$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$ is independent of $n$ 
If $n$ is a positive integer, prove that
$$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
is independent of $n$.

Taking
$$f(n)=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
I could prove that
$$f(n+25)=f(n)$$
But, I have no idea how to show
$$f(n+k)=f(n)\\\forall k\in \{1,2,\dots 24\}$$
Of course, we can check these finite number of cases by force. But is there any other elegant method to handle this?

This is an amendment of my solution inspired by fleablood's comment:
To simplify the expression $\left \lfloor \frac{n-17}{25} \right \rfloor$, let $n-17=25k+r$ i.e. $n=25k+r+17$ where $0 \le r \le 24$.
Then
$$\left\lfloor \frac{8n+13}{25}\right \rfloor=\left\lfloor \frac{8(25k+r+17)+13}{25} \right\rfloor = 8k+5+\left\lfloor \frac{8r+24}{25} \right\rfloor$$
and  $$\left\lfloor \frac{n-12-\left\lfloor \frac{n-17}{25} \right\rfloor}{3} \right\rfloor =\left\lfloor \frac{25k+r+17-12-\left\lfloor \frac{25k+r+17-17}{25}\right \rfloor}{3} \right\rfloor=8k+1+\left\lfloor \frac{r+2}{3}\right \rfloor $$
The job is done if we can prove that
$$\left\lfloor \frac{8r+24}{25}\right \rfloor - \left\lfloor \frac{r+2}{3} \right\rfloor $$ is independent of $r$ when $0 \le r \le 24$
For this, put $r=3m+h$ where $0 \le h \le 2$ and $0 \le r \le 24$.
It is not difficult to check that
$$\left\lfloor \frac{8r+24}{25}\right \rfloor = \left\lfloor \frac{r+2}{3} \right\rfloor  $$
under the said conditions.
