Find all integers $m$ such that $\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} $ How would you determine all integers $m$ such that the following is true? 
$$\frac{1}{m}=\frac{1}{\lfloor 2x \rfloor}+\frac{1}{\lfloor 5x \rfloor} .$$
Note that $\lfloor \cdot \rfloor$ means the greatest integer function. Also, $x$ must be a positive real number.

If $k + j/10 \le x < k + (j+1)/10$ where $k$ and $j$ are nonnegative integers and $j \le 9$, $\lfloor 2x \rfloor = \begin{cases} 2k & 0 \le j \le 4 \\ 2k+1 &  5 \le j \le 9 \end{cases}$ while $\lfloor 5x \rfloor = \begin{cases} 5k + j/2 & j \ \text{even} \\ 5k + (j-1)/2 & j \ \text{odd} \end{cases}$.
Going over the various cases $j = 0$ to $9$,  I find two cases where  $f(x) = \frac{1}{1/\lfloor 2x \rfloor + 1/\lfloor 5x \rfloor} $ is an integer:
1) if $j = 0$ or $1$ and $k > 0$ is divisible by 7, $f(x) = 10 (k/7)$.
2) if $k=2$ and $j=4$, $f(x) = 3$.
Case (2) comes about as follows: if $j = 4$, $f(x) = \frac{1}{1/(2k) + 1/(5k+2)} = \frac{(2k)(5k+2)}{7k+2}$.  Note that $\gcd(k,7k+2) = \gcd(k,2) = 1$ or $2 $ while $\gcd(5k+2,7k+2) = \gcd(5k+2,2k) = \gcd(k+2,2k) =1$, $2$ or $4$.  The largest possible value of the denominator that could divide the numerator is thus $2 \times 2 \times 4 = 16$, which occurs for $k=2$.
