The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$,(where $[]$ denotes the greatest integer function) belongs to the interval $(a,\frac{b}{c}]$,where $a,b,c\in N$ and $\frac{b}{c}$is in its lowest form.Find the value of $a+b+c+abc$.
I tried to solve this question.Let $\left[\frac{3}{x}\right]=t$,therefore $\left[\frac{4}{x}\right]=5-t$.
$t\leq \frac{3}{x}<t+1$......(i) and $5-t\leq \frac{4}{x}<6-t$.....(ii)
Add (i) and (ii),we get $5\leq\frac{3}{x}+\frac{4}{x}<7$
$5\leq\frac{7}{x}<7$
Then i solved $5\leq\frac{7}{x}$ and $\frac{7}{x}<7$
and took the intersection of the two solution sets and got $x\in (1,\frac{7}{5}]$,so my $a+b+c+abc=48$ but the answer is given to be 20.What is wrong in my method.Please help me.

So first of all, for $x=4$ you have $1$, for $x>4$ you have $0$. Next, your function has jumps whenever $x=3/n$ or $x=4/m$ for some integers $n,m$. Unless both of these happen at once (for instance, at $x=1$), these jumps are by one unit at a time. Can you find the closest four jumps to the left of $x=4$? (Hint: the first one is at $x=3$.)
Your actual mistake was that you essentially simplified $\left \lfloor \frac{3}{x} \right \rfloor + \left \lfloor \frac{4}{x} \right \rfloor$ into $\left \lfloor \frac{7}{x} \right \rfloor$, which is not correct.
