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?
 A: 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
$$\begin{align}a&\equiv -6 \pmod{13}\\
&\equiv 7 \pmod{13}\end{align}$$
A: 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: $a\equiv\frac{23!}{13}\pmod{13}$.
A: For every term except $1/13$, the numerator is a multiple of $13$.
The remainder is the same as the remainder of 
$1.2.3...12.14.15...23$.
Use Wilson's Theorem for $1.2...12\pmod{13}$ and $14.15...25\pmod{13}$, then adjust for 24 and 25.
'remainder' and 'rest' are synonyms for ordinary English, but not for this mathematical meaning.
A: $a$ cannot be divided by 13. Because 23!/13 is not divisible by 13 and the remainder is $7$
A: 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 $12\equiv -1\pmod{13}$.
