<s>[INST] Solving an equation that contain floor function Let suppose I have three positive integers $a, b,c$ and one unknown $x$ ($x$ is also a positive integer). Solve for the smallest $x$ that satisfies this equation
$$\left \lfloor\frac{x}{a} \right\rfloor + \left \lfloor \frac{x}{b} \right\rfloor \geq c$$
where $\lfloor x\rfloor$ is the floor function
For instance let $a = 1$, $b=1$ and $c=5$ in this case
$$\left \lfloor x \right\rfloor + \left \lfloor x \right\rfloor \geq 5$$
here the smallest $x$ is $3$.
For another case, if $a=3$ and $b=5$ and $c=4$ we obtain
$$\left \lfloor\frac{x}{3} \right\rfloor + \left \lfloor \frac{x}{5} \right\rfloor \geq 4$$
here the smallest $x$ is $9$
My question is, can we find the smallest $x$ exactly ?
 [/INST] Assume that $a,b,c$ are positive
.
Let $y=\frac{abc}{a+b}$. Then $$\frac{y}{a}+\frac{y}{b}=c.$$
Therefore $$\left \lfloor\frac{y}{a} \right\rfloor + \left \lfloor \frac{y}{b} \right\rfloor$$ will be either $c$ or $c-1$.
If it is $c$
Then $y$ is a solution. We will still have a solution when $y$ is increased, until $y$ becomes the next multiple of either $a$ or $b$.
If it is $c-1$
Then increase $y$ until it becomes the next multiple of either $a$ or $b$. This will be the solution $x$ unless it is a multiple of both $a$ and $b$ in which case there is no solution. Given that we have a solution $x$, then we can increase it as above until the next multiple of either $a$ or $b$ is reached.
Example
Solve $\left \lfloor\frac{x}{5} \right\rfloor + \left \lfloor \frac{x}{7} \right\rfloor=8.$
We obtain $y=70/3$ and then $\left \lfloor\frac{y}{5} \right\rfloor + \left \lfloor \frac{y}{7} \right\rfloor=7.$
As $y$ is increased, the next two multiples of $5$ or $7$ are $25$ and $28$. The solution is $$25\le x<28.$$
Is this the sort of practical answer you were looking for?
Summary
The solutions can be completely expressed in terms of how $\frac{abc}{a+b}$ fits between successive terms of the ordered sequence of elements of $aZ\bigcup bZ$.
A formula for the revised question
A formula for the minimal $x$ is obtained as in the above answer and can be expressed as follows.
If $\frac{abc}{a+b}$ is in $aZ\bigcup bZ$ then that is the minimal $x$.
Otherwise $x=$ min$ (a\left \lfloor\frac{bc}{a+b}\right\rfloor+a,b\left \lfloor\frac{ac}{a+b}\right\rfloor+b)$.
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