jasperE3 wrote: Let $m>1$ be an integer. Determine the number of positive integer solutions of the equation $\left\lfloor\frac xm\right\rfloor=\left\lfloor\frac x{m-1}\right\rfloor$. Let nonnegative integer $n=\left\lfloor\frac xm\right\rfloor=\left\lfloor\frac x{m-1}\right\rfloor$ So we have $mn\le x<(m-1)(n+1)$ and so $\max(0,m-n-1)$ integer values for $x$ And so answer is $\sum_{n=0}^{\infty}\max(0,m-n-1)$ $=\sum_{n=0}^{m-1}(m-n-1)$ $=\boxed{\frac{m(m-1)}2}$