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?

HINT:

Take the equation:
$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{23}=\frac{a}{23!}$
Multiply each side by $23!$:
$\frac{23!}{1}+\frac{23!}{2}+...+\frac{23!}{23}=a$
Divide each side by $13$:
$\frac{23!}{1\times13}+\frac{23!}{2\times13}+...+\frac{23!}{23\times13}=\frac{a}{13}$

There is only one non-integer element in the series: $\frac{23!}{13\times13}$.
Hence the remainder of $a$ when divided by $13$ is equal to the remainder of $\frac{23!}{13}$ when divided by $13$.
In other words: $a\equiv\frac{23!}{13}\pmod{13}$.
