Remainder of the numerator of a harmonic sum modulo 13 Let $a$ be the integer determined by 
 $$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{23}=\frac{a}{23!}.$$
Determine the remainder of $a$ when divided by 13.
Can anyone help me with this, or just give me any hint?

By Wilson's theorem, $12!\equiv -1\pmod{13}$. 
By Wolstenholme's theorem, $H_{12}\equiv 0\pmod{13}$. Since:
$$ a = \sum_{k=1}^{12}\frac{23!}{k} + (12!)(14\cdot 15\cdot\ldots\cdot 23)+\sum_{k=14}^{23}\frac{23!}{k} $$
we have:
$$ a\equiv 0-10!+0 \equiv \frac{-12!}{11\cdot 12}\equiv \frac{1}{(-2)\cdot(-1)}\equiv\frac{1}{2}\equiv\color{red}{7}\pmod{13}$$
since $11\equiv -2\pmod{13}$ and $12\equiv -1\pmod{13}$.
