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.
 A: You can solve it by cases. Let $x=n+r$, where $n$ is an integer and $r=\lfloor x\rfloor$. Then $\lfloor 2x\rfloor = \lfloor 2n+2r\rfloor = 2n+\lfloor 2r\rfloor$ and $\lfloor 5x\rfloor = \lfloor 5n+5r\rfloor = 5n+\lfloor 5r\rfloor$, and the original equation can be written as $$\frac1m = \frac1{2n+\lfloor 2r\rfloor}+\frac1{5n+\lfloor 5r\rfloor}.$$
You know that $0\le r<1$, and you can get exact values for $\lfloor 2r\rfloor$ and $\lfloor 5r\rfloor$ for $r$ in different subintervals of $[0,1)$.


*

*If $0\le r<\frac15$, $\lfloor 2r\rfloor=\lfloor 5r\rfloor=0$.  

*If $\frac15\le r <\frac25$, $\lfloor 5r\rfloor = 1$ and $\lfloor 2r\rfloor = 0$.  

*If $\frac25\le r<\frac12$, $\lfloor 5r\rfloor = 2$ and $\lfloor 2r\rfloor = 0$.  

*If $\frac12\le r<\frac35$, $\lfloor 5r\rfloor = 2$ and $\lfloor 2r\rfloor = 1$.


And there are two more intervals, which I’ll leave to you.
If $0\le r<\frac15$, the equation becomes simply $$\frac1m=\frac1{2n}+\frac1{5n}=\frac{7n}{10n^2}=\frac7{10n};$$ in order for $m$ to be an integer, $n$ must be a multiple of $7$, say $n=7k$, we get $m=10k$. In other words, this case gives us every positive multiple of $10$ as a solution.
If $\frac15\le r <\frac25$, the equation becomes $$\frac1m=\frac1{2n}+\frac1{5n+1}=\frac{7n+1}{10n^2+2n},$$ or $$m=\frac{10n^2+2n}{7n+1}.$$ Divide this out to get $$m = \frac{10}7n+\frac4{49}-\frac4{49(7n+1)} = \frac{70n+4}{49}-\frac4{49(7n+1)},$$ and multiply through by $49$ to get $$49m=70n+4-\frac4{7n+1}$$ or, with a little rearrangement, $$49m-70n-4=\frac4{7n+1}.$$ But this is impossible if $n$ and $m$ are integers, because the lefthand side is an integer, and the righthand side isn’t. Thus, there are no solutions in this case.
If $\frac25\le r<\frac12$, the equation becomes $$\frac1m=\frac1{2n}+\frac1{5n+2}=\frac{7n+2}{10n^2+4n},$$ so $$m=\frac{10n^2+4n}{7n+2}=\frac{10n}7+\frac8{49}-\frac{16}{49(7n+2)},$$ and $$49m=70n+8-\frac{16}{7n+2}.$$ This time there is one value of $n$ that makes the righthand side an integer, namely, $n=2$. Substituting $n=2$ into the last equation, we get $49m=140+8-1=147$, and $m=3$; this is the only solution in this case. (Note that had the righthand side not been a multiple of $49$ when $n=2$, there would have been no solutions in this case.)
You can continue in this fashion through the remaining three cases. There may be an easier approach, but this one is at least systematic and workable.
A: Here is an idea, the computations seem too long to actually try.
Naive approach: forget the integer part. We need then $m=\frac{10x}{7}$ or $x = \frac{7m}{10}$.
Now non-naive approach: Try $x = \frac{7m+k}{10}$ where $k$ is small enough. $0 \leq k \leq 9$ or $6$ should work depending by whatever $7m$ is module 10. A case by case analysis might work, but the computations are too long to try.
It seems to me that this idea should work, but it could simply lead to a waste of time.
BTW: is it probably easy to argue that $x = \frac{7y+k}{10} + \epsilon$, for some integers $y,k$ with $k$ "small" and $0< \epsilon <\frac{1}{10}$ and then one can argue that we can  ignore that  $\epsilon$ part.
A: 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 $k=2$.
A: Consider the reciprocal equation 
$$ m = \dfrac{\lfloor 2x \rfloor \cdot \lfloor 5x \rfloor}
             {\lfloor 2x \rfloor + \lfloor 5x \rfloor}$$
Letting $x=n+\delta$ where $0 \le \delta < 1$, we get
$$ m = \dfrac{(2n + \lfloor 2\delta \rfloor)(5n + \lfloor 5\delta \rfloor)}
             {7n + \lfloor 2\delta \rfloor + \lfloor 5\delta \rfloor}$$
We list the possible values of m with respect to  $\delta$.
\begin{array}{|c|c|c|c|c|c|}
\delta \in & \lfloor 2\delta \rfloor & \lfloor 5\delta \rfloor
           & \lfloor 2\delta \rfloor + \lfloor 5\delta \rfloor
           & \lfloor 2\delta \rfloor \cdot \lfloor 5\delta \rfloor
           & m \\
\hline
[  0, 0.2) & 0 & 0 & 0 & 0 & \dfrac{10n}{7}\\
[0.2, 0.4) & 0 & 1 & 1 & 0 & \dfrac{10n^2+2n}{7n+1}\\
[0.4, 0.5) & 0 & 2 & 2 & 0 & \dfrac{10n^2+4n}{7n+2}\\
[0.5, 0.6) & 1 & 2 & 3 & 2 & \dfrac{10n^2+9n+2}{7n+3}\\
[0.6, 0.8) & 1 & 3 & 4 & 3 & \dfrac{10n^2+11n+3}{7n+4}\\
[0.8, 1.0) & 1 & 4 & 5 & 4 & \dfrac{10n^2+13n+4}{7n+5}\\
\hline
\end{array}
Now its just a matter of finding when those rational polynomials have integer solutions.
