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

solve for x,y >0  2x⌊y⌋ = 2020  3y⌊x⌋ = 2021

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

Commented byjohn santu last updated on 13/May/20

ans : (((1010)/(673)) , ((2021)/3))

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

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

⌊x⌋=n≠0  ⌊y⌋=m≠0  2020=2x⌊y⌋≥2nm ⇒nm≤1010  2020=2x⌊y⌋<2(n+1)m ⇒(n+1)m≥1011  2021=3y⌊x⌋≥3mn⇒mn≤((2021)/3)=673  2021=3y⌊x⌋<3(m+1)n⇒n(m+1)≥674  ⇒m−n≥337  m≥337+n  mn=(337+n)n≤673  ⇒n=1 is the only solution  ⇒y=((2021)/3)  ⇒x=((1010)/(⌊((2021)/3)⌋))=((1010)/(673))

$$\lfloor{x}\rfloor={n}\neq\mathrm{0} \\ $$ $$\lfloor{y}\rfloor={m}\neq\mathrm{0} \\ $$ $$\mathrm{2020}=\mathrm{2}{x}\lfloor{y}\rfloor\geqslant\mathrm{2}{nm}\:\Rightarrow{nm}\leqslant\mathrm{1010} \\ $$ $$\mathrm{2020}=\mathrm{2}{x}\lfloor{y}\rfloor<\mathrm{2}\left({n}+\mathrm{1}\right){m}\:\Rightarrow\left({n}+\mathrm{1}\right){m}\geqslant\mathrm{1011} \\ $$ $$\mathrm{2021}=\mathrm{3}{y}\lfloor{x}\rfloor\geqslant\mathrm{3}{mn}\Rightarrow{mn}\leqslant\frac{\mathrm{2021}}{\mathrm{3}}=\mathrm{673} \\ $$ $$\mathrm{2021}=\mathrm{3}{y}\lfloor{x}\rfloor<\mathrm{3}\left({m}+\mathrm{1}\right){n}\Rightarrow{n}\left({m}+\mathrm{1}\right)\geqslant\mathrm{674} \\ $$ $$\Rightarrow{m}−{n}\geqslant\mathrm{337} \\ $$ $${m}\geqslant\mathrm{337}+{n} \\ $$ $${mn}=\left(\mathrm{337}+{n}\right){n}\leqslant\mathrm{673} \\ $$ $$\Rightarrow{n}=\mathrm{1}\:{is}\:{the}\:{only}\:{solution} \\ $$ $$\Rightarrow{y}=\frac{\mathrm{2021}}{\mathrm{3}} \\ $$ $$\Rightarrow{x}=\frac{\mathrm{1010}}{\lfloor\frac{\mathrm{2021}}{\mathrm{3}}\rfloor}=\frac{\mathrm{1010}}{\mathrm{673}} \\ $$

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