Show that the numerator of $1+\frac12 +\frac13 +\cdots +\frac1{96}$ is divisible by $97$ 
Let $\frac{x}{y}=1+\frac12 +\frac13 +\cdots +\frac1{96}$ where $\text{gcd}(x,y)=1$. Show that $97\;|\;x$. 

I try adding these together, but seems very long boring and don't think it is the right way to solving. Sorry for bad english.

What we need to do is as follows:
$$
\begin{equation}\begin{split}
1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{96} & = \Big(1+\frac{1}{96}\Big) + \Big(\frac{1}{2}+\frac{1}{95}\Big) + \ldots + \Big(\frac{1}{48} + \frac{1}{49}\Big) \\
& = \frac{97}{96} + \frac{97}{2*95} + \frac{97}{3*94} + \ldots + \frac{97}{48*49} \\
& = 97 \Big( \frac{1}{96} + \frac{1}{2*95} + \frac{1}{3*94} + \ldots + \frac{1}{48*49}\Big)
\end{split}\end{equation}
$$
Hence, the numerator is a multiple of $97$. To see that the denominator is not a multiple of $97$, note that $97$ is a prime, and the denominator contains natural numbers less than $97$ (in fact, the reduced denominator is a factor of $96$!), hence cannot possibly be a multiple of $97$. Thus, in it's reduced form, the numerator of the expression is a multiple of $97$.
In fact, Wolstenhomme's  theorem says that the numerator is a multiple of $97^2 = 9409$!
