Find real $x$ such $\lfloor x\rfloor+\lfloor 6x\rfloor=\lfloor 2x\rfloor+\lfloor 5x\rfloor=\lfloor 3x\rfloor+\lfloor 4x\rfloor$

Question

if $x$ be real numbers,solve the following equation $$\lfloor x\rfloor+\lfloor 6x\rfloor=\lfloor 2x\rfloor+\lfloor 5x\rfloor=\lfloor 3x\rfloor+\lfloor 4x\rfloor$$

Interesting $1+6=2+5=3+4$

Answer 1

The equation trivially holds for $x\in\Bbb Z$. To look for other solutions, write $x=m+y$ with $0\le y<1$. You can extract the integer $m$, which means we are left with solving the original equation for $0\le x<1$. Note that the first expression makes a jump at $\frac16,\frac13,\frac12,\frac23,\frac 56$, the scond makes a jump at $\frac15,\frac25,\frac12,\frac35,\frac45$, and the last at $\frac14,\frac13,\frac12,\frac23,\frac34$. So all you have to check is, at which of these points $\frac16,\frac15,\frac14,\frac13,\frac25,\frac12,\frac35,\frac23,\frac34,\frac45,\frac56$ all jumps "cancel" nicely; then we have equality from there (inclusive) until the next jump point (exclusive)