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Sylhet - 2014

Posted: 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.

Re: Sylhet - 2014

Posted: 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$2999$.

Re: Sylhet - 2014

Posted: 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.

Re: Sylhet - 2014

Posted: 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.