Read the following instruction carefully and answer the question that follows:
Expression $$\sum_{n=1}^{13}\frac{1}{n}$$ can also be written as $$\frac{x}{13!}$$ What would be the remainder if x is divided by 11?

Expression : $$\sum_{n=1}^{13}\frac{1}{n} = \frac{x}{13!}$$

=> $$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ...... + \frac{1}{13} = \frac{x}{13!}$$

=> $$x = \frac{13!}{1} + \frac{13!}{2} + \frac{13!}{3} + ......... + \frac{13!}{13}$$

Now, if $$x$$ is divided by 11

=> $$\frac{13! + \frac{13!}{2} + \frac{13!}{3} + ......... + \frac{13!}{13}}{11}$$

All terms are divisible by 11 except $$\frac{13!}{11}$$

$$\therefore$$ Remainder if x is divided by 11 = Remainder of $$\frac{13!}{11 \times 11}$$

= $$(10! \times 12 \times 13) \% 11$$

= $$(10 \times 1 \times 2) \% 11 = 20 \% 11 = 9$$