A Consecutivity problem

From the problem we have the equations: $n=9x+36$

$n=10y+45$

$n=11z+55$ ,which can be rewritten as : $n=9(x+4)$

$n=10(y+4)+5$

$n=11(z+5)$ . From the first and third equation we can see that n is divisible by $99$ .Notice that from the second equation we find that $n$ ends with $5$ .The smallest number $n=99k$ with these conditions will be obtained when $k=5$ and thus $n=99*5=\boxed{495}$ .