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

Let $x = 19n + r; 0\le r < 19$ then $x = 20n + (r-n)$ but
$\frac x{19} = n + \frac rn$ but $0 \le \frac rn < 1$ so $[\frac x{19}] = n$.
And if $[\frac x{20}] = [\frac {x}{19}] = n$ then as $\frac {x}20 = n + \frac {r-n}{20}$ we must have $0 \le \frac {r-n}{20}<1 $ or in other words $0 \le r-n <20$.
And any $0 \le r-n < 20; r< 19; r\ge n \ge 0$ will do (except $r=n=0$).
$r=1.... 18$ and $n=0.....r$ so the number of possible solutions are $\sum_{r=1}^{18} (r+1)= \sum_{k=2}^{19} k = \sum_{k=1}^{19} k - 1 = \frac {19*20}2 -1 = 189$.
but.... I may have an indexing error.
