parmenides51 wrote: Find all real $x > 0$ and integer $n > 0$ so that $$ \lceil x \rceil+\left\{ \frac{1}{x}\right\}= 1.005 \cdot n.$$ $x=1$ does not fit $x=m\in\mathbb Z_{>1}$ means $m+\frac 1m=\frac{201n}{200}$ and so $200m^2+200=201mn$, which has no solution $\pmod 3$ If $x\in(0,1)$, equation is $1+\left\{\frac 1x\right\}=\frac{201}{200}n$ and so, since $LHS\in(1,2)$, $n=1$ And then $\left\{\frac 1x\right\}=\frac 1{200}$ and so $x=\frac{200}{200k+1}$, whatever is $k\in\mathbb Z_{>0}$ If $x>1$ and $x\notin\mathbb Z$, let $m=\lfloor x\rfloor$ with $m\in\mathbb Z_{>0}$ Equation is $m+1+\frac 1x=\frac{201}{200}n$ and so $\frac{200}x=201n-200(m+1)$ with $x\in(m,m+1)$ So inequation $\frac{200}m>201n-200(m+1)>\frac{200}{m+1}$ This implies $m\le 199$ (else $1>201n-200(m+1)>0$, impossible) and so we have : $201n>200(m+1)+\frac{200}{m+1}>200(m+1)>201m$ and so $n>m$ $201n<200(m+1)+\frac {200}m<201(m+2)$ and so $n<m+2$ So $n=m+1$ and inequation is $\frac{200}m>m+1>\frac{200}{m+1}$, which is impossible (right side implies $m\ge 14$ while left side implies $m\le 13$) And so only solutions $\boxed{x\in\left\{\frac{200}{200k+1}\quad\forall k\in\mathbb Z_{>0}\right\}\text{ and }n=1}$