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Question Number 87009 by mr W last updated on 01/Apr/20

$$solve$$$$7\lfloor x+3{\rfloor }^2-3\lfloor x\rfloor +6=5mod11$$

Answered by MJS last updated on 01/Apr/20

$$\lfloor x+3\rfloor =\lfloor x\rfloor +3$$$$\Rightarrow$$$$7\lfloor x{\rfloor }^2+39\lfloor x\rfloor +69=11n+5;n\in \mathbb{Z}$$$$\mathrm{let}\lfloor x\rfloor =m\in \mathbb{Z}$$$$7m^2+39m-11n+64=0$$$$\Rightarrow$$$$n=\frac{7m^2+39m+64}{11}$$$$n\in \mathbb{Z}\Rightarrow m=11k+1\vee m=11k+6$$$$\Rightarrow$$$$x=11k+1+\lambda \vee x=11k+6+\lambda ;k\in \mathbb{Z}\wedge \lambda \in \mathbb{R}\wedge 0⩽\lambda <1$$

Answered by M±th+et£s last updated on 02/Apr/20

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