number theory - How to prove that the numerator is divided by 101 $1+frac{1}{2}+ldots+frac{1}{100}$

How to prove that: $1+\frac{1}{2}+\ldots+\frac{1}{100} = \frac{p}{q}, \gcd(p,q) = 1 \Rightarrow p \vdots 101$

I tried to allocate a numerator, but nothing happened, I tried to calculate the amount manually, but this also did not lead to success, I will be happy with any help.

Answer

For $1\leq n\leq 100$ let $1\leq r(n)\leq 100$ where $\frac {100!}{n}\equiv r(n)\pmod {101}.$

We have $1\leq n

Another way is to consider this in the field $F=\Bbb Z_{101}.$ Let $S=\sum_{n=1}^{100}1/n.$ Now $F$ does not have characteristic $2$, so in $F$ we have $S=\sum_{x\in F\backslash \{0\}}(x^{-1})=\sum_{y\in F\backslash \{0\}}(y)=0.$ The implication is that in $\Bbb Z$ we have $S=A/100!$ for some $A\in \Bbb Z,$ and if $101$ does not divide $A$ then in $\Bbb Z_{101}$ we would have $S\ne 0.$