Let $p, q$ and $r$ be positive real numbers such that the equation $$\lfloor pn \rfloor + \lfloor qn \rfloor + \lfloor rn \rfloor = n$$is satisfied for infinitely many positive integers $n{}$. (a) Prove that $p, q$ and $r$ are rational. (b) Determine the number of positive integers $c$ such that there exist positive integers $a$ and $b$, for which the equation $$\left \lfloor \frac{n}{a} \right \rfloor+\left \lfloor \frac{n}{b} \right \rfloor+\left \lfloor \frac{cn}{202} \right \rfloor=n$$is satisfied for infinitely many positive integers $n{}$.
Problem
Source: CAPS Match 2023 P4
Tags: floor function
sn6dh
18.08.2023 14:24
Claim 1: $p+q+r=1$.
If $p+q+r>1$, then for any $n>\frac3{p+q+r-1}$, $\lfloor pn\rfloor+\lfloor qn\rfloor+\lfloor rn\rfloor>(p+q+r)n-3>n$, and there will only be finitely many solutions.
If $p+q+r<1$, then for any $n>\frac3{1-p-q-r}$, $\lfloor pn\rfloor+\lfloor qn\rfloor+\lfloor rn\rfloor<(p+q+r)n+3<n$, and there will only be finitely many solutions.
$\therefore p+q+r=1$.
$\because\lfloor pn\rfloor+\lfloor qn\rfloor+\lfloor rn\rfloor\geq(p+q+r)n=n$, and the equation holds $\iff\lfloor pn\rfloor=pn, \lfloor qn\rfloor=qn, \lfloor rn\rfloor=rn$. $\therefore\exists n$ s.t. $pn, qn, rn$ are integers. $\Rightarrow p, q, r$ are rational. This finishes (a). Claim 2: $\lfloor pn\rfloor+\lfloor qn\rfloor+\lfloor rn\rfloor=n$ is satisfied for infinitely many $n\iff p+q+r=1$ and $p, q, r$ are rational.
($\Rightarrow$) is proved directly from the above.
($\Leftarrow$): for any rational numbers $p, q, r$ such that $p+q+r=1$, we can see that for all $n$ that are multiples of the least common multiple of the denominators of $p, q, r$, $np, nq, nr$ are integers.
$\Rightarrow\lfloor pn\rfloor+\lfloor qn\rfloor+\lfloor rn\rfloor=(p+q+r)n=n$.
$\Rightarrow\lfloor pn\rfloor+\lfloor qn\rfloor+\lfloor rn\rfloor=n$ is satisfied for infinitely many $n$.
From Claim 2, we know that $\frac1a+\frac1b+\frac c{202}=1$, and answer to (b) is the number of positive integer $c$ such that there exists positive integers $a, b$ with $\frac1a+\frac1b=\frac{202-c}{202}$, which can be calculated with no skill.