How many positive integer solutions exist for $[\frac{x}{19}]=[\frac{x}{20}]$, where $[x]$ denotes the Greatest integer function Question
How many positive integer solutions exist for $[\frac{x}{19}]=[\frac{x}{20}]$, where $[x]$ denotes the Greatest integer function
What I tried
I took the following cases one by one,
CASE $1$
$$[\frac{x}{19}]=[\frac{x}{20}]=1$$
All numbers from $20$ till $37$ should work for this, thus a total of 18 solution in this case.
CASE $2$
$$[\frac{x}{19}]=[\frac{x}{20}]=2$$
All numbers from $40$ to $56$ should work for this, thus a total of $17$ solutions in this case.
Upon continuing this process, we reach the case where there is only one possible solution.
Thus the number of cases is $18+17+16+...+2+1$ which is equal to $171$
There is also the case of $$[\frac{x}{19}]=[\frac{x}{20}]=0$$
This case will have $18$ solutions, from $1$ till $18$.
Thus the total number of solutions is $171+18$ which is $189$
I am not sure if my answer is correct (maybe I am missing a few cases).
What I am looking for is a verfication of my method and answer, and maybe a more concrete solution which will work in cases where $[\frac{x}{m}]=[\frac{x}{n}]$ where m and n are not consecutive natural numbers.
Thank you so much in advance!
Regards

$$\frac{x}{19}-1<\bigg\lfloor\frac{x}{19}\bigg\rfloor \le \frac{x}{19}$$
$$\frac{x}{20}-1<\bigg\lfloor\frac{x}{20}\bigg\rfloor \le \frac{x}{20}$$

$$\frac{x}{19}-1<\frac{x}{20}\implies x < 380$$
$$\frac{x}{20}-1<\frac{x}{19}\implies x > -380$$
