Page **1** of **1**

### Sylhet - 2014

Posted: **Tue Jul 07, 2015 2:45 am**

by **Mahfuz Sobhan**

$${x} = {m} \times ({m} + 1) \times ({m} + 2) \times ({m} + 3) \times ........ ....\times ({3m} - 1) \times {3m}$$. 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**

by **seemanta001**

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**

by **Tahmid**

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**

by **samiul_samin**

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.