XAT 2014 — QA & DI Question 20

$$\begin{gathered} \sum _{n=1}^{13}\frac{1}{n}=\frac{x}{13!} \\ \therefore x=\frac{13!}{1}+\frac{13!}{2}+\frac{13!}{3}+...+\frac{13!}{11}+\frac{13!}{12}+\frac{13!}{13} \end{gathered}$$

All the terms in x are divisible by 11 except 13!/11

$\frac{13!}{11}=1.2.3.4...10.12.13$

According to Wilson theorem,

$rem\left(\frac{(p-1)!}{p}\right)=-1$

$\frac{13!}{11}=1.2.3.4...10.12.13.=10!\left(12\right)\left(13\right)$

$\therefore rem \left(\frac{{\displaystyle \frac{13!}{11}}}{11}\right)=\frac{10!\left(12\right)\left(13\right)}{11}=(-1).1.2=-2=9$

Hence, option (d).