<s>[INST] If $ \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\;,$ Then the Remainder when $x$ is Divided by $11$ 
If $\displaystyle \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\;,$ Then the Remainder when $x$ is Divided by $11$.

$\bf{My\; Try::}$ Given $\displaystyle \sum_{r=1}^{13}\frac{1}{r} = \frac{x}{13!}\Rightarrow 13!\left(1+\frac{1}{2}+\frac{1}{3}+................+\frac{1}{13}\right) = x$
So $\displaystyle x = \left(13!+\frac{13!}{2}+\frac{13!}{3}+.................+12!\right)$
Now How Can I solve after that
Help me
Thanks
 [/INST] Imagine bringing the left side to the common denominator $13!$. The numerator $x$ is then a sum of terms of the form $\frac{13!}{r}$. All these terms are congruent to $0$ modulo $11$ (divisible by $11$) except $\frac{13!}{11}$.
So the required remainder is the same as the remainder when $(10!)(12)(13)$ is divided by $11$. 
Now by Wilson's Theorem, we have $10!\equiv -1\pmod{11}$. And $(12)(13)\equiv 2\pmod{11}$. Thus $x\equiv -2\equiv 9\pmod{11}$.
Without Wilson's Theorem, we could just calculate directly using $(10!)(12)(13)$. 
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