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?

Note that 
$$a = \frac{23!}{1} + \frac{23!}{2} + \dots + \frac{23!}{23}$$
Since each term is an integer, $a$ is an integer.

To find $a \pmod{13}$ (if that's what you're asking), note that due to the factorial term, each of these fraction has at least a factor of $13$, except $\frac{23!}{13}$.
As such, you'll have to determine
$$a \equiv 1\cdot2\cdot\dots12\cdot14\cdot15\cdot\dots23 \pmod{13}$$
which can be done quickly using Wilson's Theorem:
$$12!\equiv-1\pmod{13}$$
and
$$\begin{align}14\cdot15\cdot\dots23&\equiv 1\cdot2\cdot\dots\cdot10 \pmod{13}\\ &\equiv 12!\cdot12^{-1}\cdot{11}^{-1}\pmod{13}\\
&\equiv 12!\cdot12\cdot6\pmod{13}\\
&\equiv -1 \cdot -1\cdot6\\
&\equiv 6\end{align}$$
Multiplying both together, we get
$$\begin{align}a&\equiv -6 \pmod{13}\\
&\equiv 7 \pmod{13}\end{align}$$
