Max $m$ for $(\frac{1}{11^m}\prod_{i=1000}^{2014}i)\in N$

[Albert1]

$(\frac{1}{11^m}\prod_{i=1000}^{2014}i)\in N$
please find max($m$)

 

[kaliprasad]

$(\frac{1}{11^m}\prod_{i=1000}^{2014}i)\in N$
please find max($m$)

Spoiler

we need to find how many numbers between 1000 and 2014 are divisible by $11 ,11^2, 11^3$ so on

$\lfloor\dfrac{999}{11}\rfloor = 90$
$\lfloor\dfrac{2014}{11}\rfloor\ = 183$

$\lfloor\dfrac{999}{11^2}\lfloor\ = 8$
$\lfloor\dfrac{2014}{11^2}\rfloor\ = 16$

$\lfloor\dfrac{999}{11^3}\lfloor\ = 0$
$\lfloor\dfrac{2014}{11^3}\rfloor = 1$

so number of numbers divisible by 11 is 93 by $11^2$ is 8 and $11^3$ is 1

so highest poser is 93 + 8 +1 = 102
or m = 102