$${x} = {m} \times ({m} + 1) \times ({m} + 2) \times ({m} + 3) \times ........ ....\times ({3m}  1) \times {3m}$$. Here ${x}$ is divisible by ${3}^{k}$, if
m=1000 then find the maximum possible value of k.
Sylhet  2014

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Sylhet  2014
Last edited by Mahfuz Sobhan on Wed Jul 08, 2015 2:59 pm, edited 6 times in total.
 seemanta001
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Re: Sylhet  2014
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 $m^2(m+1)(m+2)(m+3)....(3m1)$.
So, in this case the maximum value of $k$ will be $1000^2$x$1001$x$1002$......x$2999$.
So, the maximum value if $k$ will be $x/3$.
We can also write it as $m^2(m+1)(m+2)(m+3)....(3m1)$.
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"
Re: Sylhet  2014
not $3k$ . it was ${3}^{k}$ in the main problem .Mahfuz Sobhan wrote: Here x is divisible by 3k, if
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
Use floor function to get the answer.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 $k$.