Prove that $83$ divides $p$ . 
Let $p$ and $q$ be the positive integers such that $1 - \frac{1}{2} + \frac{1}{3} - ... - \frac{1}{54} + \frac{1}{55} = \frac{p}{q}$ . Prove that $83$ divides $p$ .

What I Tried :- To me it does not actually seem like an easy problem for me. (I guess I couldn't figure out the idea)
This expression I found is same as $$\displaystyle\sum_{n=1}^{55} \frac{(-1)^{n+1}}{n}$$
And calculating this in Wolfram Alpha gives the answer to be this in the simplest form :- $$\displaystyle\sum_{n=1}^{55} \frac{(-1)^{n+1}}{n}=\frac{115328583812490186710549}{164249358725037825439200}$$
Surprisingly, I found that $83$ really divides the numerator of the fraction. Now I got puzzled, how will you show this without using Wolfram Alpha or a Calculator?
Can Anyone Help?

Given expression is
$$
1 - \frac{1}{2} + \frac{1}{3} - ... - \frac{1}{54} + \frac{1}{55} \\ 
= 1 + \frac{1}{2} + \frac{1}{3} + ...  + \frac{1}{55} - 2(\frac{1}{2} + \frac{1}{4} + ...  + \frac{1}{54} ) \\ 
= \frac{1}{28} + \frac{1}{29}  ...  + \frac{1}{55} \\
 = (\frac{1}{28} + \frac{1}{55}) + (\frac{1}{29} + \frac{1}{54}) + \ldots \\ 
= 83 (\frac{1}{28\cdot55} + \frac{1}{29\cdot54} + \ldots) = \frac{p}{q}
$$
from which it follows.
