Find the number of positive integers $n$ such that $n+2n^2+3n^3+\cdots +2005n^{2005}$ is divisible by $n-1.$

Let $n$ be the number $\underset{2006 9\text{'}s}{\underbrace{(999999999...999{)}^2}}-\underset{2006 6^{\text{'}}\mathrm{s}}{\underbrace{(666666666\dots 666{)}^2}}.$ Find the remainder when $n$ is divided by $11$.

Consider the non-decreasing sequence of positive integers

$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5,\dots \dots .$ in which the $n^{\mathrm{th}}$ positive integer appears $n$ times. The remainder when the ${2019}^{\mathrm{th}}$ term is divided by $5$ is