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Question Number 93460 by john santu last updated on 13/May/20

$$\mathrm{solve}\mathrm{for}\mathrm{x},\mathrm{y}>0$$ $$2x\lfloor \mathrm{y}\rfloor =2020$$ $$3\mathrm{y}\lfloor x\rfloor =2021$$

Commented byjohn santu last updated on 13/May/20

$$ans:(\frac{1010}{673},\frac{2021}{3})$$

Answered by mr W last updated on 16/May/20

$$\lfloor x\rfloor =n\ne 0$$ $$\lfloor y\rfloor =m\ne 0$$ $$2020=2x\lfloor y\rfloor ⩾2nm\Rightarrow nm⩽1010$$ $$2020=2x\lfloor y\rfloor <2(n+1)m\Rightarrow (n+1)m⩾1011$$ $$2021=3y\lfloor x\rfloor ⩾3mn\Rightarrow mn⩽\frac{2021}{3}=673$$ $$2021=3y\lfloor x\rfloor <3(m+1)n\Rightarrow n(m+1)⩾674$$ $$\Rightarrow m-n⩾337$$ $$m⩾337+n$$ $$mn=(337+n)n⩽673$$ $$\Rightarrow n=1istheonlysolution$$ $$\Rightarrow y=\frac{2021}{3}$$ $$\Rightarrow x=\frac{1010}{\lfloor \frac{2021}{3}\rfloor }=\frac{1010}{673}$$

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