<s>[INST] 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.
 [/INST] 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 $97^2 = 9409$!
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