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.
4 Answers
$\begingroup$Here are some hints.
Note that as $x$ increases, the sum can get no larger - it is decreasing but not strictly so, because it is sometimes constant. Also with $x=1$ the sum is equal to $7$ so we must have $x \gt 1$ for a sum as low as $5$.
Now if $x\gt 1$ we have immediately that $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]\le 2+3=5$ so we must have equality with $\left[\frac{3}{x}\right]=2$ and $\left[\frac{4}{x}\right]=3$
You should be able to complete things from there.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$Suppose that $x>1$, to get $5$ its needed to add two positive integers bcz $\displaystyle \forall x>1,\left(\lfloor \frac{3}{x}\rfloor < \lfloor \frac{4}{x}\rfloor \right)$.
So its obvious that only integer that satisfy the condition $(2,3)$ or $(1,4),$ However for the
Second couple its not find Corrosponding $x,$ bcz to have $\displaystyle \lfloor \frac{3}{x}\rfloor=1$ necessary $\displaystyle x>\frac{3}{2}$
and together with the values $\displaystyle \lfloor \frac{4}{x}\rfloor = 4\Rightarrow x<1$ Gives an empty set
Analyse Situation when $\displaystyle \lfloor \frac{3}{x}\rfloor = 2$ and $\displaystyle \lfloor \frac{4}{x}\rfloor = 3$
So we get $\displaystyle 2\leq \frac{3}{x}\leq 3$ and $\displaystyle 3\leq \frac{4}{x}\leq 4$
so solution $\displaystyle x\in \left(1,\frac{4}{3}\right]$
$\endgroup$ $\begingroup$The problem with your method is that you only established necessary conditions on $x$ but those are not necessarily sufficient conditions. E.g. if you plug in $x=\frac{7}{5}$ you will obtain $\left[\frac{15}{7}\right]+\left[\frac{20}{7}\right]=4 \ne 5$.
To find the correct interval boundaries, you may first want to get some intuition for this function.
Therefore, denote $f(x)=\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]$. If you sketch the graph of $f$ you will see that it is mononotonically decreasing (why?) and jumps from one integer to another at some points. You may note that the jumps only occur at points where $\left[\frac{3}{x}\right]$ or $\left[\frac{4}{x}\right]$ is an integer (why?).
Now try again to find the correct boundaries. (Hint: Your lower bound was already correct.)
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