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
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 $(10!)(12)(13)$.
Each term in your expression (each $\frac{13!}{k}$) is an integer, and except for the $k=11$ case is expressed as a rational fraction where the numerator contains a factor of 11 but the denominator does not. Hence, each term except $k=11$ is 0 mod 11. So the answer is going to be $$ \frac{13!}{11} \mod 11 = (13 \cdot 12 \cdot \prod_{n=1}^{10}n) \mod 11 = (2 \cdot 1 \cdot (-1)) \mod 11 = -2 \equiv 9 \mod 11 $$ Here I used the fact (Fermat) that for $p$ prime, $p! \equiv -1 \mod p$
So the remainder will be 9.
