Sum of Reciprocals and Remainder

$x = 13! \sum_{n=1}^{13} \frac{1}{n} = \sum_{n=1}^{13} \frac{13!}{n}$. All terms in the sum contain factor 11 except $\frac{13!}{11} = 13 \times 12 \times 10!$. $x \pmod{11} \equiv (13 \times 12 \times 10!) \pmod{11} \equiv (2 \times 1 \times (-1)) \pmod{11} \equiv -2 \equiv 9$.