How many positive integer solutions exist for $[\frac{x}{19}]=[\frac{x}{20}]$, where $[x]$ denotes the Greatest integer function

Question

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

Answer 1

For $x >0$, the two are equal if and only if there exists $n$ such that $n \leq \frac x{20} < \frac x{19} < n+1$. This translates to $20n \leq x < 19n+19$.

Now for a given $n$, we have $(19n+19)-20n = 19-n$. Therefore we have $19-n$ values between $20n$ and $19n+19$ for $n < 19$ which $x$ can take (note : for $n=0$ we actually can't have $x=0$ so we must subtract $1$), and none for $n \geq 19$. Thus, the answer is : $$ \sum_{n=0}^{18} (19-n) - 1 = 361 - 171 - 1 = 190-1 = 189 $$

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In general case of $\lfloor x/n \rfloor = \lfloor x/m \rfloor$ with $n > m$, we get that for $x>0$ equality works if and only if for some $N$ we have $N \leq \frac xn < \frac xm < N+1$. This translates to $nN \leq x < mN + m$.

Therefore for all $N$ such that $N \geq 0$ and $N < \frac{m}{n-m}$, we get $(mN - nN + m)$ values of $x$. So the answer would be (again accounting that $x=0$ will be counted in the $N=0$ case so we must remove it by subtracting $1$): $$ \sum_{N=0}^{\lfloor \frac{m}{n-m} - 1\rfloor} (mN - nN + m) - 1 $$

in our case, $m=19$, $n= 20$, so $\frac{m}{n-m} - 1 = 18$ ,with $mN - nN + m = 19-n$, so we are back. You can of course evaluate the above sum better ,this I leave to you.

Answer 2

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

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

Answer 3

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.