re: Sylhet - 2014

Unread post by Tahmid » Wed Jul 08, 2015 2:23 am

Mahfuz Sobhan wrote: Here x is divisible by 3k, if
not $3k$ . it was ${3}^{k}$ in the main problem .
mahfuz sobhan you can use latex to avoid this type of mistakes.
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samiul_samin
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Re: Sylhet - 2014

Unread post by samiul_samin » Mon Feb 18, 2019 4:40 pm

Mahfuz Sobhan wrote:
Tue Jul 07, 2015 2:45 am
$${x} = {m} \times ({m} + 1) \times ({m} + 2) \times ({m} + 3) \times ........ ....\times ({3m} - 1) \times {3m}$$. Here ${x}$ is divisible by ${3}^{k}$, if
$m=1000$ then find the maximum possible value of $k$.
Use floor function to get the answer.

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