Modular Arithmetic - summing from 1 to a prime Apologises for the vague title; I couldn't think of anything better to call it.
I'm currently working on the following question:

Consider the equation $\sum_{i=1}^{5} \frac{1}{i} = \frac{X}{5Y}$.  Without making explicit calculations on the left hand side, show that $X \equiv Y$ modulo $125$.

I'm unsure how to approach this really.  I've noted that $125 = 5^3$ of course, but I don't know how significant that will be.  I've tried some simple things like multiplying by $5! \cdot s$ to try and eliminate all denominators, but that doesn't seem to be of any use to me?  
Can anyone give a nudge in the right direction for me?  Thanks in advance!
 A: I think one way to look at this is by subtracting $\frac 1 5$ from both sides of the equation to yield the following:
$$\sum_{i-1}^5\frac 1 i-\frac 1 5=\frac{X}{5Y}-\frac 1 5$$
$$\sum_{i-1}^4\frac 1 i=\frac{X-Y}{5Y}$$
The lowest common denominator of the numbers $1$ through $4$ does not have $5$ as a factor since $5$ is prime. However, the denominator on the right side does have $5$ as a factor. This means that both the numerator and denominator must have been multiplied by some multiple of $5$ from the original reduced fraction. Therefore, $5 | (X-Y)$ since $X-Y$ is the numerator of the resulting fraction from this multiplication.
Now, we want to show $X \equiv Y \pmod{125}$. However, if we show that $125 | (X-Y)$, then $X-Y \equiv 0 \pmod{125}$ and thus $X \equiv Y \pmod{125}$. Therefore, we just need to prove $125 | (X-Y)$.
Now, express $\sum_{i=1}^4\frac 1 i$ as $\frac{p}{4!}$ for some $p \in \mathbb{N}$. If $25 | p$, then the numerator of the reduced fraction is also divisible by $25$ because by reducing a fraction with denominator $4!$, there is no way we can get rid of $25$ as a factor since $4!$ and $25$ are coprime.
Thus, if we show $25 | p$, then since $X-Y$ is a multiple of $5$ times the numerator of the reduced fraction, $125 | (X-Y)$ as $5*25=125$. Thus, by showing $25 | p$, we will have proven the theorem.
I have not solved this problem yet either, so I don't know if I'm leading you in the right direction, but hopefully, this will give you something to go off of. Good luck!
