How many solutions does the equation

$$
\left\lfloor\frac{x}{20}\right\rfloor=\left\lfloor\frac{x}{17}\right\rfloor
$$

have over the set of positive integers?

Here, $\lfloor a\rfloor$ denotes the greatest integer less than or equal to $a$.

(Karl Czakler)
(Karl Czakler) Let

$$
\left\lfloor\frac{x}{20}\right\rfloor=\left\lfloor\frac{x}{17}\right\rfloor=n .
$$

Then it follows that $20 n \leq x<20 n+20$ and $17 n \leq x<17 n+17$. Therefore, for all possible solutions $x$, it must hold that

$$
20 n \leq x<17 n+17
$$

For the value of $n$, we get $20 n<17 n+17$ thus $n \in\{0,1,2,3,4,5\}$.

For $n=0$, we get $1 \leq x<17$, yielding a total of 16 solutions. For $n=1$, we get $20 \leq x<34$, yielding a total of 14 solutions. For $n=2$, we get $40 \leq x<51$, yielding a total of 11 solutions, and so on. (The inequality (1) has $17 n+17-20 n=17-3 n$ solutions, and for $n=0$, the solution $x=0$ is excluded.)

Therefore, we have $16+14+11+8+5+2=56$ solutions over the set of natural numbers for this equation.