Equal Quotient in Division by $10$ and $(10+1)$

Relevant wiki: Floor Function

Two rational numbers have the same quotient if their floors are equal. Therefore, if $q$ is the quotient, we want to solve $q = \lfloor \frac{m}{10} \rfloor = \lfloor \frac{m}{11} \rfloor$ . This translates into the inequality $q \leq \frac{m}{11} < \frac{m}{10} < q + 1$ . Dividing everything by $m$ we then get the inequality $\frac{q}{m} \leq \frac{1}{11} < \frac{1}{10} < \frac{q+1}{m}$ . Because the middle part of this inequality is trivially true we can divide this into a system of two simultaneous inequalities: $\frac{q}{m} \leq \frac{1}{11}$ and $\frac{q+1}{m} > \frac{1}{10}$ .

With a little algebra these can be rewritten as $m \geq 11q, -m > -10q - 10$ and now adding both of them gives $0 > q - 10 \implies 10 > q$ . Which implies a solution exists for our system of inequalities as long as $10 > q$ so $q = 1,2,...,9$ .

We can also find explicit solutions, $m=11$ gives $q=1$ , $m=22$ gives $q=2$ , all the way up to $m=99$ giving $q=9$ so the previously proven inequality can be used as a guarantee that we do not need to check further, no solution exists for $q=10$ and up. Also, the argument used to prove my inequality generalizes easily as I never use any particular property of $10$ . If we replace $10$ with $n$ we would get $n > q$ so solutions would exist for $q=1,...,(n-1)$ and the sum of these numbers would be the $(n-1)th$ triangular number.

Lets define the properties of m as $\begin{cases} m=10\cdot q + r \\ m=(10+1)\cdot q + r^{´} \end{cases}$

Then, $(\alpha$ ): $10\cdot q + r = 11\cdot q + r^{´} \iff q = r - r^{´}$

And by the division algorithm $(\beta$ ) : $0 \leq r < 10 \\ 0 \leq r^{´} < 11 \\$

Considering $q$ must be a positive intenger and $(\alpha$ ) we get : $0<r - r^{´} \iff r > r^{´}$ By the equations of $(\beta$ ) and the new restriction, we can know: $r_{max}=9$ , $r_{min}=1$ and $r_{min}^{´}=0$

Therefore,

$q_{min} \leq q \leq q_{max} \\ r_{min}- r_{min}^{´}\leq q \leq r_{max} - r_{min}^{´} \\ 1 \leq q \leq 9$

Since every q must satisfy for at least one m, by the definition and the conditions exposed lets create a particular set of m.

For example: $(r^{´}=0$ , $r=q$ and $q\mapsto [1,9]$ $\implies m=\{11, 22, \dots\ ,99\})$

$\therefore q = [1,9]$