Question

The set of real values of '$x$' satisfying the equality $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ (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 $a+b+c+abc$.

Answer 1

For, real $x$, $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ holds.

Obviously $x>1$, otherwise the above will not hold.

For, $x>1$, the only possibility to hold the above equality is that

$\left[\frac{3}{x}\right]=2$ and $\left[\frac{4}{x}\right]=3$

$\Rightarrow 2\le \frac{3}{x}$ $<3$ and $3\le \frac{4}{x}$ $<4$

$\Rightarrow 1<x$ $\le \frac{3}{2}$ and $1<x$ $\le \frac{4}{3}$.

The smallest interval is $(1,4/3]$.

$\therefore a=1,b=4,c=3$.

$\therefore a+b+c+abc=8+12=20$.