If-x-and-y-are-positive-integers-and-2x-4y-3-then-find-16-x-256-y-

Question Number 179650 by Linton last updated on 31/Oct/22

$${If}\:{x}\:{and}\:{y}\:{are}\:{positive} \\ $$$${integers}\:{and}\:\mathrm{2}{x}−\mathrm{4}{y}\:=\:\mathrm{3} \\ $$$${then}\:{find}\:\frac{\mathrm{16}^{{x}} }{\mathrm{256}^{{y}} } \\ $$

Commented by Rasheed.Sindhi last updated on 31/Oct/22

2x_(even) −4y_(even) = 3 _(odd) Difference of two even numbers can′t be odd. 2x−4y = 3⇒x−2y=(3/2) left side is an integer while right side is non-integer rational! Contradiction.

$$\underset{{even}} {\underbrace{\mathrm{2}{x}}}−\underset{{even}} {\underbrace{\mathrm{4}{y}}}\:=\underset{{odd}} {\underbrace{\:\mathrm{3}\:}} \\ $$$$\mathcal{D}{ifference}\:{of}\:{two}\:{even}\:{numbers} \\ $$$${can}'{t}\:{be}\:{odd}.\: \\ $$$$\mathrm{2}{x}−\mathrm{4}{y}\:=\:\mathrm{3}\Rightarrow{x}−\mathrm{2}{y}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${left}\:{side}\:{is}\:{an}\:{integer}\:{while}\:{right} \\ $$$${side}\:{is}\:{non}-{integer}\:{rational}! \\ $$$${Contradiction}. \\ $$

Commented by CElcedricjunior last updated on 31/Oct/22

2x−4y=3=>x=2y+(3/2) on a ((16^x )/(256^y ))=((16^(2y+(3/2)) )/(16^(2y) ))=16^(2y−2y+(3/2)) =16^(3/2) =4^3 =>((16^x )/(256^y ))=64 ................le celebre cedric junior......

$$\mathrm{2}\boldsymbol{{x}}−\mathrm{4}\boldsymbol{{y}}=\mathrm{3}=>\boldsymbol{{x}}=\mathrm{2}\boldsymbol{{y}}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{a}}\:\frac{\mathrm{16}^{\boldsymbol{{x}}} }{\mathrm{256}^{\boldsymbol{{y}}} }=\frac{\mathrm{16}^{\mathrm{2}\boldsymbol{{y}}+\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{16}^{\mathrm{2}\boldsymbol{{y}}} }=\mathrm{16}^{\mathrm{2}\boldsymbol{{y}}−\mathrm{2}\boldsymbol{{y}}+\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{16}^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{4}^{\mathrm{3}} \\ $$$$=>\frac{\mathrm{16}^{\boldsymbol{{x}}} }{\mathrm{256}^{\boldsymbol{{y}}} }=\mathrm{64} \\ $$$$\: \\ $$$$…………….\boldsymbol{{le}}\:\boldsymbol{{celebre}}\:\boldsymbol{{cedric}}\:\boldsymbol{{junior}}…… \\ $$

Commented by Rasheed.Sindhi last updated on 31/Oct/22

((16^x )/(256^y ))=(4^(2x) /4^(4y) )=4^(2x−4y) =4^3 =64 [But x & y can′t be positive integers]

$$\frac{\mathrm{16}^{{x}} }{\mathrm{256}^{{y}} }=\frac{\mathrm{4}^{\mathrm{2}{x}} }{\mathrm{4}^{\mathrm{4}{y}} }=\mathrm{4}^{\mathrm{2}{x}−\mathrm{4}{y}} =\mathrm{4}^{\mathrm{3}} =\mathrm{64} \\ $$$$\left[\mathcal{B}{ut}\:{x}\:\&\:{y}\:{can}'{t}\:{be}\:{positive}\:{integers}\right] \\ $$

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