Numerator Modulo 2011

We only need to prove that $\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{2010}\equiv 0\pmod {2011}$

This is because if $\frac{m}{n}\equiv 0\pmod {2011}$ , then n\cdot\frac{1}{n}\equiv 1\pmod {2011}\stackrel{\times m}\implies n\cdot \frac{m}{n}\equiv 0\equiv m\pmod {2011}

Since $2011$ is prime, all the inverses are defined.

We'll prove that $\frac{1}{k}\equiv -\frac{1}{2011-k}\pmod {2011}, \forall k\in[1;2010]\cup \mathbb N$ . This, since $k, 2011-k$ are of different parity, will imply that the sum of the inverses $\equiv 0\pmod {2011}$ .

$k\frac{1}{k}\equiv -(2011-k)\frac{1}{k}\equiv 1\pmod {2011}$

But, as you can see, this means that $\frac{1}{k}\equiv -\frac{1}{2011-k}\pmod {2011}$ - we have that $\frac{1}{k}$ is the inverse of $-(2011-k)$ , which is $-\frac{1}{2011-k}$ .

We prove Wolstenholme's Theorem where if prime p > 3 and 1/1 + 1/2 + 1/3 + ... + 1/p-1 = m/n, then p | m.

We use multiplicative inverses to prove this out. Since the multiplicative inverses are unique modulo p, 1^(-1) + 2^(-1) + ... (p-1)^(-1) is congruent to 1 + 2 + ... + (p-1) modulo p and is congruent to (p-1)(p)/2. Since p > 3, p is odd and p-1 is even, hence, it is congruent to 0 modulo p implying that p | m.