Prove that neither $A$ nor $B$ is divisible by $5$ Let the sum 
$$ {1+ \frac12 + \frac13 + \frac 14+ \dots +\frac1{99} + \frac 1{100}}$$
be written as $\frac AB$, where $A$ and $B$ are positive integers with no common factors. Show that neither $A$ nor $B$ is divisible by $5$.

Using Mathematica, I found the sum is $$\frac AB=\frac {14466636279520351160221518043104131447711}{278881500918849908658135235741249214‌2272}$$

Firstly $\frac{1}{25} + \frac{1}{50} + \frac{1}{75} + \frac{1}{100} = \frac{1}{25} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) = \frac{1}{12}$.
Also $\frac{1}{25k+5} + \frac{1}{25k+10} + \frac{1}{25k+15} + \frac{1}{25k+20} = \frac{1}{5} \left( \frac{(5k+4)+(5k+1)}{(5k+1)(5k+4)} + \frac{(5k+3)+(5k+2)}{(5k+2)(5k+3)} \right)$
  $\ \ = \frac{2k+1}{(5k+1)(5k+4)} + \frac{2k+1}{(5k+2)(5k+3)}$ for any integer $k$.
Therefore $\sum_{k=1}^{100} \frac{1}{k}$ has no factor of $5$ in the reduced fraction form.
