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Question Number 80004 by jagoll last updated on 30/Jan/20

x and y any integer satisfy equation (x−2004)(x−2006)=2^y the greatest possible value of x+y

$${x}\:{and}\:{y}\:{any}\:{integer}\:{satisfy} \\ $$$${equation}\:\left({x}−\mathrm{2004}\right)\left({x}−\mathrm{2006}\right)=\mathrm{2}^{{y}} \\ $$$${the}\:{greatest}\:{possible}\:{value} \\ $$$${of}\:{x}+{y} \\ $$

Commented by jagoll last updated on 30/Jan/20

thank you mister

$${thank}\:{you}\:{mister} \\ $$

Commented by mr W last updated on 30/Jan/20

let u=x−2004 u(u−2)=2^y u^2 −2u+1=2^y +1 (u−1)^2 =(2^y +1) u=1±(√(2^y +1)) 2^y +1=(n+1)^2 say 2^y =(n+1)^2 −1=n(n+2) ⇒n=2^p , n+2=2^q ⇒(2^(p−1) +1)=2^(q−1) ⇒p=1 ⇒n=2 ⇒2^y =2×4=8 ⇒y=3 ⇒u=x−2004=1±3 ⇒x=2008 or 2002 (x+y)_(max) =2008+3=2011

$${let}\:{u}={x}−\mathrm{2004} \\ $$$${u}\left({u}−\mathrm{2}\right)=\mathrm{2}^{{y}} \\ $$$${u}^{\mathrm{2}} −\mathrm{2}{u}+\mathrm{1}=\mathrm{2}^{{y}} +\mathrm{1} \\ $$$$\left({u}−\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{2}^{{y}} +\mathrm{1}\right) \\ $$$${u}=\mathrm{1}\pm\sqrt{\mathrm{2}^{{y}} +\mathrm{1}} \\ $$$$\mathrm{2}^{{y}} +\mathrm{1}=\left({n}+\mathrm{1}\right)^{\mathrm{2}} \:{say} \\ $$$$\mathrm{2}^{{y}} =\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}={n}\left({n}+\mathrm{2}\right) \\ $$$$\Rightarrow{n}=\mathrm{2}^{{p}} ,\:{n}+\mathrm{2}=\mathrm{2}^{{q}} \:\Rightarrow\left(\mathrm{2}^{{p}−\mathrm{1}} +\mathrm{1}\right)=\mathrm{2}^{{q}−\mathrm{1}} \\ $$$$\Rightarrow{p}=\mathrm{1}\:\Rightarrow{n}=\mathrm{2}\:\Rightarrow\mathrm{2}^{{y}} =\mathrm{2}×\mathrm{4}=\mathrm{8}\:\Rightarrow{y}=\mathrm{3} \\ $$$$\Rightarrow{u}={x}−\mathrm{2004}=\mathrm{1}\pm\mathrm{3} \\ $$$$\Rightarrow{x}=\mathrm{2008}\:{or}\:\mathrm{2002} \\ $$$$\left({x}+{y}\right)_{{max}} =\mathrm{2008}+\mathrm{3}=\mathrm{2011} \\ $$

Answered by mind is power last updated on 30/Jan/20

(x−2004)(x−2006)=2^(a+b) ⇒x−2004=2^a y−2006=2^b ⇒2^a −2^b =2^b (2^(a−b) −1)=2⇒b=1⇒a−b=1⇒a=2 ⇒y=a+b=3 x=2^2 +2004=2008 x+y=2008+3=2011 its exames of 2011?

$$\left({x}−\mathrm{2004}\right)\left({x}−\mathrm{2006}\right)=\mathrm{2}^{{a}+{b}} \\ $$$$\Rightarrow{x}−\mathrm{2004}=\mathrm{2}^{{a}} \\ $$$${y}−\mathrm{2006}=\mathrm{2}^{{b}} \\ $$$$\Rightarrow\mathrm{2}^{{a}} −\mathrm{2}^{{b}} =\mathrm{2}^{{b}} \left(\mathrm{2}^{{a}−{b}} −\mathrm{1}\right)=\mathrm{2}\Rightarrow{b}=\mathrm{1}\Rightarrow{a}−{b}=\mathrm{1}\Rightarrow{a}=\mathrm{2} \\ $$$$\Rightarrow{y}={a}+{b}=\mathrm{3} \\ $$$${x}=\mathrm{2}^{\mathrm{2}} +\mathrm{2004}=\mathrm{2008} \\ $$$${x}+{y}=\mathrm{2008}+\mathrm{3}=\mathrm{2011}\:{its}\:{exames}\:{of}\:\mathrm{2011}? \\ $$

Commented by jagoll last updated on 30/Jan/20

if x = 2010 (x−2004)(x−2006)= 6×4 = 24≠2^y sir

$${if}\:{x}\:=\:\mathrm{2010}\: \\ $$$$\left({x}−\mathrm{2004}\right)\left({x}−\mathrm{2006}\right)=\:\mathrm{6}×\mathrm{4}\:=\:\mathrm{24}\neq\mathrm{2}^{{y}} \\ $$$${sir} \\ $$

Commented by mind is power last updated on 30/Jan/20

mistack soory 2^y ,y≠2^y

$${mistack}\:{soory}\:\:\:\mathrm{2}^{{y}} ,{y}\neq\mathrm{2}^{{y}} \\ $$$$ \\ $$

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