For every natural number $n,$ $n(n^2 - 1)$ is divisible by

The sum $1! + 4! + 7! + 10! + 12! + 13! + 15! + 16! + 17!$ is divisible by

If $11^{12}-11^2=k(5 \times 10^9+6 \times 10^9+33 \times 10^8+110 \times 10^7+\ldots+33)$,then $k=$

The value of the greatest positive integer $k$,such that $49^k + 1$ is a factor of $48(49^{125} + 49^{124} + \ldots + 49^2 + 49 + 1)$ is

If $2 \cdot 4^{2k+1} + 3^{3k+1} = 11t$ and $2 \cdot 4^{2k+3} + 3^{3k+4} = 11(pt + 3^q)$,where $k, t \in Z^{+}$,then $(p, q) =$

The remainder when $5^{99}$ is divided by $13$ is